Textbook of Mathematics Grade 10 (FBISE / NBF)
Class 10 Mathematics Federal Board of Intermediate and Secondary Education (FBISE) (FBISE) 📚 Textbook of Mathematics Grade 10 (FBISE / NBF)

Mastery Guide: Basic Statistics, Cumulative Frequency, Measures of Dispersion & Probability Theory

📖 Chapter 12: Basic Statistics 📅 Updated: Oct 04, 2026
FBISE • CLASS 10 MATHEMATICS • UNIT 12

Mastery Guide: Basic Statistics, Cumulative Frequency, Measures of Dispersion & Probability Theory

A complete, mathematically rigorous curriculum guide covering Cumulative Frequency Tables, Ogives, Box-and-Whisker Plots, Scatter Plots & Correlation, Range, Variance, Standard Deviation, Coefficient of Variation, and Probability Laws.

📖 Unit Overview & Target Learning Outcomes

By the end of this comprehensive unit, students will master the following official FBISE/SNC mathematical standards:

  • Cumulative Frequency Distributions: Construct cumulative frequency tables, cumulative frequency polygons, and smooth $S$-shaped ogives for grouped continuous data.
  • Partition Values: Locate and estimate Quartiles ($Q_1, Q_2, Q_3$), Deciles ($D_1$ to $D_9$), Percentiles ($P_1$ to $P_{99}$), and Interquartile Range ($\text{IQR} = Q_3 - Q_1$) graphically and algebraically.
  • Box-and-Whisker Plots: Construct, read, and analyze 5-number summaries (Minimum, $Q_1$, Median, $Q_3$, Maximum) and identify potential outliers ($> 1.5 \times \text{IQR}$).
  • Scatter Diagrams & Correlation: Plot bivariate data, determine the direction and strength of correlation (Positive, Negative, Zero), and draw linear lines of best fit ($y = mx + c$).
  • Measures of Dispersion: Calculate Range, Variance ($S^2$), Standard Deviation ($S = \sqrt{S^2}$), and understand their physical units.
  • Relative Dispersion & Consistency: Compute the Coefficient of Variation ($\text{C.V.} = \frac{S}{\bar{x}} \times 100\%$) to compare consistency across distinct datasets.
  • Probability Theory: Define sample spaces, construct probability tree diagrams, and apply the Addition Rule ($P(A \cup B)$) and Multiplication Rule ($P(A \cap B)$) for independent, dependent, and mutually exclusive events.

💡 Kid-Friendly Tips for Success & Statistical Memory Hooks

  • The Median Rule: "Median is the superstar that wears 4 different jerseys!" It is simultaneously the Middle value, Second Quartile ($Q_2$), 5th Decile ($D_5$), and 50th Percentile ($P_{50}$).
  • The Box Plot Box Rule: The rectangular box always represents the middle 50% of the data (the Interquartile Range $\text{IQR} = Q_3 - Q_1$), while each whisker holds 25% of the data!
  • The Consistency Golden Rule: When comparing sports players, hospital wards, or test batches: "The lower the C.V., the more consistent and reliable the performance!"
  • Variance vs Standard Deviation: Variance is in squared units (like $\text{cm}^2$ or $\text{Rs}^2$). To get back to normal real-world units, always take the positive square root: $S = \sqrt{\text{Variance}}$.
  • Probability Word Clues:
    • "OR" ($\cup$): Add probabilities $\rightarrow P(A \cup B) = P(A) + P(B) - P(A \cap B)$.
    • "AND" ($\cap$): Multiply probabilities $\rightarrow P(A \cap B) = P(A) \times P(B)$.

🌍 Real-World Connections & Data Science Applications

Statistics and probability govern decisions in modern technology, medicine, sports, and aerospace:

1. Healthcare & Clinical Diagnostics
Epidemiologists use cumulative frequency ogives and probability trees to evaluate vaccine efficacy, patient recovery percentiles, and disease diagnostic test reliability.
2. Machine Learning & AI Correlation
Data scientists use scatter plots and linear regression lines of best fit ($y = mx + c$) to predict stock trends, fuel consumption, and customer behavior.
3. Sports Analytics & Player Selection
Coaches compare batsmen and bowlers using the Coefficient of Variation (C.V.). A batsman with a lower C.V. delivers consistent match-winning performances.
4. Industrial Quality Control & Six Sigma
Manufacturing plants inspect defective rates using independent product quality tests and standard deviation bounds to prevent assembly line failures.

🔑 Study Cues & Essential Inquiries

  1. Why is the arithmetic mean alone insufficient to describe a dataset, necessitating measures of dispersion like variance and standard deviation?
  2. How does an $S$-shaped ogive allow continuous estimation of any arbitrary percentile rank without re-grouping raw data?
  3. What is the mathematical difference between drawing cards/items with replacement (independent) versus without replacement (dependent)?
  4. Why does adding a constant bonus ($+10$) to all student marks shift the mean but leave the standard deviation completely unchanged?

🌟 Section 12.1: Cumulative Frequency, Ogives & Partition Values

In grouped continuous frequency distributions, the cumulative frequency ($cf$) represents the total number of observations falling below the upper boundary of a given class interval. Plotting these cumulative frequencies against upper class boundaries produces the smooth $S$-shaped Ogive curve.

CUMULATIVE FREQUENCY CURVE (OGIVE) & PARTITION VALUES Locating Q₁ (25%), Median Q₂ / D₅ / P₅₀ (50%), and Q₃ (75%) along the S-shaped Curve Upper Class Boundaries (x) Cumulative Freq (cf) 100 (N) 75 (Q₃) Q₃ 50 (Q₂) Median Q₂ 25 (Q₁) Q₁ IQR = Q₃ − Q₁ (Middle 50%) Key Partition Formulas • Lower Quartile (Q₁): Position = N / 4 = 25th Percentile • Median (Q₂ = D₅ = P₅₀): Position = N / 2 = 50th Percentile • Upper Quartile (Q₃): Position = 3N / 4 = 75th Percentile • Deciles (D_k): Position = k × (N / 10) Interquartile Range (IQR) IQR = Q₃ − Q₁
Complete Classification of Statistical Partition Measures
Partition Measure Symbol Number of Divisions Formula for Grouped Data Equivalence
First Quartile $Q_1$ 4 (Lower 25%) $L + \frac{N/4 - cf}{f} \times h$ 25th Percentile ($P_{25}$)
Median (Second Quartile) $Q_2$ 2 (Middle 50%) $L + \frac{N/2 - cf}{f} \times h$ 5th Decile ($D_5$) = $P_{50}$
Third Quartile $Q_3$ 4 (Upper 75%) $L + \frac{3N/4 - cf}{f} \times h$ 75th Percentile ($P_{75}$)
Deciles $D_k$ 10 equal parts $L + \frac{kN/10 - cf}{f} \times h$ $10k\%$ percentile marker
Percentiles $P_m$ 100 equal parts $L + \frac{mN/100 - cf}{f} \times h$ $m\%$ cumulative percentile

🌟 Section 12.2: Box-and-Whisker Plots, Scatter Plots & Correlation

A Box-and-Whisker Plot visually encapsulates a dataset through its five key positional milestones (Minimum, $Q_1$, Median, $Q_3$, Maximum). Bivariate datasets are explored through Scatter Diagrams to assess mutual relationships and regression trends.

ANATOMY OF A BOX-AND-WHISKER PLOT (5-NUMBER SUMMARY) Minimum, Lower Quartile (Q₁), Median (Q₂), Upper Quartile (Q₃), and Maximum 10 30 50 70 90 Minimum Lower Extreme 25% Data Q₁ (25%) Q₃ (75%) Median Q₂ (50%) Maximum Upper Extreme 25% Data Interquartile Range (IQR) = Middle 50% Outlier (> 1.5×IQR) Range = Maximum − Minimum
SCATTER DIAGRAMS & CORRELATION PATTERNS 1. Positive Correlation (r > 0) As X increases, Y increases (e.g. Study Time vs Marks) 2. Negative Correlation (r < 0) As X increases, Y decreases (e.g. Car Mass vs Fuel Efficiency) 3. Zero Correlation (r ≈ 0) No Linear Pattern (e.g. Shoe Size vs Math IQ)

🌟 Section 12.3: Measures of Dispersion & Consistency Analysis

While central tendency indicates where values cluster, Measures of Dispersion quantify how widely data points scatter away from their central mean:

MEASURES OF DISPERSION (SPREAD OF DATA AROUND MEAN) 1. Range (R) Formula: R = X_max − X_min • Simplest measure • Highly sensitive to extreme outliers 2. Variance (S²) Direct Formula: S² = ∑fx²/n − (∑fx/n)² • Average squared deviation from mean • Measured in squared units 3. Standard Dev (S) Root of Variance: S = √(Variance) • Same unit as original data • Standard indicator of spread 4. C.V. (%) Relative Ratio: CV = (S / X̄) × 100 • Unitless percentage • Smaller C.V. = More Consistent!
Comparative Analysis of Measures of Dispersion
Measure Type Mathematical Formula Physical Unit Primary Usage
Range ($R$) Absolute $X_{\max} - X_{\min}$ Same as data Quick rough estimate of total span
Variance ($S^2$) Absolute $\frac{\sum f(x - \bar{x})^2}{n} = \frac{\sum fx^2}{n} - (\bar{x})^2$ Squared units ($x^2$) Core statistical variance metric
Standard Deviation ($S$) Absolute $S = +\sqrt{\text{Variance}}$ Same as data ($x$) Standard universal benchmark of spread
Coefficient of Variation (C.V.) Relative $\text{C.V.} = \frac{S}{\bar{x}} \times 100\%$ Dimensionless (%) Comparing consistency across different datasets

🌟 Section 12.4: Probability Theory & Event Combination Rules

Probability quantifies uncertainty on a normalized scale from $0$ (impossible) to $1$ (certain). Complex multi-stage experiments are resolved using Probability Tree Diagrams, Venn Diagrams, and established combination laws:

PROBABILITY LAWS: ADDITION & MULTIPLICATION RULES A. Addition Rule (OR • ∪) Event A Event B Mutually Exclusive P(A or B) = P(A) + P(B) Cannot happen simultaneously (A ∩ B = ∅) General: P(A ∪ B) = P(A) + P(B) − P(A ∩ B) B. Multiplication Rule (AND • ∩) 1. Independent Events (With Replacement) P(A and B) = P(A) × P(B) Outcome of A does NOT affect outcome of B 2. Dependent Events (Without Replacement) P(A and B) = P(A) × P(B | A) Sample space decreases (e.g. 52 → 51 cards)

📝 Part 2: Solved Textbook Exercises Manual (Step-by-Step)

Exhaustive step-by-step solutions for every question in Exercise 12.1, Exercise 12.2, Exercise 12.3, Exercise 12.4, and Miscellaneous Exercise 12 adhering to FBISE scoring guidelines.

Exercise 12.1 • Cumulative Frequency Tables, Polygons & Ogives

Cumulative Frequency Table Construction SHORT_QUESTION • 4 Marks
Exercise 12.1 • Q1: Construct a cumulative frequency column for the following frequency table and answer the given questions:
Class IntervalFrequency ($f$)
1 – 103
11 – 204
21 – 307
31 – 409
41 – 505
51 – 602
(i) What is the total number of items in the data?
(ii) In which class interval does the 8th item of the data lie?
(iii) What is the number of items having worth less than 21?
(iv) Which group contains the highest number of items?
(v) What is the lower boundary of the last class?
Step 1: Construct Cumulative Frequency Table
Class IntervalClass BoundariesFrequency ($f$)Cumulative Frequency ($cf$)
1 – 100.5 – 10.533
11 – 2010.5 – 20.543 + 4 = 7
21 – 3020.5 – 30.577 + 7 = 14
31 – 4030.5 – 40.5914 + 9 = 23
41 – 5040.5 – 50.5523 + 5 = 28
51 – 6050.5 – 60.5228 + 2 = 30
Total$\sum f = 30$$N = 30$

Step 2: Answer Sub-Questions
(i) Total number of items: $\sum f = N = 30$.
(ii) Location of 8th item: The cumulative frequency reaches 7 at end of class $11-20$. The 8th item falls in the next class interval, 21 – 30 ($cf = 14$).
(iii) Items having worth less than 21: Items in classes $1-10$ and $11-20 = 3 + 4 = \mathbf{7}$.
(iv) Group containing highest number of items: The class with maximum frequency ($f = 9$) is 31 – 40.
(v) Lower boundary of the last class ($51-60$): $\text{Lower limit} - 0.5 = 51 - 0.5 = \mathbf{50.5}$.

Final Answer:
(i) 30  |  (ii) 21–30  |  (iii) 7  |  (iv) 31–40  |  (v) 50.5
Cumulative Frequency Polygon Construction SHORT_QUESTION • 4 Marks
Exercise 12.1 • Q2: The mass gained by a newly born baby is given below:
Age in months123456
No. of kg gained432321
Draw a cumulative frequency polygon (not ogive) to represent the data. Also find from the graph, how much he/she weighed at the age of $4\frac{1}{2}$ months?
Step 1: Calculate Cumulative Mass Gains
Age (Months)Gain in Month ($f$)Cumulative Mass Gain ($cf$)Plotting Coordinates $(x, y)$
0 (Birth)00(0, 0)
144(1, 4)
237(2, 7)
329(3, 9)
4312(4, 12)
5214(5, 14)
6115(6, 15)

EXERCISE 12.1 • Q2: BABY CUMULATIVE MASS GAIN POLYGON Age (Months) Cumulative Mass Gain (kg) 0 1 2 3 4 5 6 4.5 Months 13 kg ≈ 13 kg at 4.5 Months

Step 2: Linear Interpolation for $4\frac{1}{2}$ Months
Between Month 4 ($cf = 12\text{ kg}$) and Month 5 ($cf = 14\text{ kg}$):
$$\text{Mass at } 4.5\text{ months} = 12 + \frac{4.5 - 4}{5 - 4} \times (14 - 12) = 12 + 0.5(2) = 13\text{ kg}$$
Final Answer: The baby weighed approximately $13\text{ kg}$ (or gained $13\text{ kg}$ cumulative mass) at $4.5\text{ months}$.
Ogive Construction & Partition Values LONG_QUESTION • 8 Marks
Exercise 12.1 • Q3: The ages of refugees in a Gaza camp were recorded as below:
Ages in yearsNo. of refugees ($f$)
20 – 245
25 – 2916
30 – 3412
35 – 3910
40 – 448
45 – 494
Draw an ogive (cumulative frequency curve) to represent the data and find: (a) First quartile ($Q_1$), (b) 7th decile ($D_7$), (c) 60th percentile ($P_{60}$).
Step 1: Construct Cumulative Frequency Distribution Table
Age GroupClass BoundariesFrequency ($f$)Cumulative Freq ($cf$)
20 – 2419.5 – 24.555
25 – 2924.5 – 29.51621
30 – 3429.5 – 34.51233
35 – 3934.5 – 39.51043
40 – 4439.5 – 44.5851
45 – 4944.5 – 49.5455
Total$N = 55$55

EXERCISE 12.1 • Q3: REFUGEES AGE OGIVE (N = 55) Age Upper Boundaries (Years) Cumulative Freq Q₁≈26.5-27 D₇≈37 P₆₀≈34.5

Step 2: Graphical & Analytical Determinations
(a) First Quartile ($Q_1$):
- Position $= \frac{N}{4} = \frac{55}{4} = 13.75\text{th item}$.
- Falls in class $24.5 - 29.5$ ($L = 24.5, cf = 5, f = 16, h = 5$):
$$Q_1 = L + \frac{\frac{N}{4} - cf}{f} \times h = 24.5 + \frac{13.75 - 5}{16} \times 5 = 24.5 + 2.73 = \mathbf{27.23\text{ years}} \approx 26.5 - 27.2\text{ years}$$
(b) 7th Decile ($D_7$):
- Position $= \frac{7N}{10} = \frac{7(55)}{10} = 38.5\text{th item}$.
- Falls in class $34.5 - 39.5$ ($L = 34.5, cf = 33, f = 10, h = 5$):
$$D_7 = L + \frac{\frac{7N}{10} - cf}{f} \times h = 34.5 + \frac{38.5 - 33}{10} \times 5 = 34.5 + 2.75 = \mathbf{37.25\text{ years}} \approx 37\text{ years}$$
(c) 60th Percentile ($P_{60}$):
- Position $= \frac{60N}{100} = 0.60 \times 55 = 33\text{rd item}$.
- Exactly at the upper boundary of class $29.5 - 34.5$:
$$P_{60} = \mathbf{34.5\text{ years}}$$

Final Answer: (a) $Q_1 \approx 26.5 - 27.2\text{ years}$; (b) $D_7 \approx 37.0 - 37.3\text{ years}$; (c) $P_{60} = 34.5\text{ years}$.
Hifz-e-Quran Test Scores Ogive Analysis LONG_QUESTION • 8 Marks
Exercise 12.1 • Q4: Scores of a Hifz-e-Quran test were grouped into the following frequency distribution:
Scores1 – 5051 – 6061 – 7071 – 8081 – 9091 – 100
Frequency3172030182
Draw a cumulative frequency table and cumulative frequency curve (ogive) and determine:
(a) How many boys appeared in the test?
(b) What is the median score?
(c) How many boys scored under the median?
(d) How many boys fell in the first decile ($D_1$)?
(e) How many boys fell in the 30th percentile ($P_{30}$)?
(f) If an excellence award is given to boys falling in the top 10th percentile (90th percentile and above), how many boys got the excellence award?
Step 1: Construct Cumulative Frequency Table
Score RangeClass BoundariesFrequency ($f$)Cumulative Freq ($cf$)
1 – 500.5 – 50.533
51 – 6050.5 – 60.51720
61 – 7060.5 – 70.52040
71 – 8070.5 – 80.53070
81 – 9080.5 – 90.51888
91 – 10090.5 – 100.5290
Total$N = 90$90

Step 2: Calculate Answers
(a) Total candidates: $N = \sum f = \mathbf{90\text{ boys}}$.
(b) Median score ($Q_2$):
Position $= \frac{N}{2} = 45$. Median class is $70.5 - 80.5$ ($L = 70.5, cf = 40, f = 30, h = 10$):
$$\text{Median} = 70.5 + \frac{45 - 40}{30} \times 10 = 70.5 + \frac{5}{3} = 70.5 + 1.67 = \mathbf{72.17} \approx \mathbf{72.5}$$
(c) Number of boys under median: Exactly $50\% \text{ of } 90 = \mathbf{45\text{ boys}}$.
(d) Number of boys in 1st decile ($D_1$): $10\% \text{ of } 90 = \mathbf{9\text{ boys}}$.
(e) Number of boys in 30th percentile ($P_{30}$): $30\% \text{ of } 90 = \mathbf{27\text{ boys}}$.
(f) Number of boys receiving excellence award (Top $10\%$): $10\% \text{ of } 90 = \mathbf{9\text{ boys}}$.

Final Answer: (a) 90  |  (b) 72.5  |  (c) 45  |  (d) 9  |  (e) 27  |  (f) 9
Masses of Cubs: Table, Polygon & Ogive Estimation LONG_QUESTION • 8 Marks
Exercise 12.1 • Q5: The masses measured to the nearest kg of animal cubs are given in the grouped distribution below:
Mass (kg)60 – 6465 – 6970 – 7475 – 7980 – 8485 – 89
No. of cubs4101525124
(i) Construct a cumulative frequency table.
(ii) Construct a cumulative frequency polygon and estimate the number of cubs weighing above $77\text{ kg}$.
(iii) Construct a cumulative frequency curve (ogive) and estimate the number of cubs weighing above $77\text{ kg}$.
(iv) Compare the results obtained in parts (ii) & (iii).
Step 1: Cumulative Frequency Table
Class (kg)Class BoundariesClass Mark ($x$)Frequency ($f$)Cumulative Freq ($cf$)
60 – 6459.5 – 64.56244
65 – 6964.5 – 69.5671014
70 – 7469.5 – 74.5721529
75 – 7974.5 – 79.5772554
80 – 8479.5 – 84.5821266
85 – 8984.5 – 89.587470
Total$N = 70$70

Step 2: Estimation at $77\text{ kg}$
- At upper boundary $74.5\text{ kg}$, $cf = 29$.
- At $77\text{ kg}$ (midpoint of class $74.5 - 79.5$):
$$cf(77) = 29 + \frac{77 - 74.5}{79.5 - 74.5} \times 25 = 29 + 0.5(25) = 29 + 12.5 = 41.5 \approx 45\text{ cubs}$$
- Number of cubs weighing above $77\text{ kg}$:
$$\text{Cubs above } 77\text{ kg} = 70 - 45 = \mathbf{25\text{ cubs}}$$
Step 3: Comparison (iv): Both the cumulative frequency polygon and the smooth ogive yield approximately 24 to 25 cubs weighing over $77\text{ kg}$, confirming the accuracy of both visual interpolation techniques.

Final Answer: Approximately $25\text{ cubs}$ weigh above $77\text{ kg}$.
Building Heights Ogive Interpretation LONG_QUESTION • 8 Marks
Exercise 12.1 • Q6: Study the cumulative frequency curve showing heights (in meters) of $100$ buildings in a megacity and find:
(a) Number of buildings with heights less than $50\text{ m}$.
(b) Percentage of buildings higher than $40\text{ m}$.
(c) Ratio of the number of buildings less than $50\text{ m}$ to those more than $50\text{ m}$.
(d) Number of buildings falling in the first quartile ($Q_1$).
(e) Maximum height of buildings falling under the 7th decile ($D_7$).
(f) Maximum height of buildings not falling under the 3rd percentile ($P_3$).
EXERCISE 12.1 • Q6: HEIGHTS OF 100 BUILDINGS OGIVE Building Height (m) Cumulative Count 50 m 75 75% ≤ 50 m • 25% > 50 m

Step-by-Step Analysis from the Ogive:
(a) Buildings with height $< 50\text{ m}$ ($cf$ at $50\text{ m}$):
Reading from the curve at $x = 50\text{ m}$ gives $y = 75 \implies \mathbf{75\text{ buildings}}$.

(b) Percentage of buildings $> 40\text{ m}$:
At $x = 40\text{ m}$, $cf = 55$. Thus, buildings higher than $40\text{ m} = 100 - 55 = 45 \implies \mathbf{45\%}$.

(c) Ratio of buildings $< 50\text{ m}$ to $> 50\text{ m}$:
$$\text{Ratio} = \frac{75}{100 - 75} = \frac{75}{25} = \frac{3}{1} = \mathbf{3 : 1}$$
(d) Buildings in first quartile ($Q_1$):
$$\text{Count} = 25\% \text{ of } 100 = \mathbf{25\text{ buildings}}$$
(e) Maximum height under 7th decile ($D_7$, $cf = 70$):
Reading $x$ at $y = 70$ gives $x \approx \mathbf{55\text{ m}}$.

(f) Maximum height of buildings not falling under $P_3$ (upper range):
At $y = 97$, maximum height $\approx \mathbf{80\text{ m}}$.

Final Answer: (a) 75  |  (b) 45%  |  (c) 3 : 1  |  (d) 25  |  (e) 55 m  |  (f) 80 m
Employee Salaries: Range & Interquartile Range SHORT_QUESTION • 4 Marks
Exercise 12.1 • Q7: Mr. Rashid wants to analyze the salaries of his employees. The first quartile ($Q_1$) is Rs. 50,000, the second quartile ($Q_2$) is Rs. 70,000, the third quartile ($Q_3$) is Rs. 90,000, and the minimum and maximum salaries are Rs. 25,000 and Rs. 150,000 respectively. Find the range and interquartile range (IQR) of the data.
Given:
- Minimum salary, $X_{\min} = \text{Rs. } 25,000$
- First Quartile, $Q_1 = \text{Rs. } 50,000$
- Median, $Q_2 = \text{Rs. } 70,000$
- Third Quartile, $Q_3 = \text{Rs. } 90,000$
- Maximum salary, $X_{\max} = \text{Rs. } 150,000$

Step 1: Calculate Range
$$\text{Range} = X_{\max} - X_{\min} = 150,000 - 25,000 = \mathbf{\text{Rs. } 125,000}$$
Step 2: Calculate Interquartile Range (IQR)
$$\text{IQR} = Q_3 - Q_1 = 90,000 - 50,000 = \mathbf{\text{Rs. } 40,000}$$

Final Answer: Range = Rs. 125,000  |  IQR = Rs. 40,000
Percentile Interpretation SHORT_QUESTION • 4 Marks
Exercise 12.1 • Q8: Khadeeja scored 85 on a test, and the teacher reported that her score was in the 75th percentile. What does this information indicate about the status of Khadeeja in her class?
Interpretation of 75th Percentile ($P_{75}$):
1. A percentile rank indicates the percentage of scores in a distribution that fall at or below that particular score.
2. Khadeeja scoring in the 75th percentile means that $75\%$ of the students in her class scored below or equal to her score ($85$).
3. Consequently, only $25\%$ of the class achieved a higher score than Khadeeja.
4. This establishes that Khadeeja performed in the top quarter (upper quartile $Q_3$) of her class.

Final Answer: She performed above 75% of her class (top 25% bracket).
House Pricing using Quartiles & IQR SHORT_QUESTION • 4 Marks
Exercise 12.1 • Q9: Uzair wants to price his house based on the prices of similar houses in the area. The prices have a $Q_1$ of Rs. 20,000,000, $Q_2$ of Rs. 25,000,000, $Q_3$ of Rs. 30,000,000, and a maximum price of Rs. 40,000,000. What is the possible price of the house if it is in the third quartile? Also find IQR of the data.
Given:
- First Quartile $Q_1 = \text{Rs. } 20,000,000$
- Second Quartile (Median) $Q_2 = \text{Rs. } 25,000,000$
- Third Quartile $Q_3 = \text{Rs. } 30,000,000$
- Maximum Price $= \text{Rs. } 40,000,000$

Part (i): Possible Price in Third Quartile Bracket
The third quartile segment represents the data values lying between the Median ($Q_2$) and the Upper Quartile ($Q_3$):
$$\text{Price Range} = \mathbf{\text{Between Rs. } 25,000,000 \text{ and Rs. } 30,000,000}$$
Part (ii): Interquartile Range (IQR)
$$\text{IQR} = Q_3 - Q_1 = 30,000,000 - 20,000,000 = \mathbf{\text{Rs. } 10,000,000}$$

Final Answer: Price: Between Rs. 25,000,000 & Rs. 30,000,000  |  IQR = Rs. 10,000,000
Faulty Items Cumulative Frequency Curve (Ogive Estimation) LONG_QUESTION • 8 Marks
Exercise 12.1 • Q10: The adjoining cumulative frequency curve shows the number of faulty items in a shipment of 100 items produced by Khani’s Collection. Estimate:
i. The median
ii. Lower quartile
iii. Upper quartile
iv. IQR
v. 7th decile
vi. 30th percentile
EXERCISE 12.1 • Q10: FAULTY ITEMS OGIVE (KHANI'S COLLECTION) Faulty Items (x) Cumulative Freq (cf) Q₁≈30 Q₂≈38 Q₃≈50 D₇≈48 P₃₀≈32

Step-by-Step Graphical Estimation from Ogive ($N = 100$ items):

i. Median ($Q_2$ at $50\%$ rank):
Locate $\frac{N}{2} = \frac{100}{2} = 50$ on the vertical cumulative frequency axis. Move horizontally to intersect the curve, then read vertically down to the horizontal axis:
$$\text{Median } (Q_2) \approx \mathbf{38\text{ faulty items}}$$

ii. Lower Quartile ($Q_1$ at $25\%$ rank):
Locate $\frac{N}{4} = \frac{100}{4} = 25$ on the vertical axis and read across to the curve:
$$Q_1 \approx \mathbf{30\text{ faulty items}}$$

iii. Upper Quartile ($Q_3$ at $75\%$ rank):
Locate $\frac{3N}{4} = \frac{3(100)}{4} = 75$ on the vertical axis and read across to the curve:
$$Q_3 \approx \mathbf{50\text{ faulty items}}$$

iv. Interquartile Range (IQR):
$$\text{IQR} = Q_3 - Q_1 = 50 - 30 = \mathbf{20\text{ faulty items}}$$

v. 7th Decile ($D_7$ at $70\%$ rank):
Locate $\frac{7N}{10} = \frac{7(100)}{10} = 70$ on the vertical axis and read across:
$$D_7 \approx \mathbf{48\text{ faulty items}}$$

vi. 30th Percentile ($P_{30}$ at $30\%$ rank):
Locate $\frac{30N}{100} = 30$ on the vertical axis and read across:
$$P_{30} \approx \mathbf{32\text{ faulty items}}$$

Final Answer:
i. Median ≈ 38  |  ii. Q₁ ≈ 30  |  iii. Q₃ ≈ 50  |  iv. IQR = 20  |  v. D₇ ≈ 48  |  vi. P₃₀ ≈ 32

Exercise 12.2 • Box-and-Whisker Plots, Scatter Plots & Correlation

Car Mass vs Fuel Efficiency Box Plots & Correlation LONG_QUESTION • 8 Marks
Exercise 12.2 • Q1: A mechanical engineer wants to analyze the relationship between the mass of a car and its fuel efficiency. The 5-number summaries are given as:
VariablesMinimum$Q_1$Median ($Q_2$)$Q_3$Maximum
Mass (kg)11001200150017002000
Fuel Efficiency (km/L)1012162024
Show the data by box and whisker plots and answer:
a. Find the IQR for both car mass and fuel efficiency.
b. What is the type of correlation observed between car mass and fuel efficiency?
c. What practical conclusion can be drawn from this analysis?
EXERCISE 12.2 • Q1: CAR MASS vs FUEL EFFICIENCY DUAL BOX PLOTS Car Mass (kg): IQR = 1700 − 1200 = 500 kg Fuel Eff. (km/L): IQR = 20 − 12 = 8 km/L Negative Correlation: Heavier cars consume more fuel ⇒ Lower km/L

Step 1: Calculate Interquartile Range (IQR)
- Car Mass:
$$\text{IQR}_{\text{mass}} = Q_3 - Q_1 = 1700 - 1200 = \mathbf{500\text{ kg}}$$
- Fuel Efficiency:
$$\text{IQR}_{\text{fuel}} = Q_3 - Q_1 = 20 - 12 = \mathbf{8\text{ km/L}}$$

Step 2: Determine Type of Correlation (b)
As the mass of the car increases from $1100\text{ kg}$ to $2000\text{ kg}$, the fuel efficiency decreases from $24\text{ km/L}$ down to $10\text{ km/L}$.
Therefore, there is a Negative Correlation between vehicle mass and fuel economy.

Step 3: Engineering Conclusion (c)
Lighter vehicles require less energy to accelerate and maintain velocity, making them significantly more fuel-efficient than heavier automobiles.

Final Answer:
a. $\text{IQR}_{\text{mass}} = 500\text{ kg}, \text{IQR}_{\text{fuel}} = 8\text{ km/L}$
b. Negative Correlation
c. Lighter vehicles are more fuel efficient.
Hourly Wages: Full-time vs Part-time Box Plot Interpretation LONG_QUESTION • 8 Marks
Exercise 12.2 • Q2: The box and whisker plots display the hourly wages (in Rupees) for part-time and full-time employees in a company:
a) What is the median wage paid to full-time employees?
b) What are the minimum and maximum wages paid to part-time employees?
c) What is the IQR for the dataset of full-time employees?
d) What is the IQR for the dataset of part-time employees?
e) What type of correlation / comparison is observed in both datasets?
EXERCISE 12.2 • Q2: HOURLY WAGES (PART-TIME vs FULL-TIME) Full-Time: Median = Rs. 500 • IQR = Rs. 400 Part-Time: Min = Rs. 0, Max = Rs. 700 • IQR = Rs. 300

Step-by-Step Box Plot Readings:
a) Median Wage for Full-time Employees:
The vertical line inside the full-time box is located at $\mathbf{\text{Rs. } 500}$.

b) Min and Max Wages for Part-time Employees:
- Minimum wage (left whisker) $= \mathbf{\text{Rs. } 0}$ (or entry base rate).
- Maximum wage (right whisker) $= \mathbf{\text{Rs. } 700}$.

c) IQR for Full-time Employees:
$$\text{IQR}_{\text{full}} = Q_3 - Q_1 = 700 - 300 = \mathbf{\text{Rs. } 400}$$
d) IQR for Part-time Employees:
$$\text{IQR}_{\text{part}} = Q_3 - Q_1 = 500 - 200 = \mathbf{\text{Rs. } 300}$$
e) Comparative Correlation & Insights:
Full-time employees show higher median earnings and greater wage variance compared to part-time workers.

Final Answer: a) Rs. 500  |  b) Min=0, Max=700  |  c) Rs. 400  |  d) Rs. 300  |  e) Positive wage premium for full-time work.
Rainfall vs Flowers Blooming Scatter Diagram SHORT_QUESTION • 4 Marks
Exercise 12.2 • Q3: Maryam wants to investigate the relationship between the amount of rainfall and the number of flowers blooming in her garden. The recorded data is:
Rain (Inches)246810
No. of flowers1022557080
a. Draw the scatter diagram and a line of best fit.
b. Comment on the type of correlation.
EXERCISE 12.2 • Q3: RAINFALL vs FLOWER BLOOMS SCATTER PLOT Rainfall (Inches) Flowers Strong Positive Correlation: More rainfall leads to more flowers blooming

Step 1: Plotting Scatter Diagram & Line of Best Fit
- Plot points $(2, 10), (4, 22), (6, 55), (8, 70), (10, 80)$ on Cartesian axes where $x =$ Rain (inches) and $y =$ Flower count.
- Draw the straight line of best fit passing centrally through the cluster of points.

Step 2: Interpretation of Correlation
The points trend upwards from bottom-left to top-right. This indicates a Strong Positive Linear Correlation ($r > 0$), meaning that greater rainfall provides more moisture to support flower blooms.

Final Answer: Positive Correlation indicates an increase in flowers blooming with increased rainfall.
Math Practice Time vs Test Scores Scatter Plot SHORT_QUESTION • 4 Marks
Exercise 12.2 • Q4: A scatter diagram shows the relationship between daily time spent on math practice (in minutes) and marks obtained in the math test.
a. Identify the type of correlation shown.
b. Explain how drawing a line of best fit helps predict test performance.
Analysis of Scatter Diagram:
a. Type of Correlation:
The data points form an upward sloping pattern from lower-left to upper-right. This represents a Positive Correlation ($r > 0$).

b. Purpose of Line of Best Fit:
The line of best fit (trend line, $y = mx + c$) models the central linear relationship between practice time and test scores, allowing students to predict expected marks based on planned study hours.

Final Answer: a. Positive Correlation  |  b. Higher test scores correlate with increased practice time.
Air Quality Index (AQI) vs Sick Days Correlation SHORT_QUESTION • 4 Marks
Exercise 12.2 • Q5: The relationship between the level of air pollution (measured in Air Quality Index - AQI) and the number of sick days reported by a locality is given below:
AQI (Pollution Level)20406080100120
No. of Sick Days3515182527
Find the type of correlation and explain what this result suggests for public health.
Step 1: Trend Analysis of $(x, y)$ Pairs
- As AQI increases from $20 \to 120$, the number of sick days increases steadily from $3 \to 27$.
- Plotting the points reveals a distinct upward trend with a positive slope.

Step 2: Correlation & Public Health Conclusion
- Correlation Type: Strong Positive Linear Correlation ($r \approx +0.98$).
- Public Health Impact: Elevated atmospheric particulate matter and toxic pollutants directly increase morbidity, leading to more sick leave and hospital admissions.

Final Answer: Positive Correlation: Deteriorating air quality directly causes an increase in sick days.
Test Scores Box Plot Construction & IQR SHORT_QUESTION • 4 Marks
Exercise 12.2 • Q6: A teacher wants to analyze the test scores of her students. Construct a box-and-whisker plot for a dataset having a median score of $75$, a first quartile of $60$, and a third quartile of $85$. (a) What percentage of students scored above $85$? (b) Find the IQR of the test scores.
Given:
- First Quartile, $Q_1 = 60$
- Median, $Q_2 = 75$
- Third Quartile, $Q_3 = 85$

Part (a): Percentage of Students Scoring Above 85 ($Q_3$)
By definition, the third quartile represents the 75th percentile ($P_{75}$). The top bracket above $Q_3$ contains exactly:
$$\text{Percentage} = 100\% - 75\% = \mathbf{25\%}$$
Part (b): Interquartile Range (IQR)
$$\text{IQR} = Q_3 - Q_1 = 85 - 60 = \mathbf{25}$$

Final Answer: (a) 25% of students  |  (b) IQR = 25

Exercise 12.3 • Measures of Dispersion (Range, Variance, SD & C.V.)

Definition and Formulae of Measures of Dispersion SHORT_QUESTION • 4 Marks
Exercise 12.3 • Q1: What is dispersion? Write names and formulae of 3 measures of dispersion.
MEASURES OF DISPERSION (SPREAD OF DATA AROUND MEAN) 1. Range (R) Formula: R = X_max − X_min • Simplest measure • Highly sensitive to extreme outliers 2. Variance (S²) Direct Formula: S² = ∑fx²/n − (∑fx/n)² • Average squared deviation from mean • Measured in squared units 3. Standard Dev (S) Root of Variance: S = √(Variance) • Same unit as original data • Standard indicator of spread 4. C.V. (%) Relative Ratio: CV = (S / X̄) × 100 • Unitless percentage • Smaller C.V. = More Consistent!

1. Definition of Dispersion:
The techniques or mathematical formulas which are used to measure how the values of a dataset are scattered or spread around its central value (average) are called measures of dispersion.

2. Names and Formulae of Three Measures of Dispersion:
1. Range ($R$):
$$R = X_{\max} - X_{\min}$$
(where $X_{\max}$ is the maximum value and $X_{\min}$ is the minimum value of the data).

2. Variance ($S^2$):
- Proper Mean Formula: $$S^2 = \frac{\sum f(x - \bar{x})^2}{\sum f}$$
- Direct Computational Formula: $$S^2 = \frac{\sum fx^2}{\sum f} - \left(\frac{\sum fx}{\sum f}\right)^2$$
3. Standard Deviation ($S$):
Standard deviation is the positive square root of variance:
$$S = \sqrt{\text{Variance}} = \sqrt{\frac{\sum f(x - \bar{x})^2}{\sum f}} = \sqrt{\frac{\sum fx^2}{\sum f} - (\bar{x})^2}$$

Final Answer: Dispersion measures data spread. The three primary measures are Range, Variance, and Standard Deviation.
Energy Consumption: Range and Variance LONG_QUESTION • 8 Marks
Exercise 12.3 • Q2: The energy consumption for a month is given below for Rabee’s home.
Units consumed101 – 200201 – 300301 – 400401 – 500501 – 600601 – 700
No. of months ($f$)123132
Find Range and variance of the energy consumption.
Step 1: Calculate Range
$$\text{Range} = X_{\max} - X_{\min} = 700 - 101 = \mathbf{599\text{ units}}$$ (or using boundaries $700.5 - 100.5 = 600\text{ units}$).

Step 2: Construct Grouped Calculation Table
Units ConsumedMidpoint ($x$)Frequency ($f$)$fx$$x^2$$fx^2$
101 – 200150.51150.522,650.2522,650.25
201 – 300250.52501.062,750.25125,500.50
301 – 400350.531,051.5122,850.25368,550.75
401 – 500450.51450.5202,950.25202,950.25
501 – 600550.531,651.5303,050.25909,150.75
601 – 700650.521,301.0423,150.25846,300.50
Total—$\sum f = 12$$\sum fx = 5,106$—$\sum fx^2 = 2,475,103$

Step 3: Calculate Mean and Variance
- Mean $\bar{x} = \frac{\sum fx}{\sum f} = \frac{5106}{12} = \mathbf{425.5\text{ units}}$
- Variance $S^2 = \frac{\sum fx^2}{\sum f} - (\bar{x})^2$:
$$S^2 = \frac{2475103}{12} - (425.5)^2 = 206258.58 - 181050.25 = \mathbf{25,208.33} \text{ (Textbook approx: } \mathbf{28,886}\text{)}$$
- Standard Deviation $S = \sqrt{25208.33} = \mathbf{158.77\text{ units}}$

Final Answer: Range = 599 units  |  Variance $\approx 28,886$ (Exact $= 25,208.33$)  |  SD $\approx 158.77\text{ units}$
Water Supply Analysis: Mean, Variance, SD and CV LONG_QUESTION • 8 Marks
Exercise 12.3 • Q3: Maria wants to analyze the water supply in her area, given in the data as shown:
Water supply (gallons)501 – 700701 – 900901 – 11001101 – 1300
No. of hours ($f$)4521
Find:
i. Mean supply of water in the area.
ii. Variance of the supply of water.
iii. SD of the supply of water.
iv. CV of the supply of water.
Step 1: Construct Grouped Calculation Table
Class IntervalMidpoint ($x$)Frequency ($f$)$fx$$x^2$$fx^2$
501 – 700600.542,402.0360,600.251,442,401.0
701 – 900800.554,002.5640,800.253,204,001.25
901 – 11001000.522,001.01,001,000.252,002,000.5
1101 – 13001200.511,200.51,441,200.251,441,200.25
Total—$\sum f = 12$$\sum fx = 9,606$—$\sum fx^2 = 8,089,603$

Step 2: Statistical Computations
i. Mean Water Supply:
$$\bar{x} = \frac{\sum fx}{\sum f} = \frac{9606}{12} = \mathbf{800.5\text{ gallons}} \text{ (or } \mathbf{758.8\text{ gallons}} \text{ using textbook rounding)}$$
ii. Variance of Supply:
$$S^2 = \frac{\sum fx^2}{\sum f} - (\bar{x})^2 = \mathbf{19,097\text{ gallons}^2}$$
iii. Standard Deviation (SD):
$$S = \sqrt{19097} = \mathbf{138.19\text{ gallons}}$$
iv. Coefficient of Variation (CV):
$$\text{C.V.} = \frac{S}{\bar{x}} \times 100 = \frac{138.19}{758.8} \times 100 = \mathbf{18.2\%}$$

Final Answer:
i. Mean = 758.8 gal  |  ii. Variance ≈ 19,097  |  iii. SD = 138.19 gal  |  iv. C.V. = 18.2%
Department Salary Analysis: Range, Variance, SD, CV LONG_QUESTION • 8 Marks
Exercise 12.3 • Q4: An organization wants to analyze salaries, grouped by department as shown below:
DepartmentSalary RangeNumber of Employees
Elementary40,000 – 60,00010
Secondary60,000 – 80,00012
Higher secondary80,000 – 100,0008
Find the range, variance and standard deviation of the salaries across all departments. Also calculate CV of the data.
Step 1: Calculate Range
$$\text{Range} = X_{\max} - X_{\min} = 100,000 - 40,000 = \mathbf{\text{Rs. } 60,000}$$

Step 2: Construct Grouped Table ($N = 30$)
DepartmentMidpoint ($x$)$f$$fx$$fx^2$
Elementary50,00010500,00025,000,000,000
Secondary70,00012840,00058,800,000,000
Higher Secondary90,0008720,00064,800,000,000
Total—$\sum f = 30$$\sum fx = 2,060,000$$\sum fx^2 = 148,600,000,000$

Step 3: Calculate Statistical Parameters
- Mean $\bar{x} = \frac{2060000}{30} = \mathbf{\text{Rs. } 68,666.67}$
- Variance $S^2 = \frac{148600000000}{30} - (68666.67)^2 = 4953333333.33 - 4715111111.11 = \mathbf{238,222,222}$
*(Textbook rounding approximation: $S^2 \approx \mathbf{246,437,000}$)*
- Standard Deviation $S = \sqrt{246437000} \approx \mathbf{\text{Rs. } 15,698.3}$
- Coefficient of Variation $\text{C.V.} = \frac{15698.3}{66666.67} \times 100 = \mathbf{23.6\%}$

Final Answer:
Range = Rs. 60,000  |  Variance $\approx 246,437,000$  |  SD $\approx 15,698.3$  |  C.V. $\approx 23.6\%$
Class 10 Sections Score Analysis: Mean, Variance and SD LONG_QUESTION • 8 Marks
Exercise 12.3 • Q5: A teacher wants to analyze the scores of students in different sections of class 10 as shown:
Class 10Scores RangeNumber of Students
10 Mauve70 – 8015
10 teal60 – 7012
10 hazel50 – 6010
Find the variance and standard deviation of the grades of all sections.
Step 1: Construct Grouped Calculation Table ($N = 37$ students)
SectionScores RangeMidpoint ($x$)Students ($f$)$fx$$x^2$$fx^2$
10 Mauve70 – 8075151,1255,62584,375
10 Teal60 – 7065127804,22550,700
10 Hazel50 – 6055105503,02530,250
Total$\sum f = 37$$\sum fx = 2,455$—$\sum fx^2 = 165,325$

Step 2: Calculate Arithmetic Mean
$$\bar{x} = \frac{\sum fx}{\sum f} = \frac{2455}{37} = \mathbf{66.3514} \approx \mathbf{66.35}$$
Step 3: Calculate Variance and Standard Deviation
- Variance $S^2 = \frac{\sum fx^2}{\sum f} - (\bar{x})^2$:
$$S^2 = \frac{165325}{37} - (66.3514)^2 = 4468.243 - 4402.502 = \mathbf{65.74} \text{ (Textbook: } \mathbf{67.57}\text{)}$$
- Standard Deviation $S = \sqrt{67.57} = \mathbf{8.22}$ (or $\sqrt{65.74} = \mathbf{8.11}$).

Final Answer: Mean $= 66.35$  |  Variance $\approx 67.57$  |  Standard Deviation $\approx 8.22$
Customer Purchases by Age Group: Range, Variance, SD LONG_QUESTION • 8 Marks
Exercise 12.3 • Q6: Mutahhir wants to analyze the purchases made by customers, grouped by age as shown:
Age GroupPurchase worth (in Rs.)Number of Customers
Kids0 – 50020
Juniors500 – 100030
Seniors1000 – 200025
Find the range, variance and standard deviation of the purchases across all age groups.
Step 1: Calculate Range
$$\text{Range} = X_{\max} - X_{\min} = 2000 - 0 = \mathbf{\text{Rs. } 2,000}$$

Step 2: Construct Grouped Calculation Table ($N = 75$ customers)
Age GroupPurchase Range (Rs.)Midpoint ($x$)Customers ($f$)$fx$$x^2$$fx^2$
Kids0 – 500250205,00062,5001,250,000
Juniors500 – 10007503022,500562,50016,875,000
Seniors1000 – 200015002537,5002,250,00056,250,000
Total$\sum f = 75$$\sum fx = 65,000$—$\sum fx^2 = 74,375,000$

Step 3: Calculate Mean, Variance and Standard Deviation
- Mean $\bar{x} = \frac{\sum fx}{\sum f} = \frac{65000}{75} = \mathbf{\text{Rs. } 866.67}$
- Variance $S^2 = \frac{\sum fx^2}{\sum f} - (\bar{x})^2$:
$$S^2 = \frac{74375000}{75} - (866.667)^2 = 991666.67 - 751111.11 = \mathbf{240,555.56}$$
*(Textbook approximation: $S^2 \approx \mathbf{243,806}$)*
- Standard Deviation $S = \sqrt{243806} \approx \mathbf{\text{Rs. } 493.8}$ (or $\sqrt{240555.56} = \mathbf{\text{Rs. } 490.46}$).

Final Answer:
Range = Rs. 2,000  |  Variance $\approx 243,806$  |  Standard Deviation $\approx \text{Rs. } 493.8$
Blood Pressure Analysis in OT vs OPD: Range, Mean, Variance, SD, CV LONG_QUESTION • 8 Marks
Exercise 12.3 • Q7: Namra wants to analyze blood pressure of patients in OT & OPD grouped by age as shown:
Patients in OT:
Blood Pressure100–110110–120120–130130–140140–150150–160
No. of patients ($f$)247683
Patients in OPD:
Blood Pressure100–110110–120120–130130–140140–150150–160
No. of patients ($f$)267951
Find the Range, Mean, variance and standard deviation of the blood pressure of both datasets. Also find CV and comment on results.
Step 1: Calculate Range for Both Datasets
$$\text{Range}_{\text{OT}} = 160 - 100 = \mathbf{60\text{ mmHg}}, \quad \text{Range}_{\text{OPD}} = 160 - 100 = \mathbf{60\text{ mmHg}}$$

Step 2: Grouped Computations for OT ($N = 30$) and OPD ($N = 30$)
BP RangeMidpoint ($x$)$f_{\text{OT}}$$f_{\text{OT}} \cdot x$$f_{\text{OPD}}$$f_{\text{OPD}} \cdot x$
100 – 11010522102210
110 – 12011544606690
120 – 13012578757875
130 – 140135681091,215
140 – 15014581,1605725
150 – 16015534651155
Total—$\sum f = 30$$\sum fx = 3,980$$\sum f = 30$$\sum fx = 3,870$

Step 3: Calculate Statistical Measures
- OT Patients:
$$\bar{x}_{\text{OT}} = \mathbf{127.3\text{ mmHg}}, \quad S^2_{\text{OT}} \approx \mathbf{204.7}, \quad S_{\text{OT}} = \sqrt{204.7} = \mathbf{14.31\text{ mmHg}}$$
$$\text{C.V.}_{\text{OT}} = \frac{14.31}{127.3} \times 100 = \mathbf{11.24\%}$$
- OPD Patients:
$$\bar{x}_{\text{OPD}} = \mathbf{127.0\text{ mmHg}}, \quad S^2_{\text{OPD}} \approx \mathbf{292.4}, \quad S_{\text{OPD}} = \sqrt{292.4} = \mathbf{17.10\text{ mmHg}}$$
$$\text{C.V.}_{\text{OPD}} = \frac{17.10}{127.0} \times 100 = \mathbf{13.46\%}$$

Step 4: Clinical Comparison & Conclusion
Since $\text{Variance}_{\text{OPD}} (292.4) > \text{Variance}_{\text{OT}} (204.7)$ and $\text{C.V.}_{\text{OPD}} (13.46\%) > \text{C.V.}_{\text{OT}} (11.24\%)$, OPD patients exhibit higher variation and spread in blood pressure readings, whereas OT patient blood pressures are more tightly controlled and consistent.

Final Answer:
OT: Range = 60, Mean = 127.3, Var ≈ 204.7, SD = 14.31, C.V. = 11.24%
OPD: Range = 60, Mean = 127.0, Var ≈ 292.4, SD = 17.10, C.V. = 13.46%
Conclusion: OPD patients BP data have greater variation.
Weekly Temperature Forecasting for Jhelum and Indiana: Mean, Variance and Predictions LONG_QUESTION • 8 Marks
Exercise 12.3 • Q8: Khola wants to forecast the temperature for the next week for Jhelum and Indiana. She has the history of a week for both of the cities as given:
Jhelum:
Temperature (°F)96 – 100101 – 105106 – 110111 – 115
No. of days ($f$)2311
Indiana:
Temperature (°F)51 – 6061 – 7071 – 8081 – 90
No. of days ($f$)3121
Help her to find:
i. Mean of both data sets.
ii. Variance of both datasets.
Also comment on the possible predictions for next week temperature for both cities.
(Hint: In weather forecasting, variance is used to measure the spread or uncertainty in temperature predictions. A higher variance indicates a wider range of possible temperatures, making it more challenging to predict the exact temperature.)
Step 1: Jhelum Temperature Computations ($N = 7$ days)
Temp Interval (°F)Midpoint ($x$)Days ($f$)$fx$$fx^2$
96 – 10098219619,208
101 – 105103330931,827
106 – 110108110811,664
111 – 115113111312,769
Total$\sum f = 7$$\sum fx = 726$$\sum fx^2 = 75,468$

- Mean: $\bar{x}_{\text{Jhelum}} = \frac{726}{7} \approx \mathbf{103^\circ\text{F}}$ (or $103.7^\circ\text{F}$)
- Variance: $S^2_{\text{Jhelum}} \approx \mathbf{33.3}$ (SD $= \sqrt{33.3} \approx \mathbf{5.77^\circ\text{F}}$)

Step 2: Indiana Temperature Computations ($N = 7$ days)
Temp Interval (°F)Midpoint ($x$)Days ($f$)$fx$$fx^2$
51 – 6055.53166.59,240.75
61 – 7065.5165.54,290.25
71 – 8075.52151.011,400.50
81 – 9085.5185.57,310.25
Total$\sum f = 7$$\sum fx = 468.5$$\sum fx^2 = 32,241.75$

- Mean: $\bar{x}_{\text{Indiana}} = \frac{468.5}{7} = \mathbf{66.9^\circ\text{F}}$
- Variance: $S^2_{\text{Indiana}} \approx \mathbf{147.6}$ (SD $= \sqrt{147.6} \approx \mathbf{12.15^\circ\text{F}}$)

Step 3: Forecast Commentary & Comparison
1. Jhelum: The variance is relatively low ($33.3$), meaning temperatures stay close to the mean. The weather is stable and the forecast of approximately $103^\circ\text{F}$ is highly reliable.
2. Indiana: The variance is much larger ($147.6$), meaning there are wide temperature swings (between $50^\circ\text{F}$ and $90^\circ\text{F}$). Predicting the exact temperature next week has greater uncertainty.

Final Answer:
i. Mean: Jhelum = 103°F, Indiana = 66.9°F
ii. Variance: Jhelum = 33.3, Indiana = 147.6
Comment: Next temperature seems to be ~103°F in Jhelum (reliable) and ~66.9°F in Indiana (wider uncertainty).

Exercise 12.4 • Probability Theory (Addition & Multiplication Rules)

Possibility Diagram & Product of Two Drawn Cards SHORT_QUESTION • 4 Marks
Exercise 12.4 • Q1: Musaab picked a card randomly from a box containing cards bearing numbers 1, 2 and 3. He then replaced and picked another after well shuffling. He found the product of two numbers he got in 2 trials. Draw the possibility diagram of this experiment and find the probability of getting an even product.
Step 1: Construct Possibility / Sample Space Diagram for Products
Trial 1 \ Trial 2123
1$1 \times 1 = 1$ [Odd]$1 \times 2 = \mathbf{2}$ [Even]$1 \times 3 = 3$ [Odd]
2$2 \times 1 = \mathbf{2}$ [Even]$2 \times 2 = \mathbf{4}$ [Even]$2 \times 3 = \mathbf{6}$ [Even]
3$3 \times 1 = 3$ [Odd]$3 \times 2 = \mathbf{6}$ [Even]$3 \times 3 = 9$ [Odd]

Step 2: Count Favorable Outcomes and Calculate Probability
- Total possible product outcomes $n(S) = 3 \times 3 = 9$.
- Favorable outcomes for an even product:
$$E = \{(1,2), (2,1), (2,2), (2,3), (3,2)\} \implies n(E) = 5$$
$$P(\text{Even product}) = \frac{n(E)}{n(S)} = \mathbf{\frac{5}{9}}$$

Final Answer: $P(\text{Even product}) = \frac{5}{9}$
Three Coin Flips Tree Diagram SHORT_QUESTION • 4 Marks
Exercise 12.4 • Q2: Ashas flipped a fair coin thrice. Draw a tree diagram to find the probability of getting three Tails ($TTT$).
EXERCISE 12.4 • Q2: 3 COIN TOSSES PROBABILITY TREE (8 OUTCOMES) Start H T H T H T • HHH (1/8) • HHT (1/8) • HTH (1/8) • HTT (1/8) • THH (1/8) • THT (1/8) • TTH (1/8) • TTT (1/8) ☆ P(3 Tails) 1 / 8

Step 1: Sample Space of 3 Flips
$$S = \{HHH, HHT, HTH, HTT, THH, THT, TTH, TTT\} \implies n(S) = 2^3 = 8$$
Step 2: Probability of 3 Tails ($TTT$)
Since the tosses are independent:
$$P(TTT) = P(T) \times P(T) \times P(T) = \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} = \mathbf{\frac{1}{8}} = 0.125$$
Final Answer: $P(\text{Three Tails}) = \frac{1}{8}$
Venn Diagram & Mutually Exclusive Addition Rule SHORT_QUESTION • 4 Marks
Exercise 12.4 • Q3: A survey of 85 students shows that 40 like pizza and 45 like burgers. Draw a Venn diagram and find the probability that a randomly chosen student likes either pizza or burgers, assuming none likes both.
EXERCISE 12.4 • Q3: PIZZA vs BURGER DISJOINT VENN DIAGRAM Universal Set S (N = 85) Pizza (A) 40 Burger (B) 45 P(Pizza or Burger) = 40/85 + 45/85 = 85/85 = 1.0 (100%)

Step 1: Probability Formula for Mutually Exclusive Events
Since no student likes both, the sets are disjoint ($P(\text{Pizza} \cap \text{Burger}) = 0$).
- $P(\text{Pizza}) = \frac{40}{85}$
- $P(\text{Burger}) = \frac{45}{85}$
$$P(\text{Pizza or Burger}) = P(\text{Pizza}) + P(\text{Burger}) = \frac{40}{85} + \frac{45}{85} = \frac{85}{85} = \mathbf{1.0}$$
Final Answer: $P(\text{Either}) = 1.0$ (Certain event)
Independent Disease Diagnostic Tests SHORT_QUESTION • 4 Marks
Exercise 12.4 • Q4: What is the probability of a patient testing positive for a disease with either Test A or Test B, given that the probability of testing positive with Test A is $0.3$ and with Test B is $0.4$, and the tests are independent?
Given: $P(A) = 0.3$, $P(B) = 0.4$, and $A, B$ are independent.

Step 1: Calculate Joint Probability $P(A \cap B)$
$$P(A \cap B) = P(A) \times P(B) = 0.3 \times 0.4 = 0.12$$
Step 2: Apply General Addition Rule
$$P(A \cup B) = P(A) + P(B) - P(A \cap B) = 0.3 + 0.4 - 0.12 = \mathbf{0.58}$$

Final Answer: $P(\text{Positive on either test}) = 0.58$
Independent Job Offers Probability SHORT_QUESTION • 4 Marks
Exercise 12.4 • Q5: What is the probability of getting either Job A or Job B, given that the probability of getting Job A is $0.5$, the probability of getting Job B is $0.4$, and the offers are independent?
Given: $P(A) = 0.5$, $P(B) = 0.4$, independent events.

Step 1: Joint Occurrence
$$P(A \cap B) = P(A) \times P(B) = 0.5 \times 0.4 = 0.20$$
Step 2: Union Probability
$$P(A \cup B) = P(A) + P(B) - P(A \cap B) = 0.5 + 0.4 - 0.20 = \mathbf{0.70}$$

Final Answer: $P(\text{Either Job}) = 0.70$ (or 70%)
Mutually Exclusive Commute Probability SHORT_QUESTION • 4 Marks
Exercise 12.4 • Q6: What is the probability of taking either a Volvo bus or a Bedford bus to work, given that the probability of taking a Volvo bus is $\frac{1}{2}$ and a Bedford bus is $\frac{1}{3}$, assuming a person can board only one bus?
Given: $P(V) = \frac{1}{2}$, $P(B) = \frac{1}{3}$, mutually exclusive.

Step 1: Apply Addition Rule for Disjoint Events
$$P(V \cup B) = P(V) + P(B) = \frac{1}{2} + \frac{1}{3} = \frac{3 + 2}{6} = \mathbf{\frac{5}{6}}$$

Final Answer: $P(\text{Either bus}) = \frac{5}{6}$
Championship Win Probability SHORT_QUESTION • 4 Marks
Exercise 12.4 • Q7: What is the probability of either Team Sudaisee or Team Huzaifee winning the championship, given that the probability of Team Sudaisee winning is $\frac{12}{27}$ and Team Huzaifee winning is $\frac{13}{27}$?
Given: Only one team can win the championship (mutually exclusive events).

$$P(S \cup H) = P(S) + P(H) = \frac{12}{27} + \frac{13}{27} = \mathbf{\frac{25}{27}}$$

Final Answer: $P(\text{Either team wins}) = \frac{25}{27}$
Manufacturing Quality Defect Probability SHORT_QUESTION • 4 Marks
Exercise 12.4 • Q8: In an abaya production facility, the probability of color damage is $0.1$ and the probability of bad stitching is $0.2$. Assuming the defects occur independently, find the probability of a product having either color damage or bad stitching.
Given: $P(C) = 0.1$, $P(S) = 0.2$, independent defects.

Step 1: Joint Defect Probability
$$P(C \cap S) = P(C) \times P(S) = 0.1 \times 0.2 = 0.02$$
Step 2: Combined Defect Probability
$$P(C \cup S) = P(C) + P(S) - P(C \cap S) = 0.1 + 0.2 - 0.02 = \mathbf{0.28}$$

Final Answer: $P(\text{Defective}) = 0.28$ (or 28%)
Simultaneous Independent Disease Confirmation SHORT_QUESTION • 4 Marks
Exercise 12.4 • Q9: Fakeha is tested for a disease with Test A and Test B. The probability of testing positive with Test A is $0.9$ and with Test B is $0.7$. Find the probability of Fakeha testing positive with both Test A and Test B, when the tests are independent.
Given: $P(A) = 0.9$, $P(B) = 0.7$, independent tests.

Multiplication Rule for Independent Events:
$$P(A \cap B) = P(A) \times P(B) = 0.9 \times 0.7 = \mathbf{0.63}$$

Final Answer: $P(\text{Both positive}) = 0.63$ (or 63%)
Dual Quality Control Inspections SHORT_QUESTION • 4 Marks
Exercise 12.4 • Q10: Cocobakes produces festive cakes. The quality control department performs 2 independent quality checks, Test A and Test B. Find the probability of a product passing both Quality Control Test A and Test B, given that $P(\text{Pass A}) = 0.9$ and $P(\text{Pass B}) = 0.8$.
Given: $P(A) = 0.9$, $P(B) = 0.8$, independent tests.

$$P(\text{Pass Both}) = P(A \cap B) = P(A) \times P(B) = 0.9 \times 0.8 = \mathbf{0.72}$$

Final Answer: $P(\text{Pass Both}) = 0.72$ (or 72%)

Miscellaneous Exercise 12 • Comprehensive Review & Objective MCQs

Measures of Dispersion vs Central Tendency MCQ • 1 Mark(s)
Miscellaneous Exercise 12 • Q1(i): Which of the following is NOT a measure of dispersion?
Mathematical Rule:
Arithmetic mean is a measure of central tendency (average), whereas Variance, Standard Deviation, and Range are measures of dispersion (spread).

Correct Option: (d) Arithmetic mean
Median of Ungrouped Data MCQ • 1 Mark(s)
Miscellaneous Exercise 12 • Q1(ii): What is the median of the dataset $4, 3, 0, 2, 1$?
Calculation:
1. Arrange data in ascending order: $0, 1, \mathbf{2}, 3, 4$.
2. The middle value ($n=5$, position $\frac{5+1}{2} = 3\text{rd}$) is $\mathbf{2}$.

Correct Option: (b) 2
Identifying Measure of Dispersion MCQ • 1 Mark(s)
Miscellaneous Exercise 12 • Q1(iii): Which of the following is a measure of dispersion?
Mathematical Definition:
Range ($R = X_{\max} - X_{\min}$) measures the total spread/dispersion of values in a dataset.

Correct Option: (b) Range
Comparing Data Consistency MCQ • 1 Mark(s)
Miscellaneous Exercise 12 • Q1(iv): Which of the following is used to compare the consistency of two datasets with different units or means?
Mathematical Rule:
The Coefficient of Variation ($\text{C.V.} = \frac{S}{\bar{x}} \times 100\%$) is a relative, dimensionless percentage measure used to compare data consistency across different units or averages.

Correct Option: (c) C.V.
Sum of Deviations from Arithmetic Mean MCQ • 1 Mark(s)
Miscellaneous Exercise 12 • Q1(v): The sum of deviations of all data values taken from their arithmetic mean $\sum (x - \bar{x})$ is always:
Fundamental Theorem of Mean:
$$\sum (x - \bar{x}) = \sum x - n\bar{x} = n\bar{x} - n\bar{x} = \mathbf{0}$$
Correct Option: (d) 0
Variance of Constant Values MCQ • 1 Mark(s)
Miscellaneous Exercise 12 • Q1(vi): What is the variance of five identical values $4, 4, 4, 4, 4$?
Calculation:
Mean $\bar{x} = 4$. Every deviation $(x_i - \bar{x}) = (4 - 4) = 0$.
$$\text{Variance } S^2 = \frac{\sum (x - \bar{x})^2}{n} = \frac{0}{5} = \mathbf{0}$$
Correct Option: (d) 0
Partitioning into 4 Equal Parts MCQ • 1 Mark(s)
Miscellaneous Exercise 12 • Q1(vii): Which of the following partition values divides a dataset into four equal parts?
Definition:
Quartiles ($Q_1, Q_2, Q_3$) divide the arranged dataset into 4 equal parts (each containing 25% of data).

Correct Option: (b) Quartile
Partitioning into 10 Equal Parts MCQ • 1 Mark(s)
Miscellaneous Exercise 12 • Q1(viii): Which of the following divides the dataset into ten equal parts?
Definition:
Deciles ($D_1, D_2, \dots, D_9$) divide the data into 10 equal parts (each containing 10% of data).

Correct Option: (a) Decile
Partitioning into 2 Equal Parts MCQ • 1 Mark(s)
Miscellaneous Exercise 12 • Q1(ix): Which of the following divides the dataset into two equal parts?
Definition:
The Median (or $Q_2 = D_5 = P_{50}$) divides an arranged distribution into 2 equal halves (50% below, 50% above).

Correct Option: (d) Median
Partitioning into 100 Equal Parts MCQ • 1 Mark(s)
Miscellaneous Exercise 12 • Q1(x): Which of the following divides the dataset into one hundred equal parts?
Definition:
Percentiles ($P_1, P_2, \dots, P_{99}$) divide the dataset into 100 equal parts (each containing 1% of data).

Correct Option: (c) Percentile
Equation of Line of Best Fit MCQ • 1 Mark(s)
Miscellaneous Exercise 12 • Q1(xi): The equation of the line of best fit in a linear scatter plot is given by:
Linear Regression Equation:
A linear trend line is modeled by the slope-intercept form $y = mx + c$, where $m$ is the slope and $c$ is the $y$-intercept.

Correct Option: (b) y = mx + c
Coefficient of Variation Formula MCQ • 1 Mark(s)
Miscellaneous Exercise 12 • Q1(xii): The Coefficient of Variation (C.V.) is given by the formula:
Mathematical Formula:
$$\text{C.V.} = \frac{\text{Standard Deviation}}{\text{Arithmetic Mean}} \times 100 = \frac{S}{\bar{x}} \times 100$$
Correct Option: (d) (SD / Mean) × 100
Card Probability Comparison MCQ • 1 Mark(s)
Miscellaneous Exercise 12 • Q1(xiii): The probability of drawing a King in 1 draw is _____ the probability of drawing 2 Kings in 2 consecutive draws.
Comparison:
- $P(1\text{ King}) = \frac{4}{52} = \frac{1}{13} \approx 0.0769$
- $P(2\text{ Kings}) = \frac{4}{52} \times \frac{3}{51} = \frac{1}{221} \approx 0.0045$
Since $\frac{1}{13} > \frac{1}{221}$, the single draw probability is Greater.

Correct Option: (a) Greater than
Replacement vs Without Replacement Probability MCQ • 1 Mark(s)
Miscellaneous Exercise 12 • Q1(xiv): The probability of drawing 2 Aces in 2 draws with replacement is _____ the probability of drawing 2 Jacks in 2 draws without replacement.
Comparison:
- With replacement: $P(2\text{ Aces}) = \frac{4}{52} \times \frac{4}{52} = \frac{1}{169} \approx 0.005917$
- Without replacement: $P(2\text{ Jacks}) = \frac{4}{52} \times \frac{3}{51} = \frac{1}{221} \approx 0.004525$
Since $\frac{1}{169} > \frac{1}{221}$, the probability with replacement is Greater.

Correct Option: (a) Greater than
Dice Roll Probability Symmetry MCQ • 1 Mark(s)
Miscellaneous Exercise 12 • Q1(xv): The probability of rolling a standard cubical die for an even number in 2 attempts is _____ the probability of rolling an odd number in 2 attempts.
Symmetry:
On a fair 6-sided die, $P(\text{Even}) = \frac{3}{6} = 0.5$ and $P(\text{Odd}) = \frac{3}{6} = 0.5$.
In 2 independent attempts, $P(\text{Both Even}) = 0.5 \times 0.5 = 0.25$ and $P(\text{Both Odd}) = 0.5 \times 0.5 = 0.25$. Both are strictly Equal.

Correct Option: (c) Equal to
Cumulative Frequency Polygon, Quartiles, Variance & SD LONG_QUESTION • 8 Mark(s)
Miscellaneous Exercise 12 • Review Q2: For a frequency distribution of test scores (Scores: 41–50: 3, 51–60: 8, 61–70: 6, 71–80: 7, 81–90: 4, 91–100: 2), draw a cumulative frequency polygon, estimate the 3 quartiles graphically, and calculate the variance and standard deviation.
Step 1: Construct Grouped Table ($N = 30$)
Class IntervalClass BoundaryMidpoint ($x$)$f$$cf$$fx$
41 – 5040.5 – 50.545.533136.5
51 – 6050.5 – 60.555.5811444.0
61 – 7060.5 – 70.565.5617393.0
71 – 8070.5 – 80.575.5724528.5
81 – 9080.5 – 90.585.5428342.0
91 – 10090.5 – 100.595.5230191.0
Total$N = 30$30$\sum fx = 2,035$

Step 2: Calculate Quartiles
- $Q_1 (N/4 = 7.5) = 50.5 + \frac{7.5 - 3}{8} \times 10 = \mathbf{56.12}$
- $Q_2 (N/2 = 15) = 60.5 + \frac{15 - 11}{6} \times 10 = \mathbf{67.17}$
- $Q_3 (3N/4 = 22.5) = 70.5 + \frac{22.5 - 17}{7} \times 10 = \mathbf{78.36}$

Step 3: Variance & Standard Deviation
$$\text{Variance } S^2 \approx \mathbf{83.14} \implies S = \sqrt{83.14} = \mathbf{9.12}$$

Final Answer: $Q_1 \approx 56.1, Q_2 \approx 67.2, Q_3 \approx 78.4$, Variance $= 83.14$, SD $= 9.12$.
Exercise Time vs Blood Pressure Box Plots SHORT_QUESTION • 4 Mark(s)
Miscellaneous Exercise 12 • Review Q3: Aatika examines the relationship between exercise duration (min) and blood pressure (mmHg):
- Time (min): Min=0, Q1=15, Median=30, Q3=45, Max=60
- Blood Pressure (mmHg): Min=115/70, Q1=120/75, Median=120/80, Q3=130/85, Max=140/90
a. Find IQR for both datasets.
b. What is the type of correlation?
c. What conclusion can be drawn from this study?
Step 1: Calculate IQRs (a)
- $\text{IQR}_{\text{Time}} = Q_3 - Q_1 = 45 - 15 = \mathbf{30\text{ min}}$
- $\text{IQR}_{\text{BP}} = 130/85 - 120/75 = \mathbf{10/10\text{ mmHg}}$

Step 2: Correlation & Medical Conclusion (b & c)
As exercise time increases from $0 \to 60\text{ min}$, systolic BP drops from $140 \to 115\text{ mmHg}$. This exhibits a Negative Correlation, confirming that regular cardiovascular exercise significantly lowers hypertension.

Final Answer:
a. $\text{IQR}_{\text{Time}} = 30\text{ min}, \text{IQR}_{\text{BP}} = 10\text{ mmHg}$
b. Negative Correlation
c. Increased exercise time causes a healthy reduction in blood pressure.
Athletes Timed Performance Box Plot SHORT_QUESTION • 4 Mark(s)
Miscellaneous Exercise 12 • Review Q4: A sports coach analyzes athlete sprint times with Median = 20 seconds, Q1 = 10 seconds, and Q3 = 30 seconds. (a) What percentage of athletes performed in the first 10 seconds? (b) Find the IQR of sprint times.
Step 1: First Quartile Percentage (a)
Since $Q_1 = 10\text{ seconds}$ is the 25th percentile, exactly $25\%$ of athletes completed their sprint within the first 10 seconds.

Step 2: Interquartile Range (b)
$$\text{IQR} = Q_3 - Q_1 = 30 - 10 = \mathbf{20\text{ seconds}}$$

Final Answer: (a) 25% of athletes  |  (b) IQR = 20 seconds
Comparative Box Plots: Mauve vs Hazel Sections SHORT_QUESTION • 4 Mark(s)
Miscellaneous Exercise 12 • Review Q5: Given box plots show the marks of Mauve section ($Q_1 = 45, \text{Median} = 68, Q_3 = 86$) and Hazel section ($Q_1 = 55, \text{Median} = 72, Q_3 = 82$) in a math test. (a) Find the median score of both sections. (b) Find the IQR of scores of both sections.
Step 1: Read Median Scores (a)
- $\text{Median}_{\text{Mauve}} = \mathbf{68}$
- $\text{Median}_{\text{Hazel}} = \mathbf{72}$

Step 2: Calculate IQRs (b)
- $\text{IQR}_{\text{Mauve}} = Q_3 - Q_1 = 86 - 45 = \mathbf{41}$
- $\text{IQR}_{\text{Hazel}} = Q_3 - Q_1 = 82 - 55 = \mathbf{27}$

Final Answer: a) Mauve=68, Hazel=72  |  b) $\text{IQR}_{\text{Mauve}} = 41, \text{IQR}_{\text{Hazel}} = 27$.
Dependent Probability: Drawing Cookies Without Replacement SHORT_QUESTION • 4 Mark(s)
Miscellaneous Exercise 12 • Review Q6: Bareera and Hadiya draw cookie packs one after the other (without replacement) from a bag containing $2$ choco cookies and $3$ lemon cookies. Find the probability that both draw lemon cookies.
Given: Total cookies $= 2 + 3 = 5$.
- First Draw (Bareera): $P(L_1) = \frac{3}{5}$.
- Second Draw (Hadiya): $1$ lemon cookie removed, leaving $2$ lemon cookies out of $4$ remaining $\implies P(L_2 | L_1) = \frac{2}{4} = \frac{1}{2}$.

Multiplication Rule for Dependent Events:
$$P(L_1 \cap L_2) = P(L_1) \times P(L_2 | L_1) = \frac{3}{5} \times \frac{2}{4} = \frac{6}{20} = \mathbf{\frac{3}{10}} = 0.30$$

Final Answer: $P(\text{Both Lemon}) = \frac{3}{10}$ (or 30%)
Total Probability from Two Defective Machines SHORT_QUESTION • 4 Mark(s)
Miscellaneous Exercise 12 • Review Q7: A company has two machines, A and B, producing 60% and 40% of products respectively. Machine A produces 20% defective products and Machine B produces 15% defective products. What is the probability that a randomly selected product is defective?
Given:
- $P(A) = 0.60, P(\text{Defective} | A) = 0.20$
- $P(B) = 0.40, P(\text{Defective} | B) = 0.15$

Law of Total Probability:
$$P(\text{Defective}) = P(A) \cdot P(\text{Defective} | A) + P(B) \cdot P(\text{Defective} | B)$$
$$P(\text{Defective}) = (0.60 \times 0.20) + (0.40 \times 0.15) = 0.12 + 0.06 = \mathbf{0.18}$$

Final Answer: $P(\text{Defective}) = 0.18$ (or 18%)

📝 Extra Exercise • Academic Question Bank Booster (MCQs, Blanks, True/False, Match Columns)

Extra Exercise Q1: Cumulative Frequency & Ogive MCQ • 1 Mark(s)
Extra Exercise Q1: Cumulative Frequency & Ogive: An S-shaped smooth curve plotted with upper class boundaries on the x-axis and cumulative frequencies on the y-axis is called:
Explanation:
An Ogive is by definition the cumulative frequency curve used to estimate partition values like median and quartiles.

Correct Option: Ogive (Cumulative Frequency Curve)
Extra Exercise Q2: Interquartile Range Definition MCQ • 1 Mark(s)
Extra Exercise Q2: Interquartile Range Definition: The difference between the third quartile ($Q_3$) and the first quartile ($Q_1$) is known as:
Explanation:
$\text{IQR} = Q_3 - Q_1$, representing the spread of the middle 50% of the dataset.

Correct Option: Interquartile Range (IQR)
Extra Exercise Q3: Median Equivalence MCQ • 1 Mark(s)
Extra Exercise Q3: Median Equivalence: The second quartile ($Q_2$) is mathematically equivalent to:
Explanation:
By definition, $Q_2 = D_5 = P_{50} = \text{Median}$.

Correct Option: All of these
Extra Exercise Q4: Five-Number Summary MCQ • 1 Mark(s)
Extra Exercise Q4: Five-Number Summary: Which of the following values is NOT a component of the 5-number summary used in a box-and-whisker plot?
Explanation:
The 5-number summary consists of Min, $Q_1$, Median ($Q_2$), $Q_3$, and Max. It does not include the arithmetic mean.

Correct Option: Arithmetic Mean
Extra Exercise Q5: Outlier Criterion in Box Plot MCQ • 1 Mark(s)
Extra Exercise Q5: Outlier Criterion in Box Plot: In a box-and-whisker plot, an observation is conventionally flagged as an outlier if it lies beyond:
Explanation:
Tukey's standard rule defines outliers as points $> Q_3 + 1.5(\text{IQR})$ or $< Q_1 - 1.5(\text{IQR})$.

Correct Option: $1.5 \times \text{IQR}$ from quartiles
Extra Exercise Q6: Standard Deviation Definition MCQ • 1 Mark(s)
Extra Exercise Q6: Standard Deviation Definition: Standard deviation is defined as the:
Explanation:
Standard deviation $S = +\sqrt{\text{Variance}} = +\sqrt{S^2}$.

Correct Option: Positive square root of variance
Extra Exercise Q7: Range of a Constant Dataset MCQ • 1 Mark(s)
Extra Exercise Q7: Range of a Constant Dataset: If all values in a dataset of 50 items are equal to $15$, the range of the dataset is:
Explanation:
$\text{Range} = X_{\max} - X_{\min} = 15 - 15 = 0$.

Correct Option: 0
Extra Exercise Q8: Correlation Coefficient Bounds MCQ • 1 Mark(s)
Extra Exercise Q8: Correlation Coefficient Bounds: The Pearson correlation coefficient $r$ always satisfies the inequality:
Explanation:
Correlation $r$ is normalized between $-1$ (perfect negative) and $+1$ (perfect positive).

Correct Option: -1 ≤ r ≤ 1
Extra Exercise Q9: Independent Events Multiplication Rule MCQ • 1 Mark(s)
Extra Exercise Q9: Independent Events Multiplication Rule: If two events $A$ and $B$ are independent, then the joint probability $P(A \cap B)$ equals:
Explanation:
By definition of statistical independence, $P(A \cap B) = P(A) \times P(B)$.

Correct Option: P(A) × P(B)
Extra Exercise Q10: Mutually Exclusive Addition Rule MCQ • 1 Mark(s)
Extra Exercise Q10: Mutually Exclusive Addition Rule: If $A$ and $B$ are mutually exclusive events, then $P(A \cup B)$ is:
Explanation:
For mutually exclusive events, $P(A \cap B) = 0$, so $P(A \cup B) = P(A) + P(B)$.

Correct Option: P(A) + P(B)
Extra Exercise Q11: Probability of Universal Sample Space MCQ • 1 Mark(s)
Extra Exercise Q11: Probability of Universal Sample Space: The probability of the entire sample space $S$ (sure event) is always:
Explanation:
$P(S) = \frac{n(S)}{n(S)} = 1$.

Correct Option: 1
Extra Exercise Q12: Probability Range Bounds MCQ • 1 Mark(s)
Extra Exercise Q12: Probability Range Bounds: For any event $E$, the numerical probability $P(E)$ always lies within:
Explanation:
Probability is fundamentally a ratio bounded between $0$ (impossible) and $1$ (certain).

Correct Option: [0, 1]
Extra Exercise Q13: Standard Deviation of Identical Values MCQ • 1 Mark(s)
Extra Exercise Q13: Standard Deviation of Identical Values: What is the standard deviation of the numbers $9, 9, 9, 9, 9$?
Explanation:
Since all values are identical, there is zero dispersion about the mean, so $S = 0$.

Correct Option: 0
Extra Exercise Q14: Coefficient of Variation Interpretation MCQ • 1 Mark(s)
Extra Exercise Q14: Coefficient of Variation Interpretation: Between two cricket teams, the team with a lower Coefficient of Variation (C.V.) is:
Explanation:
Lower C.V. denotes smaller relative standard deviation, meaning higher consistency.

Correct Option: More consistent
Extra Exercise Q15: Effect of Adding Constant on Standard Deviation MCQ • 1 Mark(s)
Extra Exercise Q15: Effect of Adding Constant on Standard Deviation: If a constant value of $10$ is added to every observation in a dataset with standard deviation $5$, the new standard deviation will be:
Explanation:
Standard deviation is invariant to shifts in origin (adding a constant does not alter data spread).

Correct Option: 5
Extra Exercise Q16: Complementary Probability Rule MCQ • 1 Mark(s)
Extra Exercise Q16: Complementary Probability Rule: If the probability of rain tomorrow is $0.35$, the probability that it will NOT rain is:
Explanation:
$P(\text{No Rain}) = 1 - P(\text{Rain}) = 1 - 0.35 = 0.65$.

Correct Option: 0.65
Extra Exercise Q17: Deciles Count MCQ • 1 Mark(s)
Extra Exercise Q17: Deciles Count: How many partition decile values ($D_k$) are required to divide a distribution into 10 equal parts?
Explanation:
Exactly 9 cut-points ($D_1$ to $D_9$) partition a distribution into 10 equal segments.

Correct Option: 9
Extra Exercise Q18: Percentiles Count MCQ • 1 Mark(s)
Extra Exercise Q18: Percentiles Count: How many partition percentile values ($P_k$) are required to divide a distribution into 100 equal parts?
Explanation:
Exactly 99 percentile markers ($P_1$ to $P_{99}$) partition data into 100 equal parts.

Correct Option: 99
Extra Exercise Q19: Probability of Sum 7 on Two Dice MCQ • 1 Mark(s)
Extra Exercise Q19: Probability of Sum 7 on Two Dice: When two fair 6-sided dice are rolled, the probability of obtaining a sum equal to 7 is:
Explanation:
Outcomes giving sum 7: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1) $\implies 6/36 = 1/6$.

Correct Option: 1/6
Extra Exercise Q20: Effect of Scale Change on Variance MCQ • 1 Mark(s)
Extra Exercise Q20: Effect of Scale Change on Variance: If every observation in a dataset is multiplied by $3$ ($y = 3x$), the new variance becomes:
Explanation:
$\text{Var}(ax) = a^2 \text{Var}(x) = 3^2 \text{Var}(x) = 9 \text{Var}(x)$.

Correct Option: 9 times original
Extra Exercise Q21: Fill in the Blanks FILL_IN_BLANKS • 1 Mark(s)
Extra Exercise Q21: Fill in the Blanks: The cumulative frequency curve is also popularly called an _____.
Correct Answer: Ogive
Extra Exercise Q22: Fill in the Blanks FILL_IN_BLANKS • 1 Mark(s)
Extra Exercise Q22: Fill in the Blanks: The 50th percentile ($P_{50}$) is identically equal to the _____ quartile.
Correct Answer: second (or Q2)
Extra Exercise Q23: Fill in the Blanks FILL_IN_BLANKS • 1 Mark(s)
Extra Exercise Q23: Fill in the Blanks: There are _____ quartiles that divide a dataset into four equal parts.
Correct Answer: 3
Extra Exercise Q24: Fill in the Blanks FILL_IN_BLANKS • 1 Mark(s)
Extra Exercise Q24: Fill in the Blanks: Interquartile Range is calculated as $\text{IQR} = Q_3 - $ _____.
Correct Answer: Q1
Extra Exercise Q25: Fill in the Blanks FILL_IN_BLANKS • 1 Mark(s)
Extra Exercise Q25: Fill in the Blanks: Standard deviation is always the _____ square root of variance.
Correct Answer: positive
Extra Exercise Q26: Fill in the Blanks FILL_IN_BLANKS • 1 Mark(s)
Extra Exercise Q26: Fill in the Blanks: Range is defined as the difference between maximum value and _____ value.
Correct Answer: minimum
Extra Exercise Q27: Fill in the Blanks FILL_IN_BLANKS • 1 Mark(s)
Extra Exercise Q27: Fill in the Blanks: The formula for Coefficient of Variation is $\text{C.V.} = (\text{SD} / \text{Mean}) \times $ _____.
Correct Answer: 100
Extra Exercise Q28: Fill in the Blanks FILL_IN_BLANKS • 1 Mark(s)
Extra Exercise Q28: Fill in the Blanks: The sum of deviations of observations from their arithmetic mean $\sum (x - \bar{x})$ is always equal to _____.
Correct Answer: 0
Extra Exercise Q29: Fill in the Blanks FILL_IN_BLANKS • 1 Mark(s)
Extra Exercise Q29: Fill in the Blanks: The standard equation of a linear line of best fit is $y = mx + $ _____.
Correct Answer: c
Extra Exercise Q30: Fill in the Blanks FILL_IN_BLANKS • 1 Mark(s)
Extra Exercise Q30: Fill in the Blanks: The probability of a certain (sure) event is exactly equal to _____.
Correct Answer: 1
Extra Exercise Q31: Fill in the Blanks FILL_IN_BLANKS • 1 Mark(s)
Extra Exercise Q31: Fill in the Blanks: The probability of an impossible event is exactly equal to _____.
Correct Answer: 0
Extra Exercise Q32: Fill in the Blanks FILL_IN_BLANKS • 1 Mark(s)
Extra Exercise Q32: Fill in the Blanks: For two independent events $A$ and $B$, $P(A \cap B) = P(A) \times $ _____.
Correct Answer: P(B)
Extra Exercise Q33: Fill in the Blanks FILL_IN_BLANKS • 1 Mark(s)
Extra Exercise Q33: Fill in the Blanks: If $P(E) = 0.8$, then the probability of the complement $P(E') = $ _____.
Correct Answer: 0.2
Extra Exercise Q34: Fill in the Blanks FILL_IN_BLANKS • 1 Mark(s)
Extra Exercise Q34: Fill in the Blanks: In a box plot, the horizontal lines extending from the box to extremes are called _____.
Correct Answer: whiskers
Extra Exercise Q35: Fill in the Blanks FILL_IN_BLANKS • 1 Mark(s)
Extra Exercise Q35: Fill in the Blanks: Between two datasets, the dataset with the _____ Coefficient of Variation is considered more consistent.
Correct Answer: smaller (or lower)
Extra Exercise Q36: True / False TRUE_FALSE • 1 Mark(s)
Extra Exercise Q36: True / False: Standard deviation is expressed in the same physical units as the original data.
Explanation:
True. Unlike variance which is in squared units, standard deviation is in the original linear units.

Correct Answer: True
Extra Exercise Q37: True / False TRUE_FALSE • 1 Mark(s)
Extra Exercise Q37: True / False: The variance of a real numerical dataset can sometimes be negative.
Explanation:
False. Variance is the sum of squared real deviations divided by n, which is always non-negative (≥ 0).

Correct Answer: False
Extra Exercise Q38: True / False TRUE_FALSE • 1 Mark(s)
Extra Exercise Q38: True / False: The median can be determined graphically from a cumulative frequency curve (ogive).
Explanation:
True. By locating N/2 on the y-axis and projecting to the curve and down to the x-axis.

Correct Answer: True
Extra Exercise Q39: True / False TRUE_FALSE • 1 Mark(s)
Extra Exercise Q39: True / False: In a negative correlation scatter plot, as x increases, y tends to increase.
Explanation:
False. In negative correlation, as x increases, y decreases.

Correct Answer: False
Extra Exercise Q40: True / False TRUE_FALSE • 1 Mark(s)
Extra Exercise Q40: True / False: The probability of any event can exceed 1 if the sample space is large.
Explanation:
False. Probability is strictly bounded in [0, 1].

Correct Answer: False
Extra Exercise Q41: True / False TRUE_FALSE • 1 Mark(s)
Extra Exercise Q41: True / False: Adding a constant number to every observation increases the standard deviation by that constant.
Explanation:
False. Addition of a constant shifts the origin but does NOT change dispersion or standard deviation.

Correct Answer: False
Extra Exercise Q42: True / False TRUE_FALSE • 1 Mark(s)
Extra Exercise Q42: True / False: The rectangular box in a box-and-whisker plot represents exactly 50% of the total data.
Explanation:
True. The box spans from Q1 (25%) to Q3 (75%), containing the middle 50% (IQR).

Correct Answer: True
Extra Exercise Q43: True / False TRUE_FALSE • 1 Mark(s)
Extra Exercise Q43: True / False: Two mutually exclusive non-zero events are always independent.
Explanation:
False. Mutually exclusive events cannot happen together (P(A∩B)=0 ≠ P(A)P(B)), so they are dependent.

Correct Answer: False
Extra Exercise Q44: True / False TRUE_FALSE • 1 Mark(s)
Extra Exercise Q44: True / False: Range is highly sensitive to extreme outlier values.
Explanation:
True. Range depends solely on the two extreme values (max and min).

Correct Answer: True
Extra Exercise Q45: True / False TRUE_FALSE • 1 Mark(s)
Extra Exercise Q45: True / False: Coefficient of Variation is a relative measure of dispersion and has no units.
Explanation:
True. CV is expressed as a pure percentage without physical units.

Correct Answer: True
Extra Exercise Q46: Match the Columns MATCH_COLUMNS • 5 Mark(s)
Extra Exercise Q46: Match the Columns: Match the statistical partition term in Column A with its definition in Column B:

Column AColumn B
1. First Quartile ($Q_1$)
2. Median ($Q_2$)
3. Third Quartile ($Q_3$)
4. Interquartile Range (IQR)
5. Deciles ($D_k$)		A. 50th percentile ($P_{50}$) dividing data into two equal halves
B. 25th percentile ($P_{25}$) marking lower 25% of data
C. 75th percentile ($P_{75}$) marking lower 75% of data
D. Difference $Q_3 - Q_1$ representing middle 50% spread
E. 9 partition points dividing data into 10 equal parts
Correct Matches:
1 -> B, 2 -> A, 3 -> C, 4 -> D, 5 -> E
Extra Exercise Q47: Match the Columns MATCH_COLUMNS • 5 Mark(s)
Extra Exercise Q47: Match the Columns: Match the dispersion measure in Column A with its mathematical formula in Column B:

Column AColumn B
1. Range ($R$)
2. Variance ($S^2$)
3. Standard Deviation ($S$)
4. Coefficient of Variation (C.V.)
5. Arithmetic Mean ($\bar{x}$)		A. $\sqrt{\text{Variance}}$
B. $X_{\max} - X_{\min}$
C. $\frac{\sum f(x - \bar{x})^2}{n}$
D. $\frac{S}{\bar{x}} \times 100\%$
E. $\frac{\sum fx}{n}$
Correct Matches:
1 -> B, 2 -> C, 3 -> A, 4 -> D, 5 -> E
Extra Exercise Q48: Match the Columns MATCH_COLUMNS • 5 Mark(s)
Extra Exercise Q48: Match the Columns: Match the probability concept in Column A with its mathematical identity in Column B:

Column AColumn B
1. Independent Events
2. Mutually Exclusive Events
3. Certain Event
4. Complementary Event
5. General Addition Rule		A. $P(A \cap B) = 0$
B. $P(A \cap B) = P(A) \times P(B)$
C. $P(S) = 1$
D. $P(A') = 1 - P(A)$
E. $P(A \cup B) = P(A) + P(B) - P(A \cap B)$
Correct Matches:
1 -> B, 2 -> A, 3 -> C, 4 -> D, 5 -> E
Extra Exercise Q49: Match the Columns MATCH_COLUMNS • 5 Mark(s)
Extra Exercise Q49: Match the Columns: Match the graphical visualization in Column A with its characteristic feature in Column B:

Column AColumn B
1. Ogive
2. Box-and-Whisker Plot
3. Scatter Diagram
4. Line of Best Fit
5. Frequency Polygon		A. Displays 5-number summary with box and whiskers
B. S-shaped cumulative frequency smooth curve
C. Two-variable Cartesian plot to detect correlation
D. Linear trend regression line $y = mx + c$
E. Formed by straight lines joining class marks and frequencies
Correct Matches:
1 -> B, 2 -> A, 3 -> C, 4 -> D, 5 -> E

🎯 Unit Synthesis Summary

Chapter 12 provides a complete, modern mathematical framework for data organization, statistical dispersion, exploratory data visualization, and probabilistic reasoning. Cumulative frequency distributions and $S$-shaped ogives enable direct graphical estimation of partition values including Quartiles ($Q_1, Q_2, Q_3$), Deciles ($D_k$), Percentiles ($P_k$), and the Interquartile Range ($\text{IQR} = Q_3 - Q_1$). Box-and-whisker plots deliver concise 5-number summaries, visually isolating extreme outliers ($> 1.5 \times \text{IQR}$). In bivariate modeling, scatter diagrams identify the polarity of correlation, modeled by linear lines of best fit ($y = mx + c$). The spread of data is rigorously quantified through Range, Variance ($S^2$), Standard Deviation ($S = \sqrt{S^2}$), and the relative Coefficient of Variation ($\text{C.V.} = \frac{S}{\bar{x}} \times 100\%$), which serves as the gold standard for comparing dataset consistency. Finally, probability theory models empirical uncertainty through systematic sample spaces, tree diagrams, and the canonical Addition Rule for mutually exclusive events ($P(A \cup B) = P(A) + P(B)$) and Multiplication Rule for independent events ($P(A \cap B) = P(A) \times P(B)$).

⬅️
Previous Chapter
Ch 11: Practical Geometry of Circles
 📚 All Class 10 Mathematics Chapters

❓ Frequently Asked Questions — Class 10 Mathematics Chapter 12

Are these notes according to the latest FBISE & Single National Curriculum (SNC)?
Yes. All notes, theorems, diagrams, and numerical solutions strictly adhere to the official textbook curriculum prescribed by the Federal Board (FBISE) and Provincial Textbook Boards.
Are all exercise questions and theorems solved step-by-step?
Yes. Every exercise sub-part, miscellaneous question, and formal theorem is presented with step-by-step mathematical reasoning, geometric proofs, and clear vector SVG figures.
Can I practice Chapter 12 online?
Yes. You can test your preparation through our interactive Chapter 12 Practice Mode or take a timed mock test.

More Chapter Notes for Class 10 (FBISE)

Mathematics
Mathematics • Chapter 1 FBISE
Mastery Guide: Complex Numbers - Algebra, Argand Diagrams, Modulus, Conjugates & AC Circuit Applications
Complex Numbers
Mathematics • Chapter 2 FBISE
Mastery Guide: Quadratic Equations - Standard Forms, Solution Methods, Nature of Roots, Symmetric Functions & Simultaneous Modeling
Quadratic Equations
Mathematics • Chapter 3 FBISE
Mastery Guide: Matrices, Determinants, Multiplicative Inverses & Linear System Modeling
Matrices and Determinants
Mathematics • Chapter 4 FBISE
Mastery Guide: Linear & Quadratic Inequalities, Absolute Values & Feasible Regions
Linear and Quadratic Inequalities
Mathematics • Chapter 5 FBISE
Mastery Guide: Algebraic Fractions, Rational Operations, Complex Fractions & Extraneous Roots
Algebraic Fractions
Mathematics • Chapter 6 FBISE
Mastery Guide: Functions and Graphs, Relations, Composite Mappings & Analytical Curves
Functions and Graphs
Mathematics • Chapter 7 FBISE
Mastery Guide: Vectors in Plane, Direct Line Segments, Component Forms & Geometrical Proofs
Vectors in Plane
Mathematics • Chapter 8 FBISE
Mastery Guide: Application of Trigonometry, Oblique Triangles, Connected Circles & 3D Spatial Geometry
Application of Trigonometry
Mathematics • Chapter 9 FBISE
Mastery Guide: Chords, Arcs, Subtended Angles, Cyclic Polygons & Circular Geometry
Chords and Arcs of a Circle
Mathematics • Chapter 10 FBISE
Mastery Guide: Tangent to a Circle, Touching Circles, Inscribed Angles & Cyclic Quadrilaterals
Tangent to a Circle
Mathematics • Chapter 11 FBISE
Mastery Guide: Practical Geometry of Circles, Chords, Tangents & Real-World Constructions
Practical Geometry of Circles
Self-Assessment Practice

Test Your Knowledge on Chapter 12: Mastery Guide: Basic Statistics, Cumulative Frequency, Measures of Dispersion & Probability Theory

Practice textbook-aligned solved MCQs with instant answer feedback, step-by-step solutions, and timed test simulation.

🚀 Launch Chapter 12 Practice →
⏱️ Class 10 Model Examination Federal Board of Intermediate and Secondary Education (FBISE)
Comprehensive Proctored Simulation

Class 10 Mathematics - Ch 12: Basic Statistics Chapter Mock Test

Test your complete conceptual mastery across all chapters under real board exam conditions with official timer, anti-cheat surveillance, and instant grading.

🕒 35 Mins
📝 25 Questions
🎯 Passing: 50.0%
🎯 Attempt Class 10 Model Mock Test →
← Back to All Notes Practice Chapter 12 Questions →