Class 9 Mathematics Federal Board of Intermediate and Secondary Education (FBISE) (FBISE) 📚 Model Textbook of Mathematics Grade 9 (FBISE / NBF)

Mastery Guide: Basic Statistics - Frequency Distributions, Histograms, Central Tendencies & Probability

📖 Chapter 11: Basic Statistics 📅 Updated: Sep 19, 2026

FBISE Class 9 Mathematics • Chapter 11

Mastery Guide: Basic Statistics & Probability

Single National Curriculum (SNC) • Frequency Distributions, Histograms & Polygons, Measures of Central Tendency (Mean, Median, Mode, Weighted Mean), Dispersion & Probability Matrices

📖 1. Unit Overview & Target Learning Outcomes

Statistics is the scientific discipline of collecting, organizing, analyzing, interpreting, and presenting quantitative data. In this comprehensive final unit of Grade 9 Mathematics, students master the systematic transformation of raw empirical scores into grouped frequency tables, graphical visualization via Histograms (including unequal class intervals) and Frequency Polygons, mathematical derivation of the Three Measures of Central Tendency (Arithmetic Mean, Median, and Mode for both ungrouped and grouped distributions), Weighted Arithmetic Means, Dispersion fundamentals, and Classical/Empirical Probability.

🎯 Core Learning Outcomes & Competencies:

Data Organization: Construct grouped frequency distributions using Tally Marks, Class Limits, exact Class Boundaries, and Class Marks.
Graphical Data Science: Draw Histograms using Frequency Density ($\text{FD} = \frac{\text{Frequency}}{\text{Class Width}}$) for unequal intervals, and closed Frequency Polygons with zero-frequency bounding classes.
Arithmetic Mean: Compute mean via Direct Method ($\bar{x} = \frac{\sum fx}{\sum f}$), Short-Cut Method ($A + \frac{\sum fd}{\sum f}$), and Step-Deviation Coding Method ($A + (\frac{\sum fu}{\sum f})h$).
Positional & Modal Averages: Compute Grouped Median ($\tilde{x} = l + \frac{h}{f}(\frac{n}{2} - c)$) and Grouped Mode ($\hat{x} = l + \frac{f_m - f_1}{2f_m - f_1 - f_2} \times h$).
Empirical Relation & Weighted Mean: Apply Pearson's empirical formula ($\text{Mode} = 3\,\text{Median} - 2\,\text{Mean}$) and Weighted Mean ($\bar{x}_w = \frac{\sum wx}{\sum w}$).
Probability & Expected Values: Calculate Single Event Probabilities ($P(E) = \frac{n(E)}{n(S)}$), Complementary Rules ($P(E') = 1 - P(E)$), and Expected Frequency ($n \times p$).

💡 2. Kid-Friendly Tips for Success & Memory Hooks

🧠 The "3 M's" Quick Identification

• Mean: The fair share balance point (Add all, divide by count).
• Median: The middle man on the highway (Must sort in order first!).
• Mode: The most popular kid in class (Highest frequency).

⚠️ The Grouped Median Trap ("c")

In $\tilde{x} = l + \frac{h}{f}(\frac{n}{2} - c)$, students often pick $c$ from the median row! NEVER do this! Always pick $c$ from the PREVIOUS class cumulative frequency.

📊 Unequal Histograms • Frequency Density

When bar widths are unequal, you cannot use raw frequency on the y-axis! You must use Frequency Density ($\text{FD} = \frac{\text{Frequency}}{\text{Width}}$). That way, Bar Area = Frequency!

🎲 Probability Boundary Rule

A probability is ALWAYS a number between $0$ and $1$ inclusive: $0 \le P(E) \le 1$. If your calculation yields a negative value or $> 1$, stop immediately and verify your fraction!

🌍 3. Real-World Connections & Applications

🏏 Cricket & Sports Analytics (Batting Average)

A cricketer's batting average is the arithmetic mean of total runs divided by completed dismissals. Selection committees utilize weighted averages and standard deviation to analyze batsman reliability across differing pitches.

🧬 Medical Genetics & Clinical Trials

Gregor Mendel used empirical relative frequencies to discover genetic inheritance laws ($3:1$ ratio). Clinical trials calculate expected positive responses and efficacy rates using binomial probability distributions.

👟 Shoe Manufacturing & Fashion Retail (Mode)

Shoe manufacturers do not mass-produce the mean shoe size ($8.37$) because shoes must be whole numbers. Retail inventory is optimized using the Mode (size 9 or 8) to maximize stock turnover.

📈 National Census & Household Income (Median)

Economists evaluate national wealth using Median Household Income rather than the mean, as a small group of extreme billionaires distorts the mean, while the median reflects typical citizens accurately.

🔑 4. Study Cues & Essential Inquiries

Why do we convert Class Limits into Class Boundaries?
Class limits like $10-19$ and $20-29$ have a gap of $1$ unit. A measurement of $19.5$ cannot be categorized! Subtracting $0.5$ from the lower limit and adding $0.5$ to the upper limit creates seamless continuous boundaries ($9.5 - 19.5, 19.5 - 29.5$) where histogram bars touch without gaps.
When does Pearson's Empirical Formula ($\text{Mode} = 3\,\text{Median} - 2\,\text{Mean}$) hold true?
It holds for moderately skewed unimodal frequency distributions. In a perfectly symmetrical bell-shaped distribution, $\text{Mean} = \text{Median} = \text{Mode}$.
What is the difference between Experimental (Relative) and Theoretical Probability?
Theoretical probability is derived from mathematical symmetry assuming equally likely outcomes ($P(\text{Head}) = 0.5$). Relative frequency is computed from empirical trial counts ($\frac{\text{Heads observed}}{\text{Total tosses}}$). By the Law of Large Numbers, as trials increase, relative frequency converges to theoretical probability.

📊 5. Master Statistical Tables, Formulas & Visual Matrices

5.1 Frequency Distribution & Cumulative Frequency Table

Grouping 80 raw student scores into regular classes with class width $h = 10$:

📊 Table 1: Grouped Frequency & Cumulative Frequency Distribution (Tally Bar Method)
Class Interval	Tally Marks	Frequency ($f$)	Cumulative Frequency ($cf$)
$40 - 49$	卌 \|\|	7	7
$50 - 59$	卌卌 \|\|\|\|	14	21
$60 - 69$	卌卌卌卌 \|\|	22	43
$70 - 79$	卌卌卌 \|\|	17	60
$80 - 89$	卌卌 \|\|	12	72
$90 - 99$	卌 \|\|\|	8	80
Total	—	$\sum f = 80$	$N = 80$

5.2 Class Limits, Boundaries & Class Marks (Midpoints)

To make discrete class limits continuous for histograms and central tendency formulas, calculate adjustment $d/2 = 0.5$:

📐 Table 2: Construction of Exact Class Boundaries and Class Marks
Class Limits (Discrete)	Adjustment Factor ($\frac{d}{2}$)	Class Boundaries (Continuous)	Midpoint / Class Mark ($x$)	Width ($h$)
$10 - 19$	$\frac{20-19}{2} = 0.5$	$9.5 - 19.5$	$x_1 = \frac{10+19}{2} = 14.5$	10
$20 - 29$	$0.5$	$19.5 - 29.5$	$x_2 = \frac{20+29}{2} = 24.5$	10
$30 - 39$	$0.5$	$29.5 - 39.5$	$x_3 = \frac{30+39}{2} = 34.5$	10
$40 - 49$	$0.5$	$39.5 - 49.5$	$x_4 = \frac{40+49}{2} = 44.5$	10
$50 - 59$	$0.5$	$49.5 - 59.5$	$x_5 = \frac{50+59}{2} = 54.5$	10

5.3 Less-Than and More-Than Cumulative Frequency Table

Cumulative frequency tables determine percentiles, quartiles, and median thresholds:

📈 Table 3: Less-Than & More-Than Cumulative Frequency Table
Class Boundaries	Frequency ($f$)	Less-Than Cumulative Freq ($cf$)	More-Than Cumulative Freq	Percentage Percentile ($P_k$)
$0.5 - 10.5$	5	5	50	10.0%
$10.5 - 20.5$	12	17	45	34.0%
$20.5 - 30.5$	18	35	33	70.0%
$30.5 - 40.5$	10	45	15	90.0%
$40.5 - 50.5$	5	50	5	100.0%
Total	$\sum f = 50$	—	—	—

5.4 Histogram Data Table with Unequal Class Intervals

When class interval widths vary ($h = 5, 10, 15$), calculate Frequency Density so that Rectangle Area $= h \times \text{FD} = \text{Frequency}$:

📊 Table 4: Histogram Computation for Unequal Class Intervals (Frequency Density)
Class Interval	Class Boundaries	Frequency ($f$)	Class Width ($h$)	Frequency Density ($\text{FD} = \frac{f}{h}$)	Bar Area ($h \times \text{FD}$)
$0 - 5$	$0 - 5$	5	5	$\frac{5}{5} = 1.00$	$5 \times 1.0 = 5$
$5 - 15$	$5 - 15$	12	10	$\frac{12}{10} = 1.20$	$10 \times 1.2 = 12$
$15 - 30$	$15 - 30$	16	15	$\frac{16}{15} \approx 1.07$	$15 \times 1.07 = 16$
$30 - 40$	$30 - 40$	10	10	$\frac{10}{10} = 1.00$	$10 \times 1.0 = 10$
$40 - 45$	$40 - 45$	2	5	$\frac{2}{5} = 0.40$	$5 \times 0.4 = 2$
Total	—	$\sum f = 45$	—	—	Total Area $= 45$

5.5 Central Tendency Formulas & Selection Criteria

🎯 Table 5: Comprehensive Master Formula Matrix for Central Tendencies
Measure	Ungrouped Formula	Grouped Formula	Key Parameters	Best Used When
Arithmetic Mean ($\bar{x}$)	$\bar{x} = \frac{\sum x}{n}$	$\bar{x} = \frac{\sum fx}{\sum f}$	$x = \text{midpoint}$, $f = \text{frequency}$	Data is symmetric with no extreme outliers.
Short-Cut Mean (Deviation)	$\bar{x} = A + \frac{\sum d}{n}$	$\bar{x} = A + \frac{\sum fd}{\sum f}$	$A = \text{assumed mean}$, $d = x - A$	Large values need manual simplification.
Step-Deviation Mean (Coding)	$\bar{x} = A + \left(\frac{\sum u}{n}\right)h$	$\bar{x} = A + \left(\frac{\sum fu}{\sum f}\right)h$	$u = \frac{x - A}{h}$, $h = \text{class width}$	Equal interval grouped distributions.
Median ($\tilde{x}$)	Odd $n: \left(\frac{n+1}{2}\right)\text{th}$ Even $n: \frac{\text{mid}_1 + \text{mid}_2}{2}$	$\tilde{x} = l + \frac{h}{f}\left(\frac{n}{2} - c\right)$	$l = \text{lower boundary of median class}$ $c = cf \text{ of PREVIOUS class}$	Data has extreme skewed outliers (Income, Wealth).
Mode ($\hat{x}$)	Most frequent value	$\text{Mode} = l + \frac{f_m - f_1}{2f_m - f_1 - f_2} \times h$	$f_m = \text{modal freq}, f_1 = \text{preceding}, f_2 = \text{following}$	Categorical data & manufacturing (Shoes, Ready-wear).
Weighted Mean ($\bar{x}_w$)	$\bar{x}_w = \frac{\sum wx}{\sum w}$	$\bar{x}_w = \frac{\sum wfx}{\sum wf}$	$w = \text{assigned importance/credit hours}$	Exams, CGPA, Price Index calculations.

5.6 Grouped Arithmetic Mean: Direct, Short-Cut ($d$) & Coding ($u$) Methods

Complete calculation table verifying that all three arithmetic mean methods yield identical results:

🧮 Table 6: Worked Arithmetic Mean Calculation Table (Direct, Short-Cut & Coding Methods)
Class Interval	Midpoint ($x$)	Frequency ($f$)	Product ($fx$)	Deviation ($d = x - 35$)	$fd$	Coded ($u = \frac{x-35}{10}$)	$fu$
$10 - 19$	14.5	4	58.0	$-20$	$-80$	$-2$	$-8$
$20 - 29$	24.5	8	196.0	$-10$	$-80$	$-1$	$-8$
$30 - 39$	34.5	12	414.0	$0$	$0$	$0$	$0$
$40 - 49$	44.5	10	445.0	$+10$	$+100$	$+1$	$+10$
$50 - 59$	54.5	6	327.0	$+20$	$+120$	$+2$	$+12$
Totals	—	$\sum f = 40$	$\sum fx = 1440$	—	$\sum fd = 60$	—	$\sum fu = 6$

1. Direct Method: $\bar{x} = \frac{\sum fx}{\sum f} = \frac{1440}{40} = \mathbf{36.0}$
2. Short-Cut Method (Assumed Mean $A = 34.5$): $\bar{x} = A + \frac{\sum fd}{\sum f} = 34.5 + \frac{60}{40} = 34.5 + 1.5 = \mathbf{36.0}$
3. Step-Deviation Coding Method ($h = 10$): $\bar{x} = A + (\frac{\sum fu}{\sum f})h = 34.5 + (\frac{6}{40})10 = 34.5 + 1.5 = \mathbf{36.0}$

5.7 Grouped Median Calculation Work-Table

⚖️ Table 7: Grouped Median Calculation Work-Table with Median Class Identification
Class Limits	Class Boundaries	Frequency ($f$)	Cumulative Frequency ($cf$)	Median Class Status ($\frac{n}{2} = 25$)
$20 - 29$	$19.5 - 29.5$	5	5	Contains values $1 - 5$
$30 - 39$	$29.5 - 39.5$	12	17	Contains values $6 - 17$ ($c = 17$)
$40 - 49$	$39.5 - 49.5$	18	35	🎯 MEDIAN CLASS ($l=39.5, f=18$)
$50 - 59$	$49.5 - 59.5$	10	45	Contains values $36 - 45$
$60 - 69$	$59.5 - 69.5$	5	50	Contains values $46 - 50$
Total	—	$n = 50$	—	Median $\tilde{x} = 39.5 + \frac{10}{18}(25 - 17) = \mathbf{43.94}$

5.8 Grouped Mode Calculation Work-Table

🏆 Table 8: Grouped Mode Calculation Work-Table with Modal Frequency Mapping
Class Limits	Class Boundaries	Frequency ($f$)	Modal Parameter Role	Mode Step Breakdown
$60 - 64$	$59.5 - 64.5$	2	—	—
$65 - 69$	$64.5 - 69.5$	6	$f_1 = 6$	Preceding Modal Frequency
$70 - 74$	$69.5 - 74.5$	15	$f_m = 15$	🎯 MODAL CLASS ($l=69.5, h=5$)
$75 - 79$	$74.5 - 79.5$	9	$f_2 = 9$	Following Modal Frequency
$80 - 84$	$79.5 - 84.5$	4	—	—
Mode Formula		$\text{Mode} = 69.5 + \left(\frac{15 - 6}{2(15) - 6 - 9}\right) \times 5 = 69.5 + \left(\frac{9}{15}\right) \times 5 = 69.5 + 3.0 = \mathbf{72.5}$

5.9 Measures of Dispersion: Variance & Standard Deviation

Computation table for population/sample spread around the mean:

📏 Table 9: Variance ($\sigma^2$) and Standard Deviation ($\sigma$) Work-Table
Class Interval	Midpoint ($x$)	Frequency ($f$)	$fx$	$(x - \bar{x})$	$(x - \bar{x})^2$	$f(x - \bar{x})^2$	$x^2$	$fx^2$
$10 - 14$	12	2	24	$-10$	100	200	144	288
$15 - 19$	17	4	68	$-5$	25	100	289	1156
$20 - 24$	22	6	132	$0$	0	0	484	2904
$25 - 29$	27	5	135	$+5$	25	125	729	3645
$30 - 34$	32	3	96	$+10$	100	300	1024	3072
Totals	—	$\sum f = 20$	$\sum fx = 455$ ($\bar{x} = 22.75$)	—	—	$\sum f(x-\bar{x})^2 = 725$	—	$\sum fx^2 = 11065$

• Variance ($\sigma^2$): $\sigma^2 = \frac{\sum f(x - \bar{x})^2}{\sum f} = \frac{725}{20} = \mathbf{36.25}$
• Standard Deviation ($\sigma$): $\sigma = \sqrt{\text{Variance}} = \sqrt{36.25} \approx \mathbf{6.02}$

5.10 Probability Sample Spaces (2-Dice 36 Outcomes Matrix)

🎲 Table 10: Classical 2-Dice Sample Space (36 Equally-Likely Outcomes & Sum Probabilities)
Die 1 \ Die 2	Die 2 = 1	Die 2 = 2	Die 2 = 3	Die 2 = 4	Die 2 = 5	Die 2 = 6
Die 1 = 1	$(1,1) \to \text{Sum } 2$	$(1,2) \to \text{Sum } 3$	$(1,3) \to \text{Sum } 4$	$(1,4) \to \text{Sum } 5$	$(1,5) \to \text{Sum } 6$	$(1,6) \to \text{Sum } 7$
Die 1 = 2	$(2,1) \to \text{Sum } 3$	$(2,2) \to \text{Sum } 4$	$(2,3) \to \text{Sum } 5$	$(2,4) \to \text{Sum } 6$	$(2,5) \to \text{Sum } 7$	$(2,6) \to \text{Sum } 8$
Die 1 = 3	$(3,1) \to \text{Sum } 4$	$(3,2) \to \text{Sum } 5$	$(3,3) \to \text{Sum } 6$	$(3,4) \to \text{Sum } 7$	$(3,5) \to \text{Sum } 8$	$(3,6) \to \text{Sum } 9$
Die 1 = 4	$(4,1) \to \text{Sum } 5$	$(4,2) \to \text{Sum } 6$	$(4,3) \to \text{Sum } 7$	$(4,4) \to \text{Sum } 8$	$(4,5) \to \text{Sum } 9$	$(4,6) \to \text{Sum } 10$
Die 1 = 5	$(5,1) \to \text{Sum } 6$	$(5,2) \to \text{Sum } 7$	$(5,3) \to \text{Sum } 8$	$(5,4) \to \text{Sum } 9$	$(5,5) \to \text{Sum } 10$	$(5,6) \to \text{Sum } 11$
Die 1 = 6	$(6,1) \to \text{Sum } 7$	$(6,2) \to \text{Sum } 8$	$(6,3) \to \text{Sum } 9$	$(6,4) \to \text{Sum } 10$	$(6,5) \to \text{Sum } 11$	$(6,6) \to \text{Sum } 12$
Probability Summary		$P(\text{Sum}=7) = \frac{6}{36} = \frac{1}{6}$, $P(\text{Sum}\ge 10) = \frac{6}{36} = \frac{1}{6}$, $P(\text{Doubles}) = \frac{6}{36} = \frac{1}{6}$, Total Outcomes $n(S) = 6 \times 6 = 36$.

📝 6. Complete Solved Textbook Exercises & Examination Question Bank

Below is the complete, step-by-step solved solution manual for every textbook exercise problem (Exercise 11.1, Exercise 11.2, Exercise 11.3, Review Exercise 11, and Extra booster questions) with structured data tables, mathematical derivations, and FBISE scoring rubrics.

Exercise 11.1 • Solved Exercise

Exercise 11.1 Q1 Frequency Distribution Analysis

Exercise 11.1 Q1 The given table shows number of regular readers of a library who have completely read books:

Class Interval (books)	Frequency ($f$, No. of readers)
1 - 10	5
11 - 20	4
21 - 30	8
31 - 40	9
41 - 50	2
51 - 60	2

Answer the following:

(i) What is the total number of readers in the data?

(ii) Which group contains highest number of readers?

(iii) Which group contains least number of readers?

(iv) What is the lower limit of the last class?

(v) What is the lower boundary of the last class?

(vi) What is the size of the class interval?

(vii) Find class marks (mid points) of all groups.

Detailed Step-by-Step Solution:

• Step 1 (Total Readers): Sum frequencies $\sum f = 5 + 4 + 8 + 9 + 2 + 2 = \mathbf{30}$.

• Step 2 (Highest Group): Maximum frequency is $9$, belonging to class 31 - 40.

• Step 3 (Least Group): Minimum frequency is $2$, shared by classes 41 - 50 and 51 - 60.

• Step 4 (Lower Limit of Last Class): Last class is $51 - 60$, so lower limit is 51.

• Step 5 (Lower Boundary of Last Class): $\text{LCL} - 0.5 = 51 - 0.5 = \mathbf{50.5}$.

• Step 6 (Class Interval Size): $h = 11 - 1 = \mathbf{10}$.

• Step 7 (Class Marks Table):

Class	Midpoint ($x = \frac{\text{Lower}+\text{Upper}}{2}$)
1 - 10	5.5
11 - 20	15.5
21 - 30	25.5
31 - 40	35.5
41 - 50	45.5
51 - 60	55.5

Exercise 11.1 Q2 Grouped Frequency Table Construction

Exercise 11.1 Q2 Number of gratitude cards made by 80 students of class 9 is given below:

79, 60, 74, 59, 55, 98, 61, 67, 89, 71, 71, 46, 63, 66, 69, 42, 75, 62, 71, 77, 78, 65, 87, 57, 78, 91, 82, 73, 65, 94, 48, 87, 62, 81, 63, 66, 65, 49, 45, 51, 69, 56, 84, 93, 63, 60, 68, 51, 73, 54, 50, 88, 76, 93, 48, 70, 40, 76, 95, 57, 63, 94, 82, 54, 89, 64, 77, 94, 72, 69, 51, 56, 67, 88, 81, 70, 81, 54, 66, 87.

(i) Prepare a frequency table using classes 40-49, 50-59, etc.

(ii) Add cumulative frequency column in the table.

(iii) How many students made less than 70 cards?

(iv) What percent of students made less than 50 cards?

(v) Which group has the greatest frequency?

(vi) What is the size of the class interval?

Detailed Step-by-Step Solution:

• Complete Frequency & Cumulative Frequency Table:

Class Interval	Tally Marks	Frequency ($f$)	Cumulative Frequency ($cf$)
40 - 49	卌 \|\|	7	7
50 - 59	卌卌 \|\|\|\|	14	21
60 - 69	卌卌卌卌 \|\|	22	43
70 - 79	卌卌卌 \|\|	17	60
80 - 89	卌卌 \|\|	12	72
90 - 99	卌 \|\|\|	8	80
Total	—	$\sum f = 80$	$N = 80$

• Step 2 (Less than 70 cards): Cumulative frequency up to class $60-69$ is 43 students.

• Step 3 (Percent less than 50 cards): Class $40-49$ has $7$ students. Percentage $= \frac{7}{80} \times 100\% = \mathbf{8.75\%}$.

• Step 4 (Greatest Frequency Group): Class 60 - 69 (frequency $22$).

• Step 5 (Class Interval Size): $h = 50 - 40 = \mathbf{10}$.

Exercise 11.1 Q3 Discrete Frequency Distribution

Exercise 11.1 Q3 The number of medals won by 45 players in a certain sports festival is given as:

0, 2, 1, 0, 1, 2, 3, 5, 6, 3, 2, 1, 3, 4, 2, 6, 1, 5, 2, 4, 3, 0, 1, 2, 3, 0, 0, 2, 3, 4, 1, 5, 6, 2, 4, 5, 1, 3, 4, 6, 2, 3, 1, 2, 5.

Prepare a discrete frequency distribution. Also make a column of cumulative frequencies.

Detailed Step-by-Step Solution:

• Discrete Frequency & Cumulative Frequency Table:

Medals Won ($x$)	Tally Marks	Frequency ($f$)	Cumulative Frequency ($cf$)
0	卌	5	5
1	卌 \|\|\|	8	13
2	卌卌	10	23
3	卌 \|\|\|	8	31
4	卌	5	36
5	卌	5	41
6	\|\|\|\|	4	45
Total	—	$\sum f = 45$	$N = 45$

Exercise 11.1 Q4 Frequency Polygon Construction

Exercise 11.1 Q4 The ages of workers in a factory were recorded as below:

Ages in Years	20 - 24	25 - 29	30 - 34	35 - 39	40 - 44	45 - 49
No. of workers ($f$)	5	16	12	10	8	4

Draw a frequency polygon to represent the data.

Detailed Step-by-Step Solution:

• Frequency Polygon Coordinate Table:

Class Interval	Midpoint ($x$)	Frequency ($f$)	Plotted Point $(x, f)$
Preceding Class	17	0	$(17, 0)$
20 - 24	22	5	$(22, 5)$
25 - 29	27	16	$(27, 16)$
30 - 34	32	12	$(32, 12)$
35 - 39	37	10	$(37, 10)$
40 - 44	42	8	$(42, 8)$
45 - 49	47	4	$(47, 4)$
Succeeding Class	52	0	$(52, 0)$

• Plotting Rule: Connect points $(17,0), (22,5), (27,16), (32,12), (37,10), (42,8), (47,4), (52,0)$ consecutively with straight line segments to form a closed polygon touching the horizontal axis.

Exercise 11.1 Q5 Histogram with Unequal Class Intervals

Exercise 11.1 Q5 Represent the following data by a histogram:

Class Interval	0 - 5	5 - 15	15 - 30	30 - 40	40 - 45
Frequency ($f$)	5	12	16	10	2

Detailed Step-by-Step Solution:

• Frequency Density Calculation Table:

Class Interval	Class Width ($h$)	Frequency ($f$)	Frequency Density ($\text{FD} = \frac{f}{h}$)	Histogram Height
0 - 5	$5 - 0 = 5$	5	$\frac{5}{5} = \mathbf{1.00}$	1.00
5 - 15	$15 - 5 = 10$	12	$\frac{12}{10} = \mathbf{1.20}$	1.20
15 - 30	$30 - 15 = 15$	16	$\frac{16}{15} \approx \mathbf{1.07}$	1.07
30 - 40	$40 - 30 = 10$	10	$\frac{10}{10} = \mathbf{1.00}$	1.00
40 - 45	$45 - 40 = 5$	2	$\frac{2}{5} = \mathbf{0.40}$	0.40

• Histogram Construction: Plot continuous boundaries $0, 5, 15, 30, 40, 45$ on x-axis and Frequency Density on y-axis.

Exercise 11.1 Q6 Histogram of Unequal Intervals (Savings Certificates)

Exercise 11.1 Q6 In a saving group, there are 400 members, and the number of savings certificates held by them is shown in the following table. Construct a histogram of the distribution of saving certificates:

Class Interval	1 – 50	51 – 100	101 – 150	151 – 200	201 – 300	301 – 400	401 – 500
No. of members ($f$)	10	15	30	40	120	100	85

Detailed Step-by-Step Solution:

• Frequency Density Work-Table:

Class Interval	Class Boundaries	Width ($h$)	Frequency ($f$)	Frequency Density ($\text{FD} = \frac{f}{h}$)
1 – 50	0.5 – 50.5	50	10	$\frac{10}{50} = \mathbf{0.20}$
51 – 100	50.5 – 100.5	50	15	$\frac{15}{50} = \mathbf{0.30}$
101 – 150	100.5 – 150.5	50	30	$\frac{30}{50} = \mathbf{0.60}$
151 – 200	150.5 – 200.5	50	40	$\frac{40}{50} = \mathbf{0.80}$
201 – 300	200.5 – 300.5	100	120	$\frac{120}{100} = \mathbf{1.20}$
301 – 400	300.5 – 400.5	100	100	$\frac{100}{100} = \mathbf{1.00}$
401 – 500	400.5 – 500.5	100	85	$\frac{85}{100} = \mathbf{0.85}$

• Histogram Construction: Draw adjacent bars along the class boundaries with heights corresponding to the calculated Frequency Densities.

Exercise 11.1 Q7 Interpreting Frequency Polygon

Exercise 11.1 Q7 The frequency polygon shows per-capita income of states recorded in thousand dollars with plotted points: $(10, 8), (14, 13), (16, 10), (18, 5), (20, 2), (24, 0)$:

(i) What is the total number of states?

(ii) What is the number of states which have the least per-capita income?

(iii) What is the number of states having highest per-capita income?

(iv) What percent of states have per-capita income $14000 and $16000?

(v) How many states have per-capita income $18000 and above?

(vi) Represent the frequency polygon by a frequency table.

Detailed Step-by-Step Solution:

• Step 1 (Total States): Sum frequencies $\sum f = 8 + 13 + 10 + 5 + 2 = \mathbf{38}$ states.

• Step 2 (Least Income): Lowest income point ($10$k) has 8 states.

• Step 3 (Highest Income): Highest active income level ($20$k) has 2 states.

• Step 4 (Percent for $14k & $16k): Total $= 13 + 10 = 23 \implies \frac{23}{38} \times 100\% \approx \mathbf{60.53\%}$.

• Step 5 ($18k and above): Total $= 5 + 2 = \mathbf{7}$ states.

• Step 6 (Frequency Table):

Per-Capita Income ($)	No. of States ($f$)
$10,000	8
$14,000	13
$16,000	10
$18,000	5
$20,000	2
Total	38

Exercise 11.1 Q8 Histogram & Polygon (Masses of Boys)

Exercise 11.1 Q8 The masses measured to nearest kg of 50 boys are given in the frequency distribution:

Mass (kg)	60 – 64	65 – 69	70 – 79	80 – 89	90 – 94	95 – 99
Frequency ($f$)	2	6	12	14	10	6

(i) Construct a histogram.

(ii) Construct a frequency polygon.

Detailed Step-by-Step Solution:

• Frequency Density & Coordinate Table:

Class Limits	Class Boundaries	Width ($h$)	Freq ($f$)	FD ($\frac{f}{h}$)	Midpoint ($x$)
60 – 64	59.5 – 64.5	5	2	0.40	62.0
65 – 69	64.5 – 69.5	5	6	1.20	67.0
70 – 79	69.5 – 79.5	10	12	1.20	74.5
80 – 89	79.5 – 89.5	10	14	1.40	84.5
90 – 94	89.5 – 94.5	5	10	2.00	92.0
95 – 99	94.5 – 99.5	5	6	1.20	97.0

Exercise 11.1 Q9 Histogram Construction (Price Ranges)

Exercise 11.1 Q9 At a sale, number of items placed according to price range is given below:

Price (Rs)	600 – 999	1000 – 1999	2000 – 2499	2500 – 2999	3000 – 3499
Frequency ($f$)	50	70	75	65	70

Construct a histogram.

Detailed Step-by-Step Solution:

• Frequency Density Work-Table:

Price Range (Rs)	Class Boundaries	Width ($h$)	Frequency ($f$)	Frequency Density ($\text{FD} = \frac{f}{h}$)
600 – 999	599.5 – 999.5	400	50	$\frac{50}{400} = \mathbf{0.125}$
1000 – 1999	999.5 – 1999.5	1000	70	$\frac{70}{1000} = \mathbf{0.070}$
2000 – 2499	1999.5 – 2499.5	500	75	$\frac{75}{500} = \mathbf{0.150}$
2500 – 2999	2499.5 – 2999.5	500	65	$\frac{65}{500} = \mathbf{0.130}$
3000 – 3499	2999.5 – 3499.5	500	70	$\frac{70}{500} = \mathbf{0.140}$

Exercise 11.2 • Solved Exercise

Exercise 11.2 Q1 Arithmetic Mean by Definition

Exercise 11.2 Q1 Find arithmetic mean for the following data by definition $\bar{x} = \frac{\sum x}{n}$:

(i) $x = 2, 4, 6, 8, 10, 12$

(ii) $y = 3, 4, -1, 7, -8, -5, 0$

(iii) $z = 0, 4, 8, 12, 16, 20, 24, 28$

(iv) $u = 3.1, 4.2, 5.3, 6.4, 7.5, 8.6, 9.7, 10.8$

(v) $v = 5, 5, 5, 5, 5, 5, 5, 5$

Detailed Step-by-Step Solution:

• Part (i): $\bar{x} = \frac{2+4+6+8+10+12}{6} = \frac{42}{6} = \mathbf{7}$.

• Part (ii): $\bar{y} = \frac{3+4-1+7-8-5+0}{7} = \frac{0}{7} = \mathbf{0}$.

• Part (iii): $\bar{z} = \frac{0+4+8+12+16+20+24+28}{8} = \frac{112}{8} = \mathbf{14}$.

• Part (iv): $\bar{u} = \frac{3.1+4.2+5.3+6.4+7.5+8.6+9.7+10.8}{8} = \frac{55.6}{8} = \mathbf{6.95}$.

• Part (v): $\bar{v} = \frac{5 \times 8}{8} = \mathbf{5}$.

Exercise 11.2 Q2 Comparing Batsman Performance via Mean

Exercise 11.2 Q2 Following are the scores made by two batsmen A and B in a series of 10 innings:

A: 12, 15, 6, 73, 7, 19, 199, 36, 84, 29

B: 47, 12, 76, 48, 4, 51, 37, 48, 13, 0

Find arithmetic mean of scores of both players and state who is better as run getter?

Detailed Step-by-Step Solution:

• Step 1 (Mean of Batsman A):

$$\sum x_A = 12 + 15 + 6 + 73 + 7 + 19 + 199 + 36 + 84 + 29 = 480$$

$$\bar{x}_A = \frac{480}{10} = \mathbf{48}$$

• Step 2 (Mean of Batsman B):

$$\sum x_B = 47 + 12 + 76 + 48 + 4 + 51 + 37 + 48 + 13 + 0 = 336$$

$$\bar{x}_B = \frac{336}{10} = \mathbf{33.6}$$

• Step 3 (Comparison): Since $\bar{x}_A (48) > \bar{x}_B (33.6)$, Batsman A is a better run getter.

Exercise 11.2 Q3 Weighted Mean of Student Marks

Exercise 11.2 Q3 Marks of four students of class 9 in four subjects out of 100 with assigned weights are:

Subjects	Weight ($w$)	Student A	Student B	Student C	Student D
English	4	75	80	65	70
Urdu	3	82	74	78	85
Mathematics	4	90	85	92	88
Science	3	86	90	80	75
Total Weights	$\sum w = 14$	—	—	—	—

Find weighted arithmetic mean for each student and determine the highest performer.

Detailed Step-by-Step Solution:

• Weighted Mean Work-Table:

Student	Calculation $\sum wx$	Sum $\sum wx$	Weighted Mean $\bar{x}_w = \frac{\sum wx}{\sum w}$
Student A	$4(75) + 3(82) + 4(90) + 3(86)$	$300 + 246 + 360 + 258 = 1164$	$\frac{1164}{14} = \mathbf{83.14}$
Student B	$4(80) + 3(74) + 4(85) + 3(90)$	$320 + 222 + 340 + 270 = 1152$	$\frac{1152}{14} = \mathbf{82.29}$
Student C	$4(65) + 3(78) + 4(92) + 3(80)$	$260 + 234 + 368 + 240 = 1102$	$\frac{1102}{14} = \mathbf{78.71}$
Student D	$4(70) + 3(85) + 4(88) + 3(75)$	$280 + 255 + 352 + 225 = 1112$	$\frac{1112}{14} = \mathbf{79.43}$

• Conclusion: Student A achieved the highest weighted performance with an average of 83.14.

Exercise 11.2 Q4 Grouped Arithmetic Mean (Fuel Consumption)

Exercise 11.2 Q4 Following frequency distribution represents number of liters of fuel consumed against distance in km:

Distance (km)	10 - 14	15 - 19	20 - 24	25 - 29	30 - 34
Fuel in Liters ($f$)	4	6	10	8	2

Calculate arithmetic mean of the fuel consumed.

Detailed Step-by-Step Solution:

• Grouped Mean Work-Table:

Distance (km)	Midpoint ($x$)	Frequency ($f$)	$fx$
10 - 14	12	4	48
15 - 19	17	6	102
20 - 24	22	10	220
25 - 29	27	8	216
30 - 34	32	2	64
Total	—	$\sum f = 30$	$\sum fx = 650$

• Calculation:

$$\bar{x} = \frac{\sum fx}{\sum f} = \frac{650}{30} = \mathbf{21.67\text{ km}}$$

Exercise 11.2 Q5 Grouped Arithmetic Mean (Detergent Packs)

Exercise 11.2 Q5 Find arithmetic mean from the following data of number of detergent packs sold against its mass in grams:

Mass (grams)	200 – 249	250 – 299	300 – 349	350 – 399	400 – 449
No. of packs ($f$)	15	25	35	18	7

Detailed Step-by-Step Solution:

• Grouped Mean Work-Table:

Mass (g)	Midpoint ($x$)	Frequency ($f$)	$fx$
200 – 249	224.5	15	3367.5
250 – 299	274.5	25	6862.5
300 – 349	324.5	35	11357.5
350 – 399	374.5	18	6741.0
400 – 449	424.5	7	2971.5
Total	—	$\sum f = 100$	$\sum fx = 31300$

• Calculation:

$$\bar{x} = \frac{\sum fx}{\sum f} = \frac{31300}{100} = \mathbf{313.0\text{ grams}}$$

Exercise 11.2 Q6 Median of Ungrouped Datasets

Exercise 11.2 Q6 Find median for the following datasets:

(i) $1, 4, 2, 5, 3, 7, 6$

(ii) $\pm 2, \pm 4, \pm 6, \pm 8$

(iii) $0, 1, -2, -3, 4, 5, 6, 3$

(iv) $4, 3, 1, -3, 2, -3, 3, 4, 1$

(v) $4, 4, 4, 4, 4, 4$

Detailed Step-by-Step Solution:

• Part (i): Sorted: $1, 2, 3, 4, 5, 6, 7$ ($n=7$, odd). $\text{Median} = 4\text{th value} = \mathbf{4}$.

• Part (ii): Values: $-8, -6, -4, -2, 2, 4, 6, 8$ ($n=8$, even). $\text{Median} = \frac{-2 + 2}{2} = \mathbf{0}$.

• Part (iii): Sorted: $-3, -2, 0, 1, 3, 4, 5, 6$ ($n=8$, even). $\text{Median} = \frac{1 + 3}{2} = \frac{4}{2} = \mathbf{2}$.

• Part (iv): Sorted: $-3, -3, 1, 1, 2, 3, 3, 4, 4$ ($n=9$, odd). $\text{Median} = 5\text{th value} = \mathbf{2}$.

• Part (v): Sorted: $4, 4, 4, 4, 4, 4$ ($n=6$, even). $\text{Median} = \frac{4+4}{2} = \mathbf{4}$.

Exercise 11.2 Q7 Median of Discrete Distribution (Circle Radii)

Exercise 11.2 Q7 In practical geometry students constructed circles of radii between 0cm and 2cm:

Radius $x$ (cm)	0.5	0.8	1.0	1.2	1.5	1.8	2.0
No. of students ($f$)	3	7	12	15	8	4	1

Find median radius of the constructed circles.

Detailed Step-by-Step Solution:

• Cumulative Frequency Table:

Radius ($x$)	Frequency ($f$)	Cumulative Frequency ($cf$)
0.5	3	3
0.8	7	10
1.0	12	22
1.2	15	37 (Contains $\frac{n+1}{2} = 25.5\text{th}$ value)
1.5	8	45
1.8	4	49
2.0	1	50
Total	$N = 50$	—

• Step 1: Median rank $= \frac{N + 1}{2} = \frac{51}{2} = 25.5\text{th value}$.

• Step 2: Looking at cumulative frequency, the $25.5\text{th}$ value falls in radius $1.2\text{ cm}$.

• Answer: $\text{Median Radius} = \mathbf{1.2\text{ cm}}$.

Exercise 11.2 Q8 Grouped Median Calculation (Old Age House)

Exercise 11.2 Q8 At an old age house, people of different age groups are living as shown in table:

Age Group (Years)	50 - 59	60 - 69	70 - 79	80 - 89	90 - 99
No. of People ($f$)	8	15	22	12	3

Find median age of the people.

Detailed Step-by-Step Solution:

• Grouped Median Work-Table:

Age Group	Class Boundaries	Frequency ($f$)	Cumulative Frequency ($cf$)
50 - 59	49.5 - 59.5	8	8
60 - 69	59.5 - 69.5	15	23 ($c = 23$)
70 - 79	69.5 - 79.5	22	45 (🎯 Median Class, $\frac{n}{2} = 30$)
80 - 89	79.5 - 89.5	12	57
90 - 99	89.5 - 99.5	3	60
Total	—	$n = 60$	—

• Calculation:

$$\text{Median Class: } \frac{n}{2} = \frac{60}{2} = 30 \implies \text{Class } 69.5 - 79.5$$

$$l = 69.5,\; h = 10,\; f = 22,\; c = 23$$

$$\tilde{x} = l + \frac{h}{f}\left(\frac{n}{2} - c\right) = 69.5 + \frac{10}{22}(30 - 23) = 69.5 + \frac{70}{22} \approx \mathbf{72.68\text{ Years}}$$

Exercise 11.2 Q9 Mode of Raw Datasets

Exercise 11.2 Q9 Find mode for the following data:

(i) $1, 2, 3, 4, 5, 6$

(ii) $2, 4, 2, 3, 2, 5, 3, 2, 5, 4, 2$

(iii) $120, 130, 140, 225, 125, 225, 120$

(iv) $2, 4, 3, 5, 5, 3, 4, 2$

Detailed Step-by-Step Solution:

• Part (i): Every number occurs exactly once with frequency 1. Therefore, No mode exists.

• Part (ii): 2 occurs 5 times, 3 occurs 2 times, 4 occurs 2 times, 5 occurs 2 times. Maximum frequency is 5 for Mode = 2.

• Part (iii): 120 occurs twice, 225 occurs twice, others once. The data is bimodal with Modes = 120 and 225.

• Part (iv): 2, 3, 4, and 5 all appear with equal frequency 2. Therefore, all four are modes (Multimodal / No unique mode).

Exercise 11.2 Q10 Mean, Median & Mode of Discrete Table

Exercise 11.2 Q10 Find mean, median and mode for the following discrete distribution:

Value ($x$)	1	2	3	4	5	6
Frequency ($f$)	2	5	9	6	3	1

Detailed Step-by-Step Solution:

• Combined Work-Table:

$x$	$f$	$fx$	$cf$
1	2	2	2
2	5	10	7
3	9 (Highest $f$)	27	16 (Contains $\frac{N+1}{2} = 13.5\text{th}$)
4	6	24	22
5	3	15	25
6	1	6	26
Total	$\sum f = 26$	$\sum fx = 84$	—

• 1. Mean: $\bar{x} = \frac{\sum fx}{\sum f} = \frac{84}{26} \approx \mathbf{3.23}$

• 2. Median: $\frac{N+1}{2} = 13.5\text{th value} \implies \mathbf{3}$

• 3. Mode: Value with highest frequency ($9$) is $\mathbf{3}$.

Exercise 11.2 Q11 Grouped Mode Calculation

Exercise 11.2 Q11 Find modal value for the following frequency distribution:

Class Interval	10 - 19	20 - 29	30 - 39	40 - 49	50 - 59
Frequency ($f$)	3	7	15	8	2

Detailed Step-by-Step Solution:

• Grouped Mode Work-Table:

Class Interval	Class Boundaries	Frequency ($f$)	Modal Parameter Role
10 - 19	9.5 - 19.5	3	—
20 - 29	19.5 - 29.5	7	$f_1 = 7$ (Preceding)
30 - 39	29.5 - 39.5	15	🎯 $f_m = 15$ (Modal Class, $l = 29.5$)
40 - 49	39.5 - 49.5	8	$f_2 = 8$ (Following)
50 - 59	49.5 - 59.5	2	—

• Calculation:

$$\text{Mode} = l + \frac{f_m - f_1}{2f_m - f_1 - f_2} \times h$$

$$\text{Mode} = 29.5 + \frac{15 - 7}{2(15) - 7 - 8} \times 10 = 29.5 + \left(\frac{8}{15}\right) \times 10 = 29.5 + 5.33 = \mathbf{34.83}$$

Exercise 11.3 • Solved Exercise

Exercise 11.3 Q1 Single Event Probability (Word 'ALLAH')

Exercise 11.3 Q1 A letter is chosen randomly from the word 'ALLAH'. Find the probability of getting:

(i) a vowel

(ii) an H

(iii) an L

(iv) a consonant

Detailed Step-by-Step Solution:

• Total letters in 'ALLAH' $n(S) = 5$ (Letters: A, L, L, A, H).

• (i) A vowel: Vowels are {A, A} $\implies n(E_1) = 2$. $P(E_1) = \mathbf{\frac{2}{5}}$.

• (ii) An H: Letter {H} $\implies n(E_2) = 1$. $P(E_2) = \mathbf{\frac{1}{5}}$.

• (iii) An L: Letters {L, L} $\implies n(E_3) = 2$. $P(E_3) = \mathbf{\frac{2}{5}}$.

• (iv) A consonant: Consonants are {L, L, H} $\implies n(E_4) = 3$. $P(E_4) = \mathbf{\frac{3}{5}}$.

Exercise 11.3 Q2 Probability in Quality Inspection (Garment Factory)

Exercise 11.3 Q2 A garment factory shipped an order which contained 7000 jackets, 2000 sweaters, and 3000 trousers in which 7 trousers, 5 jackets and 9 sweaters are faulty. If the quality inspector unpacked one item at random, find the probability that it is:

(i) a trouser

(ii) not a jacket

(iii) a faulty item

Detailed Step-by-Step Solution:

• Total items in sample space: $n(S) = 7000 + 2000 + 3000 = 12,000$.

• (i) A trouser: $n(T) = 3000$. $P(T) = \frac{3000}{12000} = \mathbf{\frac{1}{4}}$ ($0.25$).

• (ii) Not a jacket: Items other than jackets $= 2000 + 3000 = 5000$. $P(J') = \frac{5000}{12000} = \mathbf{\frac{5}{12}}$.

• (iii) A faulty item: Total faulty $= 5 + 9 + 7 = 21$. $P(F) = \frac{21}{12000} = \mathbf{\frac{7}{4000}}$ ($0.00175$).

Exercise 11.3 Q3 Probability on Spinner Wheel (8 Flower Sectors)

Exercise 11.3 Q3 A number wheel is divided into 8 equal sectors labelled as: pansy, lily, orchid, tulip, jasmine, rose, marigold, sunflower. Maazz spins the wheel once. Find the probability that pointer:

(i) Stops at rose

(ii) Stops at a 4 letter flower name

(iii) Does not stop at marigold

(iv) Stops at a 3 letter flower name

(v) Stops at a flower name

Detailed Step-by-Step Solution:

• Total sectors $n(S) = 8$.

• (i) Stops at rose: $1$ sector $\implies P = \mathbf{\frac{1}{8}}$.

• (ii) 4-letter flower name: 'lily' (4), 'rose' (4) $\implies 2$ sectors. $P = \frac{2}{8} = \mathbf{\frac{1}{4}}$.

• (iii) Does not stop at marigold: $8 - 1 = 7$ sectors. $P = \mathbf{\frac{7}{8}}$.

• (iv) 3-letter flower name: No 3-letter flower name exists $\implies P = \mathbf{0}$ (Impossible event).

• (v) Stops at a flower name: All 8 sectors are flowers $\implies P = \frac{8}{8} = \mathbf{1}$ (Certain event).

Exercise 11.3 Q4 Probability with Decagonal Die

Exercise 11.3 Q4 A decagonal die labelled 4, 4, 4, 4, 5, 5, 6, 7, 8, 8 is rolled once. Find the probability of getting:

(i) a 4

(ii) an even number

(iii) a multiple of 4

(iv) not a 7

(v) an odd number

(vi) a prime number

(vii) LCM of 4 and 8

(viii) HCF of 4 and 8

(ix) factor of 12

Detailed Step-by-Step Solution:

• Total faces $n(S) = 10$. Sample space: $\{4, 4, 4, 4, 5, 5, 6, 7, 8, 8\}$.

• (i) A 4: 4 faces $\implies P = \frac{4}{10} = \mathbf{\frac{2}{5}}$.

• (ii) An even number: $\{4, 4, 4, 4, 6, 8, 8\} \to 7$ faces $\implies P = \mathbf{\frac{7}{10}}$.

• (iii) A multiple of 4: $\{4, 4, 4, 4, 8, 8\} \to 6$ faces $\implies P = \frac{6}{10} = \mathbf{\frac{3}{5}}$.

• (iv) Not a 7: $10 - 1 = 9$ faces $\implies P = \mathbf{\frac{9}{10}}$.

• (v) An odd number: $\{5, 5, 7\} \to 3$ faces $\implies P = \mathbf{\frac{3}{10}}$.

• (vi) A prime number: $\{5, 5, 7\} \to 3$ faces $\implies P = \mathbf{\frac{3}{10}}$.

• (vii) LCM of 4 and 8: $\text{LCM}(4, 8) = 8$. Number of $8\text{'s} = 2 \implies P = \frac{2}{10} = \mathbf{\frac{1}{5}}$.

• (viii) HCF of 4 and 8: $\text{HCF}(4, 8) = 4$. Number of $4\text{'s} = 4 \implies P = \frac{4}{10} = \mathbf{\frac{2}{5}}$.

• (ix) Factor of 12: Factors among faces are $4, 6$. Total $= 4 + 1 = 5$ faces $\implies P = \frac{5}{10} = \mathbf{\frac{1}{2}}$.

Exercise 11.3 Q5 Probability of 1-Digit Whole Numbers

Exercise 11.3 Q5 A one digit whole number is chosen at random ($S = \{0, 1, 2, 3, 4, 5, 6, 7, 8, 9\}, n(S)=10$). Find the probability that it is:

(i) less than 5

(ii) greater than 10

(iii) not the largest 1 digit number

(iv) additive identity of whole numbers

(v) HCF of 3 and 5

(vi) multiplicative identity of real numbers

(vii) not a prime number

(viii) factor of 9

Detailed Step-by-Step Solution:

• $S = \{0, 1, 2, 3, 4, 5, 6, 7, 8, 9\}$, $n(S) = 10$.

• (i) Less than 5: $\{0, 1, 2, 3, 4\} \to 5$ outcomes $\implies P = \frac{5}{10} = \mathbf{\frac{1}{2}}$.

• (ii) Greater than 10: Empty set $\implies P = \mathbf{0}$.

• (iii) Not the largest 1 digit number: All except $9 \to 9$ outcomes $\implies P = \mathbf{\frac{9}{10}}$.

• (iv) Additive identity of whole numbers: Number is $0 \to 1$ outcome $\implies P = \mathbf{\frac{1}{10}}$.

• (v) HCF of 3 and 5: $\text{HCF}(3, 5) = 1 \to 1$ outcome $\implies P = \mathbf{\frac{1}{10}}$.

• (vi) Multiplicative identity: Number is $1 \to 1$ outcome $\implies P = \mathbf{\frac{1}{10}}$.

• (vii) Not a prime number: Primes are $\{2, 3, 5, 7\}$. Non-primes are $\{0, 1, 4, 6, 8, 9\} \to 6$ outcomes $\implies P = \frac{6}{10} = \mathbf{\frac{3}{5}}$.

• (viii) Factor of 9: Factors of 9 in set are $\{1, 3, 9\} \to 3$ outcomes $\implies P = \mathbf{\frac{3}{10}}$.

Exercise 11.3 Q6 Relative Frequency and Expected Head/Tail

Exercise 11.3 Q6 A coin is tossed 10 times with results: Head: 6, Tail: 4.

Complete relative frequencies and answer:

(i) What is the expected frequency of getting head, if it was tossed 520 times?

(ii) What is the expected frequency of getting tail, if it was tossed 305 times?

Detailed Step-by-Step Solution:

• Step 1 (Relative Frequencies):

• Head: $\frac{6}{10} = \mathbf{0.6}$

• Tail: $\frac{4}{10} = \mathbf{0.4}$

• Step 2 (Expected Heads in 520 tosses):

$$\text{Expected Heads} = n \times p = 520 \times 0.6 = \mathbf{312}$$

• Step 3 (Expected Tails in 305 tosses):

$$\text{Expected Tails} = n \times p = 305 \times 0.4 = \mathbf{122}$$

Exercise 11.3 Q7 Expected Frequencies on Die Rolling Experiment

Exercise 11.3 Q7 A die is rolled 120 times with outcomes:

Number on Die	1	2	3	4	5	6
Frequency ($f$)	18	22	20	25	15	20

Find the relative frequency of each outcome and compare with expected frequency.

Detailed Step-by-Step Solution:

• Relative vs Expected Frequency Table:

Outcome	Observed ($f$)	Relative Frequency ($\frac{f}{120}$)	Theoretical $P(E)$	Expected Count ($n \times p$)
1	18	$\frac{18}{120} = 0.150$	$\frac{1}{6} \approx 0.167$	$120 \times \frac{1}{6} = 20$
2	22	$\frac{22}{120} = 0.183$	$\frac{1}{6} \approx 0.167$	$120 \times \frac{1}{6} = 20$
3	20	$\frac{20}{120} = 0.167$	$\frac{1}{6} \approx 0.167$	$120 \times \frac{1}{6} = 20$
4	25	$\frac{25}{120} = 0.208$	$\frac{1}{6} \approx 0.167$	$120 \times \frac{1}{6} = 20$
5	15	$\frac{15}{120} = 0.125$	$\frac{1}{6} \approx 0.167$	$120 \times \frac{1}{6} = 20$
6	20	$\frac{20}{120} = 0.167$	$\frac{1}{6} \approx 0.167$	$120 \times \frac{1}{6} = 20$
Total	120	1.000	1.000	120

Exercise 11.3 Q8 Complementary Events & Expected Selections

Exercise 11.3 Q8 There are 6 girls sections of class 9 in a public school, namely Teal, orchid, mauve, hazel, zaffre and denim. Only one girl is to be chosen randomly from all sections for national science olympiad.

(a) Find the probability of selecting a girl from hazel section.

(b) Find the probability of not selecting a girl from hazel section.

(d) If 6 girls are to be selected, find expected frequency of selecting from orchid section.

(e) If 60 girls are to be selected, what number of girls is expected to be chosen from Teal?

Detailed Step-by-Step Solution:

• Total sections $n(S) = 6$.

• (a) $P(\text{Hazel}) = \mathbf{\frac{1}{6}}$.

• (b) $P(\text{Not Hazel}) = 1 - \frac{1}{6} = \mathbf{\frac{5}{6}}$.

• (c) $P(\text{Hazel}) + P(\text{Not Hazel}) = \frac{1}{6} + \frac{5}{6} = \frac{6}{6} = \mathbf{1}$ (Verified).

• (d) Expected from orchid in 6 selections $= 6 \times \frac{1}{6} = \mathbf{1\text{ girl}}$.

• (e) Expected from Teal in 60 selections $= 60 \times \frac{1}{6} = \mathbf{10\text{ girls}}$.

Review Exercise 11 • Solved Review Exercise

Review Exercise 11 Q1 (1) Basic Statistics & Probability MCQs

Review Exercise 11 Q1 (1) Which of the following is a class mark of the interval (10 – 15)?

(A) 10

(B) 12.5

(C) 15

(D) 16

Detailed Step-by-Step Solution:

• Step 1: Analyze the question: Which of the following is a class mark of the interval (10 – 15)?

• Step 2: Class mark is the average of class limits: $\frac{10+15}{2} = \frac{25}{2} = \mathbf{12.5}$.

• Correct Answer: [B] 12.5

Review Exercise 11 Q1 (2) Basic Statistics & Probability MCQs

Review Exercise 11 Q1 (2) What is size of class interval (4 – 7)?

(A) 4

(B) 5

(C) 6

(D) 7

Detailed Step-by-Step Solution:

• Step 1: Analyze the question: What is size of class interval (4 – 7)?

• Step 2: Inclusive size of class interval $4-7$ is $7 - 4 + 1 = \mathbf{4}$.

• Correct Answer: [A] 4

Review Exercise 11 Q1 (3) Basic Statistics & Probability MCQs

Review Exercise 11 Q1 (3) Which of the following is chart of adjacent rectangles?

(A) bar graph

(B) frequency polygon

(C) histogram

(D) ogive

Detailed Step-by-Step Solution:

• Step 1: Analyze the question: Which of the following is chart of adjacent rectangles?

• Step 2: A histogram consists of a set of adjacent rectangles whose areas are proportional to frequencies.

• Correct Answer: [C] histogram

Review Exercise 11 Q1 (4) Basic Statistics & Probability MCQs

Review Exercise 11 Q1 (4) Which of the following is measure of central tendency?

(A) variance

(B) standard deviation

(C) range

(D) arithmetic mean

Detailed Step-by-Step Solution:

• Step 1: Analyze the question: Which of the following is measure of central tendency?

• Step 2: Arithmetic mean is a primary measure of central tendency (location).

• Correct Answer: [D] arithmetic mean

Review Exercise 11 Q1 (5) Basic Statistics & Probability MCQs

Review Exercise 11 Q1 (5) Which of the following is a formula of arithmetic mean for grouped data?

(A) $\frac{\sum x}{n}$

(B) $l + \frac{h}{f}(\frac{n}{2} - c)$

(C) $\frac{\sum fx}{\sum f}$

(D) $l + \frac{f_m - f_1}{2f_m - f_1 - f_2} \times h$

Detailed Step-by-Step Solution:

• Step 1: Analyze the question: Which of the following is a formula of arithmetic mean for grouped data?

• Step 2: Grouped arithmetic mean is given by $\mathbf{\frac{\sum fx}{\sum f}}$.

• Correct Answer: [C] $\frac{\sum fx}{\sum f}$

Review Exercise 11 Q1 (6) Basic Statistics & Probability MCQs

Review Exercise 11 Q1 (6) If arithmetic mean of 25 values is 10, then what is sum of values?

(A) 250

(B) 125

(C) 25

(D) 2.5

Detailed Step-by-Step Solution:

• Step 1: Analyze the question: If arithmetic mean of 25 values is 10, then what is sum of values?

• Step 2: $\sum x = n \times \bar{x} = 25 \times 10 = \mathbf{250}$.

• Correct Answer: [A] 250

Review Exercise 11 Q1 (7) Basic Statistics & Probability MCQs

Review Exercise 11 Q1 (7) What is median of the data 4, 3, 0, 2, 1?

(A) 0

(B) 2

(C) 3

(D) 4

Detailed Step-by-Step Solution:

• Step 1: Analyze the question: What is median of the data 4, 3, 0, 2, 1?

• Step 2: Arranged data: $0, 1, 2, 3, 4$. The middle value ($3\text{rd}$ term) is 2.

• Correct Answer: [B] 2

Review Exercise 11 Q1 (8) Basic Statistics & Probability MCQs

Review Exercise 11 Q1 (8) Which measure of central tendency can have more than one value?

(A) weighted mean

(B) median

(C) mode

(D) arithmetic mean

Detailed Step-by-Step Solution:

• Step 1: Analyze the question: Which measure of central tendency can have more than one value?

• Step 2: A dataset can have multiple modes (bimodal, trimodal, etc.). Thus mode can have more than one value.

• Correct Answer: [C] mode

Review Exercise 11 Q1 (9) Basic Statistics & Probability MCQs

Review Exercise 11 Q1 (9) The probability of getting M in 'MUHAMMAD' is:

(A) 1/8

(B) 3/8

(C) 3/5

(D) none of these

Detailed Step-by-Step Solution:

• Step 1: Analyze the question: The probability of getting M in 'MUHAMMAD' is:

• Step 2: Total letters in 'MUHAMMAD' = 8. Frequency of M = 3. $P(M) = \mathbf{\frac{3}{8}}$.

• Correct Answer: [B] 3/8

Review Exercise 11 Q1 (10) Basic Statistics & Probability MCQs

Review Exercise 11 Q1 (10) Probability of picking an ace from a well shuffled pack of 52 playing cards is:

(A) 1/52

(B) 1/13

(C) 4/13

(D) none of these

Detailed Step-by-Step Solution:

• Step 1: Analyze the question: Probability of picking an ace from a well shuffled pack of 52 playing cards is:

• Step 2: Number of aces = 4. $P(\text{Ace}) = \frac{4}{52} = \mathbf{\frac{1}{13}}$.

• Correct Answer: [B] 1/13

Review Exercise 11 Q1 (11) Basic Statistics & Probability MCQs

Review Exercise 11 Q1 (11) A normal fair die is rolled 6000 times. The expected number of 5 is:

(A) 100

(B) 5000

(C) 1000

(D) none of these

Detailed Step-by-Step Solution:

• Step 1: Analyze the question: A normal fair die is rolled 6000 times. The expected number of 5 is:

• Step 2: $P(5) = 1/6$. $\text{Expected} = 6000 \times \frac{1}{6} = \mathbf{1000}$.

• Correct Answer: [C] 1000

Review Exercise 11 Q1 (12) Basic Statistics & Probability MCQs

Review Exercise 11 Q1 (12) A fair coin is tossed 500 times. Expected number of tails is:

(A) 100

(B) 250

(C) 1000

(D) none of these

Detailed Step-by-Step Solution:

• Step 1: Analyze the question: A fair coin is tossed 500 times. Expected number of tails is:

• Step 2: $P(\text{Tail}) = 1/2$. $\text{Expected} = 500 \times 0.5 = \mathbf{250}$.

• Correct Answer: [B] 250

Review Exercise 11 Q1 (13) Basic Statistics & Probability MCQs

Review Exercise 11 Q1 (13) In a group of 5 people, 4 like peach juice. The expected no. of people in a population of 1200 who like peach juice is:

(A) 240

(B) 600

(C) 960

(D) none of these

Detailed Step-by-Step Solution:

• Step 1: Analyze the question: In a group of 5 people, 4 like peach juice. The expected no. of people in a population of 1200 who like peach juice is:

• Step 2: $P = 4/5$. $\text{Expected} = 1200 \times \frac{4}{5} = \mathbf{960}$.

• Correct Answer: [C] 960

Review Exercise 11 Q1 (14) Basic Statistics & Probability MCQs

Review Exercise 11 Q1 (14) How many times Haani should toss a fair coin if he expects to get 100 tails?

(A) 100

(B) 200

(C) 1000

(D) none of these

Detailed Step-by-Step Solution:

• Step 1: Analyze the question: How many times Haani should toss a fair coin if he expects to get 100 tails?

• Step 2: $n \times 0.5 = 100 \implies n = \frac{100}{0.5} = \mathbf{200}$.

• Correct Answer: [B] 200

Review Exercise 11 Q1 (15) Basic Statistics & Probability MCQs

Review Exercise 11 Q1 (15) An event which can never happen is called:

(A) sure event

(B) certain event

(C) possible event

(D) impossible event

Detailed Step-by-Step Solution:

• Step 1: Analyze the question: An event which can never happen is called:

• Step 2: An event that has zero chance of occurrence is called an impossible event ($P=0$).

• Correct Answer: [D] impossible event

Review Exercise 11 Q1 (16) Basic Statistics & Probability MCQs

Review Exercise 11 Q1 (16) Two such events whose probabilities are 1/2 each are called:

(A) likely

(B) equally likely

(C) unlikely

(D) none of these

Detailed Step-by-Step Solution:

• Step 1: Analyze the question: Two such events whose probabilities are 1/2 each are called:

• Step 2: Events having identical probability of occurrence are termed equally likely events.

• Correct Answer: [B] equally likely

Review Exercise 11 Q1 (17) Basic Statistics & Probability MCQs

Review Exercise 11 Q1 (17) The probability of a certain event is:

(A) 0

(B) 1/2

(C) 3/4

(D) 1

Detailed Step-by-Step Solution:

• Step 1: Analyze the question: The probability of a certain event is:

• Step 2: A certain (sure) event has probability 1.

• Correct Answer: [D] 1

Review Exercise 11 Q1 (18) Basic Statistics & Probability MCQs

Review Exercise 11 Q1 (18) The probability of an event can take the value:

(A) 0

(B) 1

(C) 1/2

(D) all of these

Detailed Step-by-Step Solution:

• Step 1: Analyze the question: The probability of an event can take the value:

• Step 2: Probabilities range between $0$ and $1$ inclusive, so $0, 1/2, 1$ are all valid.

• Correct Answer: [D] all of these

Review Exercise 11 Q1 (19) Basic Statistics & Probability MCQs

Review Exercise 11 Q1 (19) The probability of an event cannot take the value:

(A) 0

(B) 1

(C) 2

(D) all of these

Detailed Step-by-Step Solution:

• Step 1: Analyze the question: The probability of an event cannot take the value:

• Step 2: Probability cannot exceed 1. Thus 2 cannot be a probability.

• Correct Answer: [C] 2

Review Exercise 11 Q1 (20) Basic Statistics & Probability MCQs

Review Exercise 11 Q1 (20) The probability of an impossible event is:

(A) 0

(B) 1

(C) 1/2

(D) all of these

Detailed Step-by-Step Solution:

• Step 1: Analyze the question: The probability of an impossible event is:

• Step 2: The probability of an impossible event is strictly 0.

• Correct Answer: [A] 0

Review Exercise 11 Q2 Mean, Median, Mode of Discrete Table

Review Exercise 11 Q2 For the following data, find mean, mode and median:

$x$	5	10	15	20	25	30
Frequency ($f$)	4	6	10	8	5	2

Detailed Step-by-Step Solution:

• Combined Calculation Table:

$x$	$f$	$fx$	$cf$
5	4	20	4
10	6	60	10
15	10 (Peak $f$)	150	20 (Contains $\frac{N+1}{2} = 18\text{th}$)
20	8	160	28
25	5	125	33
30	2	60	35
Total	$\sum f = 35$	$\sum fx = 575$	—

• 1. Mean: $\bar{x} = \frac{575}{35} \approx \mathbf{16.43}$

• 2. Median: $\frac{35+1}{2} = 18\text{th value} \implies \mathbf{15}$

• 3. Mode: Value with maximum frequency ($10$) is $\mathbf{15}$.

Review Exercise 11 Q3 Grouped Median Calculation

Review Exercise 11 Q3 Find the median for the following frequency distribution:

Marks	30 – 39	40 – 49	50 – 59	60 – 69	70 – 79
No. of Students ($f$)	6	14	20	12	8

Detailed Step-by-Step Solution:

• Grouped Median Work-Table:

Marks	Class Boundaries	Frequency ($f$)	Cumulative Frequency ($cf$)
30 – 39	29.5 – 39.5	6	6
40 – 49	39.5 – 49.5	14	20 ($c = 20$)
50 – 59	49.5 – 59.5	20	40 (🎯 Median Class, $\frac{n}{2} = 30$)
60 – 69	59.5 – 69.5	12	52
70 – 79	69.5 – 79.5	8	60
Total	—	$n = 60$	—

• Calculation:

$$\tilde{x} = l + \frac{h}{f}\left(\frac{n}{2} - c\right) = 49.5 + \frac{10}{20}(30 - 20) = 49.5 + 5.0 = \mathbf{54.5}$$

Review Exercise 11 Q4 Missing Value from Arithmetic Mean

Review Exercise 11 Q4 The arithmetic mean of 10 values is 35.5. If nine values are 20, 23, 37, 48, 29, 33, 45, 40, 45, find the tenth value.

Detailed Step-by-Step Solution:

• Step 1 (Total Sum of 10 values):

$$\sum x = n \times \bar{x} = 10 \times 35.5 = 355$$

• Step 2 (Sum of given 9 values):

$$\text{Sum of 9 values} = 20 + 23 + 37 + 48 + 29 + 33 + 45 + 40 + 45 = 320$$

• Step 3 (Find 10th value):

$$\text{Tenth value } x_{10} = 355 - 320 = \mathbf{35}$$

Review Exercise 11 Q5 Comprehensive Grouped Mean, Median & Mode

Review Exercise 11 Q5 Find mean, median and mode for the following frequency distribution:

Class Limits	20 – 24	25 – 29	30 – 34	35 – 39	40 – 44
Frequency ($f$)	4	8	14	10	4

Detailed Step-by-Step Solution:

• Master Statistical Calculation Table:

Class	Boundaries	Mid ($x$)	Freq ($f$)	$fx$	$cf$
20 – 24	19.5 – 24.5	22	4	88	4
25 – 29	24.5 – 29.5	27	8 ($f_1$)	216	12 ($c$)
30 – 34	29.5 – 34.5	32	14 ($f_m$)	448	26 (Median & Modal Class)
35 – 39	34.5 – 39.5	37	10 ($f_2$)	370	36
40 – 44	39.5 – 44.5	42	4	168	40
Totals	—	—	$\sum f = 40$	$\sum fx = 1290$	—

• 1. Mean: $\bar{x} = \frac{1290}{40} = \mathbf{32.25}$

• 2. Median: $\tilde{x} = 29.5 + \frac{5}{14}(20 - 12) = 29.5 + \frac{40}{14} = 29.5 + 2.86 = \mathbf{32.36}$

• 3. Mode: $\text{Mode} = 29.5 + \frac{14 - 8}{2(14) - 8 - 10} \times 5 = 29.5 + \frac{6}{10} \times 5 = 29.5 + 3.0 = \mathbf{32.50}$

Review Exercise 11 Q6 Probability Word Problem (Dart Game)

Review Exercise 11 Q6 Affan is learning to play dart. He plays it 20 times each day and probability of hitting bull's eye is 1/10. Find:

(a) expected number of hits in 25 days.

(b) expected number of misses in 25 days.

Detailed Step-by-Step Solution:

• Total dart throws in 25 days: $n = 25 \times 20 = 500$ throws.

• (a) Expected Hits:

$$P(\text{Hit}) = \frac{1}{10} \implies \text{Expected Hits} = 500 \times \frac{1}{10} = \mathbf{50}$$

• (b) Expected Misses:

$$P(\text{Miss}) = 1 - \frac{1}{10} = \frac{9}{10} \implies \text{Expected Misses} = 500 \times \frac{9}{10} = \mathbf{450}$$

Review Exercise 11 Q7 Expected Value Word Problem (School Attendance)

Review Exercise 11 Q7 Shifaa usually gets late for her school 2 days out of 6 working days of a week. What is the expected no. of days of her late arrival in 4 weeks?

Detailed Step-by-Step Solution:

• Step 1 (Probability of getting late): $P = \frac{2}{6} = \frac{1}{3}$.

• Step 2 (Total working days in 4 weeks): $n = 4 \times 6 = 24$ days.

• Step 3 (Expected late days):

$$\text{Expected Late Days} = n \times P = 24 \times \frac{1}{3} = \mathbf{8\text{ days}}$$

Review Exercise 11 Q8 Expected Value Word Problem (Math Assessments)

Review Exercise 11 Q8 As'ha tries her best to score 100% in each maths assessment, but the probability of getting this is 0.8. Find the expected number of assessments in which she will score 100% out of total 20 assessments.

Detailed Step-by-Step Solution:

• Step 1: Total assessments $n = 20$.

• Step 2: Probability of scoring 100% $p = 0.8$.

• Step 3: $\text{Expected Assessments} = n \times p = 20 \times 0.8 = \mathbf{16\text{ assessments}}$.

Review Exercise 11 Q9 Probability in Health Diagnostics (Covid-19 Reports)

Review Exercise 11 Q9 In a lab shelf there are 50 reports of COVID-19. Among these 13 are females and 2 of the females are COVID positive. If a report is picked at random, find:

(a) The probability of getting a female report

(b) The probability of getting a male report

Detailed Step-by-Step Solution:

• Total reports $n(S) = 50$.

• (a) Female Report: $n(\text{Female}) = 13 \implies P = \frac{13}{50} = \mathbf{0.26}$.

• (b) Male Report: $n(\text{Male}) = 50 - 13 = 37 \implies P = \frac{37}{50} = \mathbf{0.74}$.

• (c) Female COVID Negative Report: Total females $= 13$, positive $= 2 \implies$ negative females $= 13 - 2 = 11$. $P = \frac{11}{50} = \mathbf{0.22}$.

Extra Objective & Concept Boosters

Extra Exercise Q1 Empirical Relation Calculation

Extra Exercise Q1: For a moderately skewed distribution, if the Mean is 24 and the Median is 20, what is the value of the Mode according to Pearson's empirical relation?

(A) 12

(B) 16

(C) 18

(D) 22

Detailed Step-by-Step Solution:

• Step 1: Empirical formula: $\text{Mode} = 3\,\text{Median} - 2\,\text{Mean}$.

• Step 2: Substitute $\text{Median} = 20$ and $\text{Mean} = 24$:

$$\text{Mode} = 3(20) - 2(24) = 60 - 48 = \mathbf{12}$$

• Correct Answer: [A] 12

Extra Exercise Q2 Cumulative Frequency Properties

Extra Exercise Q2: The cumulative frequency of the last class in a grouped frequency distribution is always equal to:

(A) The class width $h$

(B) The total number of observations $\sum f$

(C) The arithmetic mean

(D) The frequency of the first class

Detailed Step-by-Step Solution:

• Step 1: Cumulative frequency accumulates all class frequencies consecutively from first to last.

• Step 2: Therefore, the cumulative frequency of the final class is exactly equal to the sum of all frequencies $\sum f = n$.

• Correct Answer: [B] The total number of observations $\sum f$

Extra Exercise Q3 Histogram Frequency Density Principle

Extra Exercise Q3: When constructing a histogram with unequal class widths, the height of each rectangle must be proportional to:

(A) Class frequency

(B) Class width

(C) Frequency Density

(D) Cumulative frequency

Detailed Step-by-Step Solution:

• Step 1: For unequal class widths, $\text{Area} \propto \text{Frequency} = \text{Height} \times \text{Width}$.

• Step 2: Thus $\text{Height} \propto \frac{\text{Frequency}}{\text{Width}} = \text{Frequency Density}$.

• Correct Answer: [C] Frequency Density

Extra Exercise Q4 Complementary Probability Rule

Extra Exercise Q4: If the probability of an event happening is $P(E) = 0.35$, what is the probability of the complementary event $P(E')$?

(A) 0.35

(B) 0.65

(C) -0.35

(D) 1.35

Detailed Step-by-Step Solution:

• Step 1: Complementary probability rule: $P(E') = 1 - P(E)$.

• Step 2: $P(E') = 1 - 0.35 = \mathbf{0.65}$.

• Correct Answer: [B] 0.65

Extra Exercise Q5 Theoretical Probability Axioms

Extra Exercise Q5: Which of the following numbers CANNOT represent the probability of an event?

(A) 0.001

(B) 75%

(C) 5/4

(D) 2/3

Detailed Step-by-Step Solution:

• Step 1: The probability of any event must satisfy $0 \le P(E) \le 1$.

• Step 2: $5/4 = 1.25 > 1$, which violates the fundamental axiom of probability.

• Correct Answer: [C] 5/4

Extra Exercise Q6 Symmetrical Distribution Properties

Extra Exercise Q6: In a perfectly symmetrical bell-shaped unimodal distribution, the relationship between Mean, Median, and Mode is:

(A) Mean > Median > Mode

(B) Mean < Median < Mode

(C) Mean = Median = Mode

(D) Mode = 3 Median + 2 Mean

Detailed Step-by-Step Solution:

• Step 1: In a symmetrical distribution, the central balance point, the 50th percentile, and the peak frequency coincide at the same central coordinate.

• Step 2: Hence, $\text{Mean} = \text{Median} = \text{Mode}$.

• Correct Answer: [C] Mean = Median = Mode

Extra Exercise Q7 Statistical Concepts True or False

Extra Exercise Q7: State True or False with brief mathematical reasoning:

(i) The sum of deviations of values from their arithmetic mean is always zero ($\sum (x - \bar{x}) = 0$).

(ii) The median is heavily affected by extreme outlier values in a dataset.

(iii) The probability of any event can be a negative fraction if the sample space is small.

(iv) In a frequency polygon, the area under the polygon is equal to the total area of the corresponding histogram.

Detailed Step-by-Step Solution:

• (i) True: By definition of mean, $\sum (x - \bar{x}) = \sum x - n\bar{x} = n\bar{x} - n\bar{x} = 0$.

• (ii) False: The median is a positional measure determined by the middle rank, making it highly robust against extreme outliers.

• (iii) False: By probability axioms, $P(E) \ge 0$ for all events; probabilities cannot be negative.

• (iv) True: The triangular areas added outside the histogram bars exactly balance the triangular areas excluded from the bars.

Extra Exercise Q8 Statistical Terminology Blanks

Extra Exercise Q8: Fill in the blanks:

(i) The ratio of class frequency to class width is called ______________.

(ii) The class mark of the interval $a - b$ is calculated by the formula ______________.

(iii) The sum of probabilities of all elementary outcomes in a sample space equals ______________.

(iv) When different data values carry varying levels of importance, we compute the ______________ mean.

Detailed Step-by-Step Solution:

• (i) Frequency Density

• (ii) $\mathbf{\frac{a + b}{2}}$

• (iii) 1 (or Unity)

• (iv) Weighted

Extra Exercise Q9 Statistical Measures Column Matching

Extra Exercise Q9: Match each statistical term in Column A with its corresponding mathematical definition in Column B:

Column A (Statistical Measure)	Column B (Mathematical Formula / Meaning)
(1) Frequency Density	(A) Positional value dividing data into two equal halves
(2) Class Mark (Midpoint)	(B) $\frac{\text{Class Frequency}}{\text{Class Width}}$
(3) Median	(C) $3\,\text{Median} - 2\,\text{Mean}$
(4) Empirical Mode	(D) $\frac{\text{Lower Limit} + \text{Upper Limit}}{2}$
(5) Complementary Probability	(E) $1 - P(E)$

Detailed Step-by-Step Solution:

• Matching Solution Table:

Column A	Match	Column B Definition
(1) Frequency Density	(B)	$\frac{\text{Class Frequency}}{\text{Class Width}}$
(2) Class Mark (Midpoint)	(D)	$\frac{\text{Lower Limit} + \text{Upper Limit}}{2}$
(3) Median	(A)	Positional value dividing data into two equal halves
(4) Empirical Mode	(C)	$3\,\text{Median} - 2\,\text{Mean}$
(5) Complementary Probability	(E)	$1 - P(E)$

🎯 7. Unit Synthesis & Analytical Summary

Chapter 11 synthesizes descriptive data analytics and mathematical probability. Raw ungrouped measurements are converted into structured grouped distributions with exact continuous boundaries. Central tendencies provide three complementary lenses: the arithmetic mean calculates algebraic balance, the median locates the outlier-resistant 50th-percentile center, and the mode pinpoints peak frequency concentrations. Combined with frequency density histograms, frequency polygons, dispersion measures, and probability matrices, students possess the complete mathematical toolkit for scientific reasoning and higher statistics.