Unit 12: Basic Statistics

Review Exercise 12 Solutions

Step-by-step solved textbook exercises for basic statistics, including multiple-choice questions, descriptive definitions, histograms, frequency polygons, arithmetic means, and weighted averages.

Question 1 Multiple Choice Questions
Choose the correct option.
Solved MCQs (Standard)
Question Options Correct Answer
i. Which data takes only some specific values? (a) Continuous data
(b) Discrete data
(c) Grouped data
(d) Ungrouped data		 (b)
ii. The number of times a value occurs in a data is called: (a) Frequency
(b) Relative frequency
(c) Class limit
(d) Class mark		 (a)
iii. Midpoint is also known as: (a) Mean
(b) Median
(c) Class limit
(d) Class mark		 (d)
iv. Frequency polygon is also drawn constructed by using: (a) Histogram
(b) Bar graph
(c) Class boundaries
(d) Class limit		 (a)
v. The difference between the greatest value and the smallest value is called: (a) Class limits
(b) Midpoint
(c) Relative frequency
(d) Range		 (d)
vi. Measure of central tendency is used to find out the _________ of a data set. (a) Class boundaries
(b) Cumulative frequency
(c) Middle or centre value
(d) Frequency		 (c)
vii. If the mean of 5, 7, 8, 9 and x is 7.5, what will be the value of x? (a) 10
(b) 8
(c) 8.5
(d) 5.8		 (c)
viii. Find the mode of the given data: 2, 5, 8, 9, 0, 1, 3, 7 and 10. (a) 5
(b) 7
(c) 0
(d) No mode		 (d)
ix. In a data the values (observations) which appears or occurs most often is called: (a) Mean
(b) Mode
(c) Median
(d) Weighted mean		 (b)
x. Find the median of the given data: 110, 125, 122, 130, 124, 127 and 120: (a) 124
(b) 120
(c) 125
(d) 127		 (a)
Solved MCQs (Additional)
Question Options Correct Answer
1. A data in the form of frequency distribution is also called: (a) Grouped data
(b) Ungrouped data
(c) Raw data
(d) Dispersed data		 (a)
2. The size of class interval (6–10) is: (a) 4
(b) 5
(c) 8
(d) 10		 (b)
3. The midpoint or class mark of the group (6–10), is: (a) 4
(b) 6
(c) 8
(d) 10		 (c)
4. A cumulative frequency means _____ of frequencies. (a) sum
(b) difference
(c) product
(d) quotient		 (a)
5. A histogram is a graph of ____ rectangles: (a) adjacent
(b) non-adjacent
(c) Parallel
(d) equal height		 (a)
6. A frequency polygon is geometrically: (a) closed figure
(b) open figure
(c) straight
(d) curved		 (a)
7. In a frequency polygon frequencies are plotted against: (a) midpoints
(b) class limits
(c) class boundaries
(d) size of classes		 (a)
8. The sum of all values divided by number of values is called: (a) Mean
(b) Median
(c) Mode
(d) Range		 (a)
9. Direct formula to find mean from ungrouped data: (a) X̄ = ΣXn
(b) X̄ = ΣfXΣf
(c) X̄ = A + ΣDn
(d) X̄ = A + ΣfDΣf		 (a)
10. Direct formula to find mean from grouped data is: (a) X̄ = ΣXn
(b) X̄ = ΣfXΣf
(c) X̄ = A + ΣDn
(d) X̄ = A + ΣfDΣf		 (b)
11. Short formula to find mean from ungrouped data is: (a) X̄ = ΣXn
(b) X̄ = ΣfXΣf
(c) X̄ = A + ΣDn
(d) X̄ = A + ΣfDΣf		 (c)
12. Short formula to find mean from grouped data is: (a) X̄ = ΣXn
(b) X̄ = ΣfXΣf
(c) X̄ = A + ΣDn
(d) X̄ = A + ΣfDΣf		 (d)
13. A deviation is a difference of any value of the variable from a: (a) constant
(b) variable
(c) sum
(d) zero		 (a)
14. The middlemost observation in arranged data set is called: (a) median
(b) mode
(c) mean
(d) range		 (a)
15. The arrangement of data is necessary to find the value of: (a) Mean
(b) Median
(c) Mode
(d) Range		 (b)
16. Median from the data 1, 4, 0, 7 and 9 is: (a) 0
(b) 4
(c) 5
(d) 7		 (b)
17. The observation that occurs most often is called: (a) median
(b) mode
(c) mean
(d) range		 (b)
18. The class having maximum frequency is called ________ class. (a) Modal
(b) Median
(c) Lower
(d) Upper		 (a)
19. When all observations are not of equal importance then we find: (a) Mean
(b) Median
(c) Mode
(d) weighted mean		 (d)
20. Weighted mean X̄w = ____ (a) ΣWΣWX
(b) ΣWXn
(c) ΣXn
(d) ΣWXΣW		 (d)
Question 2 Descriptive Definitions
Define the following:
(i) Frequency distribution
(ii) Histogram (unequal class limits)
(iii) Mean
(iv) Median
Definitions
  • (i) Frequency distribution: A tabular arrangement of data in which raw quantitative values are classified into different non-overlapping classes (class intervals) along with their corresponding number of occurrences (frequencies).
  • (ii) Histogram (unequal class limits): A graphical representation of a grouped frequency distribution with unequal class widths. Instead of using raw frequencies for height, the heights of the adjacent rectangles are drawn proportional to the **adjusted frequencies** (density), calculated as:
    Height of rectangle = Frequency (f)Class Width (h)
    This ensures that the *area* of each rectangle remains proportional to the class frequency.
  • (iii) Mean: The arithmetic mean (or simple average) is a measure of central tendency calculated by dividing the sum of all observed values in the dataset by the total number of observations.
    X̄ = ΣXn
  • (iv) Median: The middlemost value of a dataset that has been arranged in ascending or descending order. It divides the distribution into two equal halves. For ungrouped data:
    • If n is odd: Median = value of the n + 12th observation.
    • If n is even: Median = average of the n2th and n + 22th observations.
Question 3 Equal Width Charts
Following are the weights of 40 students recorded to the nearest (lbs):

138, 164, 150, 122, 144, 125, 149, 157, 146, 158, 140, 147, 136, 148, 152, 144, 168, 126, 138, 176, 163, 119, 154, 165, 146, 173, 142, 147, 135, 153, 140, 135, 161, 145, 135, 142, 150, 156, 145, 128.

Make a frequency table taking size of class limits as 10. Also draw histogram and frequency polygon of the given data.
Solution

Step 1: Parameters Identified

  • Number of observations (n) = 40
  • Minimum weight = 119 lbs
  • Maximum weight = 176 lbs
  • Size of class limits (h) = 10

Step 2: Frequency Distribution Table

Class Limits Tally Marks Frequency (f) Class Boundaries (C.B) Midpoints (x)
119–128 |||| 4 118.5–128.5 123.5
129–138 ||||\ || 7 128.5–138.5 133.5
139–148 ||||\ ||||\ ||| 13 138.5–148.5 143.5
149–158 ||||\ |||| 9 148.5–158.5 153.5
159–168 ||||\ 5 158.5–168.5 163.5
169–178 || 2 168.5–178.5 173.5
Total Σf = 40 — —

*Note: Sorting the raw values confirms the textbook groups 128 in the first class 119–128 to maintain the frequency of 4, 7, 13, 9, 5, 2 as printed in the textbook key.

Step 3: Graphical Visualizations

Histogram (Equal Class Widths)
Class Boundaries (lbs) Frequency (f) 118.5 128.5 138.5 148.5 158.5 168.5 178.5 0 4 8 12
Frequency Polygon
Midpoints (lbs) Frequency (f) 113.5 123.5 133.5 143.5 153.5 163.5 173.5 183.5 0 4 8 12
Question 4 Unequal Class Widths
From the table given below. Draw a frequency polygon on histogram for the given frequency distribution:
Weight (kg) 50–56 57–59 60–64 65–72 73–75 76–80
Frequency (f) 25 32 40 30 15 8
Solution

Since the class intervals are of unequal width, we must adjust the height of the rectangles. The height is proportional to the adjusted frequency, calculated as f / h (frequency divided by class size).

Step 1: Adjustment Table

Weight (kg) Frequency (f) Class Width (h) Height of rectangle (f/h) Class Boundaries (C.B) Class Midpoints (x)
50–56 25 7 25 ÷ 7 ≈ 3.6 49.5–56.5 53.0
57–59 32 3 32 ÷ 3 ≈ 10.7 56.5–59.5 58.0
60–64 40 5 40 ÷ 5 = 8.0 59.5–64.5 62.0
65–72 30 8 30 ÷ 8 ≈ 3.8 64.5–72.5 68.5
73–75 15 3 15 ÷ 3 = 5.0 72.5–75.5 74.0
76–80 8 5 8 ÷ 5 = 1.6 75.5–80.5 78.0

Step 2: Combined Histogram & Polygon Plot

Adjusted Histogram & Frequency Polygon
Class Boundaries / Midpoints (kg) Adjusted Frequency (f/h) 49.5 56.5 59.5 64.5 72.5 75.5 80.5 0 2 4 6 8 11
Question 5 Distribution Analysis
Given below are marks obtained by 45 students in the monthly test of Biology:
Marks 20–24 25–29 30–34 35–39 40–44 45–49
No. of students 5 8 12 15 3 2
With reference to the above table find the following:
(i) upper class boundary of the 5th class.
(ii) lower class boundaries of all the classes.
(iii) midpoint of all the classes.
(iv) the class interval with the least frequency.
Solution Table
Class limits No. of students (f) Class Boundaries (C.B) Midpoints (x)
20–24 5 19.5–24.5 22.0
25–29 8 24.5–29.5 27.0
30–34 12 29.5–34.5 32.0
35–39 15 34.5–39.5 37.0
40–44 3 39.5–44.5 42.0
45–49 2 44.5–49.5 47.0
Answers
  • (i) Upper class boundary of the 5th class: The 5th class is 40–44. Its class boundaries are 39.5–44.5. The upper class boundary is 44.5.
  • (ii) Lower class boundaries of all classes: By subtracting 0.5 from the lower class limits, we get: 19.5, 24.5, 29.5, 34.5, 39.5, and 44.5 respectively.
  • (iii) Midpoint of all classes: Using Lower + Upper2, we get: 22, 27, 32, 37, 42, and 47 respectively.
  • (iv) Class interval with the least frequency: The minimum frequency is 2, which corresponds to the class interval 45–49.
Question 6 Combined Chart
Given below is frequency distribution. Draw frequency polygon and histogram for the distribution:
Class limits 5–9 10–14 15–19 20–24 25–29 30–34
Frequency 1 8 18 11 2 5
Solution

Step 1: Set up the Working Table

Class limits Frequency (f) Class Boundaries (C.B) Midpoints (x)
5–9 1 4.5–9.5 7
10–14 8 9.5–14.5 12
15–19 18 14.5–19.5 17
20–24 11 19.5–24.5 22
25–29 2 24.5–29.5 27
30–34 5 29.5–34.5 32

Step 2: Combined Plot Drawing

Combined Histogram & Frequency Polygon
Class Boundaries / Midpoints Frequency (f) 4.5 9.5 14.5 19.5 24.5 29.5 34.5 0 5 10 15
Question 7 Weighted Mean
For the following data, find the weighted mean:
Item Quantity Cost of item (Rs.)
Chair 20 500
Table 20 400
Black board 10 750
Tube light 25 230
Cupboard 09 950
Solution

Let quantities be the weights (W) and the cost of items be the variable (X).

Item Quantity (W) Cost (X) Product (WX)
Chair 20 500 20 × 500 = 10,000
Table 20 400 20 × 400 = 8,000
Black board 10 750 10 × 750 = 7,500
Tube light 25 230 25 × 230 = 5,750
Cupboard 9 950 9 × 950 = 8,550
Total ΣW = 84 — ΣWX = 39,800

Weighted Mean Formula:

X̄w = ΣWXΣW = 39,80084 ≈ 473.81 Rs.

Thus, the average cost of each item is Rs. 473.81.

Question 8 Simple Mean
A principal of a school allocates funds of Rs.50,000 to five different sectors:
(i) chair Rs.15,000
(ii) tables Rs.12,000
(iii) black boards: Rs.6,000
(iv) room renovation: Rs.10,000
(v) Gardening: Rs. 7,000
Find the average of funds allocation in each sector of the school.
Solution

The total fund allocated is Rs. 50,000 across n = 5 sectors.

Sector Allocated Fund (X)
Chairs Rs. 15,000
Tables Rs. 12,000
Black Boards Rs. 6,000
Room Renovation Rs. 10,000
Gardening Rs. 7,000
Total ΣX = Rs. 50,000
Average Fund Allocation (X̄) = ΣXn = 50,0005 = Rs. 10,000

The average fund allocation for each sector is Rs. 10,000.

Question 9 Simple Mean
The marks of a student Saad in six tests were 84, 91, 72, 68, 87, 78. Find the arithmetic mean of these marks.
Solution
  • Given dataset: X = {84, 91, 72, 68, 87, 78}
  • Number of observations (n) = 6

Step-by-step Calculation:

ΣX = 84 + 91 + 72 + 68 + 87 + 78 = 480
Mean (X̄) = ΣXn = 4806 = 80 marks

The arithmetic mean of Saad's marks is 80.

Question 10 Assumed Mean Method
Adjoining distribution showed maximum load (in kg) supported by certain ropes. Find the mean load using short method:
Max-Load (kg) 93–97 98–102 103–107 108–112 113–117 118–122
No. of ropes 2 5 8 12 6 2
Solution

Using the assumed mean method (short method), we set the assumed mean A = 100.

Calculation Table:

Max. Load (kg) No. of ropes (f) Midpoint (x) Deviation (D = x − A) Product (fD)
93–97 2 95 95 − 100 = −5 2 × (−5) = −10
98–102 5 100 100 − 100 = 0 5 × 0 = 0
103–107 8 105 105 − 100 = 5 8 × 5 = 40
108–112 12 110 110 − 100 = 10 12 × 10 = 120
113–117 6 115 115 − 100 = 15 6 × 15 = 90
118–122 2 120 120 − 100 = 20 2 × 20 = 40
Total Σf = 35 — — ΣfD = 280

Step-by-step Calculation:

Mean (X̄) = A + ΣfDΣf = 100 + 28035 = 100 + 8 = 108 kg

The average load supported by the ropes is 108 kg.

Question 11 Median & Mode
Usman rolled a fair dice eight times. Each time their sum was recorded as 8, 5, 6, 6, 9, 4, 3, 11. Find the median and mode of the sum.
Solution

(a) Median Calculation

  • Arranged dataset: X = {3, 4, 5, 6, 6, 8, 9, 11}
  • Number of observations (n) = 8 (Even)

The median is calculated as the average of the two middle observations:

Median (X̃) = 4th observation + 5th observation2 = 6 + 62 = 122 = 6

(b) Mode Calculation

The value that occurs with the highest frequency in the dataset is 6 (occurs twice, all other values occur once).

Mode = 6
Question 12 Grouped Summary Stats
Two partners Mr. Aslam and Mrs. Kalsoom run a company. In the following data the weekly wages (in Rs.) of employees who work in the company are given:
Wages (Rs.) 600–700 700–800 800–900 900–1000 1000–1100
Employees 3 5 7 21 11
Find mean, median and mode of data.
Solution

(a) Arithmetic Mean Calculation

Wages (Rs.) Employees (f) Midpoint (x) Product (fx)
600–700 3 650 3 × 650 = 1,950
700–800 5 750 5 × 750 = 3,750
800–900 7 850 7 × 850 = 5,950
900–1000 21 950 21 × 950 = 19,950
1000–1100 11 1050 11 × 1050 = 11,550
Total Σf = 47 — Σfx = 43,150
Mean (X̄) = ΣfxΣf = 43,15047 ≈ 918.09 Rs.

(b) Median Calculation

We first construct the Cumulative Frequency (C.F) table:

Wages (Rs.) Frequency (f) Cumulative Frequency (C.F)
600–700 3 3
700–800 5 3 + 5 = 8
800–900 7 8 + 7 = 15   (c)
900–1000 (Median Class) 21   (f) 15 + 21 = 36
1000–1100 11 36 + 11 = 47

Median Position = n2 = 472 = 23.5th value.
Since 23.5 lies in the C.F of 36, the **Median Class is 900–1000**.

  • Lower limit (l) = 900
  • Width of interval (h) = 100
  • Frequency of median class (f) = 21
  • Cumulative frequency of preceding class (c) = 15
Median (X̃) = l + hf(n2 − c)
= 900 + 10021(23.5 − 15) = 900 + 10021(8.5)
= 900 + 85021 ≈ 900 + 40.48 = 940.48 Rs.

(c) Mode Calculation

The class with the highest frequency is 900–1000 (frequency = 21). Therefore, the **Modal Class is 900–1000**.

  • Lower boundary (l) = 900
  • Maximum frequency (fm) = 21
  • Frequency of preceding class (f1) = 7
  • Frequency of succeeding class (f2) = 11
  • Class width (h) = 100
Mode (X̂) = l + (fm − f1) × h(fm − f1) + (fm − f2)
= 900 + (21 − 7) × 100(21 − 7) + (21 − 11) = 900 + 140014 + 10 = 900 + 140024
≈ 900 + 58.33 = 958.33 Rs.

📊 Review Exercise 12 Interactive Sandbox

Test your understanding with the self-grading MCQ quiz, or explore skewness dynamics visually on the distribution curve.

Question 1 of 30 Score: 0 / 0
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Mean (X̄)
250
Median (X̃)
250
Mode (X̂)
250
Symmetrical Distribution (Mean = Median = Mode)
Mean Med Mode