Unit 12: Basic Statistics

Exercise 12.1 Solutions

Step-by-step solved textbook exercises for frequency distribution, tally marks, histograms, and frequency polygons (including equal and unequal class widths).

Question 1 Frequency Table Analysis
The following distribution represents the scores achieved by a group of chemistry students in the chemistry laboratory:
Scores 24–28 29–33 34–38 39–43 44–48 49–53 Total
No. of students 3 6 12 23 15 6 65
Answer the following sub-questions:
  • (i) What is the upper limit of the last class?
  • (ii) What is the lower limit of the class 39–43?
  • (iii) What is the midpoint of the class (34–38)?
  • (iv) What are the class frequencies of the classes 29–33 and 44–48?
  • (v) What is the size of the class limits in the above frequency distribution?
  • (vi) In which class or groups does the minimum number of students fall?
  • (vii) What is the lower limit of the class having 15 as its class frequency?
  • (viii) What is the number of students having scores between 24 and 43?
Solution
  • (i) Upper limit of the last class: The last class interval is 49–53. Its upper limit is 53.
  • (ii) Lower limit of class 39–43: The lower limit of this class interval is 39.
  • (iii) Midpoint of the class (34–38): Midpoint = 34 + 382 = 722 = 36.
  • (iv) Class frequencies of 29–33 and 44–48: The frequency of 29–33 is 6 and the frequency of 44–48 is 15.
  • (v) Size of the class limits: The size of the class interval (difference between successive lower limits, e.g., 29 − 24) is 5.
  • (vi) Class with the minimum number of students: The class with the lowest frequency is 24–28 (contains only 3 students).
  • (vii) Lower limit of the class with frequency 15: The class having frequency 15 is 44–48. Its lower limit is 44.
  • (viii) Number of students having scores between 24 and 43: Sum of frequencies from class 24–28 up to class 39–43: Sum = 3 + 6 + 12 + 23 = 44 students.
Question 2 Tally Bar Method
For a school staff, the following expenditures (rupees in hundred) are required for the repair of chairs:

145, 152, 153, 156, 158, 160, 146, 152, 155, 159, 161, 163, 165, 147, 148, 151, 154, 156, 158, 160, 144, 167, 151, 150, 152, 149, 145, 153, 152, 155.

Prepare a frequency distribution by tally bar method using 3 as the size of class limits and also write down what are the frequencies of the last three classes?
Solution

Parameters identified:

  • Number of observations (n) = 30
  • Minimum value = 144
  • Maximum value = 167
  • Class size (h) = 3
Class Limits Tally Marks Frequency (f)
144–146 |||| 4
147–149 ||| 3
150–152 ||||\ || 7
153–155 ||||\ 5
156–158 |||| 4
159–161 |||| 4
162–164 | 1
165–167 || 2
Total Σf = 30

Frequencies of the last three classes: The last three classes are 159–161, 162–164, and 165–167. Their frequencies are 4, 1, and 2 respectively.

Question 3 Frequency Polygon
Given below are the weights in kg of 30 students of a high school:

30, 33, 24, 21, 15, 39, 37, 44, 42, 33, 33, 28, 29, 32, 31, 28, 26, 32, 34, 35, 38, 36, 41, 30, 35, 41, 23, 26, 18, 34.

Taking 5 as the size of the class limit, prepare a frequency table and construct a frequency polygon.
Solution

Step 1: Set up the Frequency Table (Class Interval = 5)

Class Limits Tally Marks Frequency (f) Midpoint (x)
15–19 || 2 17
20–24 ||| 3 22
25–29 ||||\ 5 27
30–34 ||||\ ||||\ 10 32
35–39 ||||\ | 6 37
40–44 |||| 4 42
Total Σf = 30 —

Step 2: Prepare Coordinates for Frequency Polygon

To close the polygon at both ends, we add two dummy classes with zero frequency: one before (10–14, Midpoint = 12) and one after (45–49, Midpoint = 47).

Midpoints (Weights in kg) Frequency (No. of students) 0 2 4 6 8 10 12 17 22 27 32 37 42 47 (17, 2) (22, 3) (27, 5) (32, 10) (37, 6) (42, 4)
Question 4 Histogram & Boundaries
A group of Grade-10 students obtained the following marks out of 100 marks in an English test:

58, 59, 58, 33, 40, 58, 45, 46, 43, 45, 45, 50, 52, 49, 50, 57, 52, 55, 49, 50, 62, 49, 48, 44, 42, 47, 46, 47, 46, 53, 40, 44.

Classify the data into a frequency distribution by (direct method) taking 6 as the size of class limit. Also find the class limit with least class frequency and construct a histogram for the data.
Solution

Step 1: Set up the Frequency Distribution Table

Number of observations (n) = 32. Smallest value = 33, Largest = 62. Class interval size = 6.

Class Limits Class Boundaries Values Included Frequency (f)
33–38 32.5–38.5 33 1
39–44 38.5–44.5 40, 40, 42, 43, 44, 44 6
45–50 44.5–50.5 45, 45, 45, 46, 46, 46, 47, 47, 48, 49, 49, 49, 50, 50, 50 15
51–56 50.5–56.5 52, 52, 53, 55 4
57–62 56.5–62.5 57, 58, 58, 58, 59, 62 6
Total Σf = 32

Least Frequency Class Limit: The class interval with the minimum frequency is 33–38 (frequency = 1).

Step 2: Construct the Histogram

Class boundaries are plotted on the horizontal axis and frequencies on the vertical axis. Since class widths are equal (width = 6), the heights of rectangles are proportional to the frequencies.

Class Boundaries (English test marks) Frequency (No. of students) 0 3 6 9 12 15 32.5 38.5 44.5 50.5 56.5 62.5 1 6 15 4 6
Question 5 Combined Graph
From the table given below. Draw a frequency polygon on a histogram for the given frequency distribution:
Weight (kg) 10–14 15–19 20–24 25–29 30–34 35–39
Frequency (f) 06 17 23 30 22 13
Solution

To plot a combined frequency polygon on top of a histogram, we calculate the class boundaries for the histogram and midpoints for the polygon.

Class Limits Class Boundaries Frequency (f) Midpoint (x)
10–14 9.5–14.5 6 12
15–19 14.5–19.5 17 17
20–24 19.5–24.5 23 22
25–29 24.5–29.5 30 27
30–34 29.5–34.5 22 32
35–39 34.5–39.5 13 37

Dummy classes for polygon: We anchor the polygon at midpoints 7 (before) and 42 (after) with zero frequency.

Class Boundaries & Midpoints (kg) Frequency 0 5 10 15 20 25 30 9.5 14.5 19.5 24.5 29.5 34.5 39.5
Question 6 Discrete Distribution
The following data shows the number of heads in an experiment of 50 sets of tossing a coin 5 times. Make a discrete frequency distribution from the information:

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

Solution

Since the variable "Number of Heads" takes only specific integer values (0, 1, 2, 3, 4, 5), this is a discrete frequency distribution. We count the occurrences directly.

No. of Heads (x) Tally Marks Frequency (f)
0 ||||\ 5
1 ||||\ || 7
2 ||||\ |||| 9
3 ||||\ ||||\ |||| 14
4 ||||\ |||| 9
5 ||||\ | 6
Total Σf = 50
Question 7 Unequal Class Widths
The marks obtained by the students of Grade-10 in a mathematics test were grouped into the following frequency distribution. Draw a histogram for the distribution:
Marks 35–37 38–44 45–54 55–61 62–67 68–72
Frequency (f) 2 12 16 13 9 3
Solution

Since the class intervals have **unequal widths**, we cannot plot raw frequencies on the vertical axis. Instead, we must plot the **adjusted frequency** (height of rectangle = fh), representing frequency density.

Class Limits Class Boundaries Frequency (f) Class Width (h) Adjusted Height (fh)
35–37 34.5–37.5 2 37.5 − 34.5 = 3 2 ÷ 3 = 0.67
38–44 37.5–44.5 12 44.5 − 37.5 = 7 12 ÷ 7 = 1.71
45–54 44.5–54.5 16 54.5 − 44.5 = 10 16 ÷ 10 = 1.60
55–61 54.5–61.5 13 61.5 − 54.5 = 7 13 ÷ 7 = 1.86
62–67 61.5–67.5 9 67.5 − 61.5 = 6 9 ÷ 6 = 1.50
68–72 67.5–72.5 3 72.5 − 67.5 = 5 3 ÷ 5 = 0.60
Class Boundaries (Marks) Adjusted Frequency (f/h) 0.0 0.5 1.0 1.5 2.0 34.5 37.5 44.5 54.5 61.5 67.5 72.5
Question 8 Unequal Combined Graph
Make a frequency polygon on a histogram for the following grouped data:
Marks 5–8 8–12 12–20 20–25 25–27 27–32
Frequency (f) 2 12 25 32 14 5
Solution

Since class widths are unequal, we calculate Class Width (h) and Adjusted Height (fh). Class limits are continuous, so boundaries match limits.

Class limits Frequency (f) Midpoint (x) Width (h) Adjusted Height (fh)
5–8 2 6.5 3 2 ÷ 3 = 0.67
8–12 12 10.0 4 12 ÷ 4 = 3.00
12–20 25 16.0 8 25 ÷ 8 = 3.13
20–25 32 22.5 5 32 ÷ 5 = 6.40
25–27 14 26.0 2 14 ÷ 2 = 7.00
27–32 5 29.5 5 5 ÷ 5 = 1.00

Anchors for polygon: Add dummy class 2–5 (Midpoint = 3.5, height = 0) and 32–37 (Midpoint = 34.5, height = 0).

Class Boundaries & Midpoints Adjusted Frequency (f/h) 0 1 2 3 4 5 6 7 5 8 12 20 25 27 32

📊 Dynamic Frequency Distribution & Chart Builder

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Class Limits Class Boundaries Tally Marks Frequency (f) Midpoint (x)
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