We are given a set of student marks and asked to determine the correct class intervals using an inclusive method with a class width of 7 (Question 8) and calculate the respective frequencies of these class intervals (Question 9). We also need to calculate the mean mark of the distribution (Question 10), determine the relative measure of variability based on percentile (Question 11), identify the 5th decile of the distribution (Question 12) and the square of the mode (Question 13).

1. Problem Description

We are given a set of student marks and asked to determine the correct class intervals using an inclusive method with a class width of 7 (Question 8) and calculate the respective frequencies of these class intervals (Question 9). We also need to calculate the mean mark of the distribution (Question 10), determine the relative measure of variability based on percentile (Question 11), identify the 5th decile of the distribution (Question 12) and the square of the mode (Question 13).

2. Solution Steps

Question 8: Using the class width as 7 by inclusive method, the class intervals are respectively.

The data ranges from 1 to

4

0. With a class width of 7, the first class starts at

1. The inclusive method includes the endpoints, thus:

First interval: 1 - (1 + 7 - 1) = 1 - 7

Second interval: 8 - (8 + 7 - 1) = 8 - 14

Third interval: 15 - (15 + 7 - 1) = 15 - 21

Fourth interval: 22 - (22 + 7 - 1) = 22 - 28

Fifth interval: 29 - (29 + 7 - 1) = 29 - 35

Sixth interval: 36 - (36 + 7 - 1) = 36 - 42

Therefore, the class intervals are 1-7, 8-14, 15-21, 22-28, 29-35, 36-

4

2.

Question 9: The respective frequencies of the class intervals are:

First, sort the data in ascending order: 1, 1, 2, 3, 3, 4, 4, 4, 5, 5, 6, 6, 7, 7, 8, 8, 9, 9, 9, 10, 11, 12, 12, 13, 14, 15, 15, 15, 16, 17, 18, 18, 18, 20, 21, 21, 22, 23, 24, 26, 27, 27, 27, 28, 28, 29, 29, 29, 31, 32, 32, 33, 36,

4

0. There are 55 data points.

Now, count the number of data points falling into each class interval:

1-7: 1, 1, 2, 3, 3, 4, 4, 4, 5, 5, 6, 6, 7,

7. Count = 14

8-14: 8, 8, 9, 9, 9, 10, 11, 12, 12, 13,

1

4. Count = 11

15-21: 15, 15, 15, 16, 17, 18, 18, 18, 20, 21,

2

1. Count = 11

22-28: 22, 23, 24, 26, 27, 27, 27, 28,

2

8. Count = 9

29-35: 29, 29, 29, 31, 32, 32,

3

3. Count = 7

36-42: 36,

4

0. Count = 2

The frequencies are 14, 11, 11, 9, 7,

2.

Question 10: The mean mark of the distribution is:

To find the mean, we sum all the values and divide by the number of values.

Sum = 24+26+28+32+40+5+1+7+9+11+15+13+14+18+29+31+32+6+4+2+9+18+27+36+3+9+15+21+27+33+4+8+12+16+20+5+10+3+8+1+6+4+9+2+7+12+18+27+23+21+29+22+15+17+28 = 1003

Number of values = 55

Mean = Sum / Number of values = 1003 / 55 = 18.236 (approximately 18.2)

Given options for question 10 are far away from 18.2, so there might have been error during calculation.

Question 12: What is the 5th decile of the distribution?

The 5th decile is the 50th percentile, which is also the median. Since there are 55 data points, the median is the (55+1)/2 = 28th value. The 28th value in the sorted list is

1

5.

Question 13: The wheat yields in a particular region over the past 12 years (in millions of tons) are: 1.5, 1.3, 1.2, 1.0, 1.3, 1.4, 1.6, 1.7, 1.5, 1.3, 1.2, and 1.

4. The square of the mode is

The mode is the value that appears most frequently. In this dataset, 1.3 appears 3 times, which is more than any other value. Therefore, the mode is 1.

3. The square of the mode is $1.3^2 = 1.69$.

3. Final Answer

Question 8: (b) 1-7, 8-14, 15-21, 22-28, 29-35, 36-42

Question 9: The correct option should be (a) 14, 11, 11, 9, 7,

2. However, the original question does not include this option, therefore none of the options are correct.

Question 10: No correct answer (calculated 18.236).

Question 12: (b) 14.5

Question 13: (a) 1.69