Question

The scores on a 100-mark test of a sample of 80 students in a large school are given in the table.

SCORE: 59-63 63-67 67-71 71-75 75-79 79-83 83-87
# OF STUDENTS: 6 10 18 24 10 8 4

a) Find the mean and standard deviation of the scores of all students.

b) A bonus of 13 points is to be added to these scores. What is the new value of the mean and standard deviation?

Answer 1

(a) Mean = 72.1 Standard deviation = 6.14

(b) New Mean = 85.1 New Standard deviation = 6.14

Explanation:

We are given the scores on a 100-mark test of a sample of 80 students in a large school;

Score No. of students (f) X X*f
`(X-Xbar)^(2)`
`f*(X-Xbar)^(2)`

59 - 63 6 61 366 123.21 739.26

63 - 67 10 65 650 50.41 504.1

67 - 71 18 69 1242 9.61 172.98

71 - 75 24 73 1752 0.81 19.44

75 - 79 10 77 770 24.01 240.1

79 - 83 8 81 648 79.21 633.68

83 - 87 4 85 340 166.41 665.64

∑f = 80 ∑X*f = 5768 Total = 2975.2

(a) Mean of the above data, X bar =
`(\sum Xf)/(\sum f)` =
`(5768)/(80)` = 72.1

Standard deviation, s =
`\sqrt{(\sum f*(X-Xbar)^(2) )/(\sum f -1) }` =
`\sqrt{(2975.2 )/(80 -1) }` = 6.14

Therefore, mean and standard deviation of the scores of all students are 72.1 and 6.14 respectively.

(b) Now, a bonus of 13 points is to be added to these scores.

So, Mean will also increases by 13 points and due to this;

New Mean = 72.1 + 13 = 85.1

Adding 13 points to each score will not affect standard deviation and due to which New standard deviation is same as before of 6.14.