Question

A professor of History is teaching a section of 100 students. Her first exam’s grade distribution is as follows. Calculate the standard deviation for this grouped data.

Exam grades	Frequency
45 to < 50	1
50 to < 55	2
55 to < 60	6
60 to < 65	19
65 to < 70	12
70 to < 75	22
75 to < 80	12
80 to < 85	13
85 to < 90	11
90 to < 95	0
95 to < 100	2

Answer 1

Solution

Back-up Theory

Let

xi be the mid-point (class mark) of the ith class interval and fi be the corresponding class frequency, i = 1 to k, k being the number of classes.

Then,

Mean (Average), µ, = {Σ(i = 1, k)(x_i.f_i)}/{Σ(i = 1, k)(f_i)} ..…………………………. (1)

Variance, σ² = [Σ(i = 1, k){f_i.(x_i – µ)²}]/{Σ(i = 1, k)(f_i)} .........................................(2a)

or equivalently

[{Σ(i = 1, k){f_i.(x_i)²}/{Σ(i = 1, k)(f_i)}] – µ² …………………....…………………….. (2b)

Standard deviation (SD), σ = sqrt(Variance) …………………..........………….. (3)

Now, to work out the solution,

Final answer is given below. Details of Calculations follow at the end.

Standard deviation = 10.15 Answer

Details of Calculations

Class Number(i)	Class Limits	Frequency (fi)	Class Mark(xi)	fi.xi	fi.xi2
1	45 to < 50	1	47.5	47.5	2256.25
2	50 to < 55	2	52.5	105.0	5512.50
3	55 to < 60	6	57.5	345.0	19837.50
4	60 to < 65	19	62.5	1187.5	74218.75
5	65 to < 70	12	67.5	810.0	54675.00
6	70 to < 75	22	72.5	1595.0	115637.50
7	75 to < 80	12	77.5	930.0	72075.00
8	80 to < 85	13	82.5	1072.5	88481.25
9	85 to < 90	11	87.5	962.5	84218.75
10	90 to < 95	0	92.5	0.0	0.00
11	95 to < 100	2	97.5	195.0	19012.50
Total		100		7250.0	535925.00

Mean = 7250/100 = 72.5	Vide (1) of Back-up Theory
Variance = (5359.25) – 72.52= 103	Vide (2b) of Back-up Theory
Standard Deviation = √103 = 10.15	Vide (3) of Back-up Theory

DONE