9. Find the mean, variance and standard deviation using short-cut method.

Height in cms 70-75 75-80 80-85 85-90 90-95 95-100 100-105 105-110 110-115
No. of students  3 4 7 7 15 9 6 6 3

 

 

Answers (1)

Let the assumed mean, A = 92.5 and h = 5

Height

in cms

Frequency

f_i

Midpoint

x_i

\dpi{100} y_i = \frac{x_i-A}{h} y_i^2 f_iy_i f_iy_i^2
70-75 3 72.5 -4 16 -12 48
75-80 4 77.5 -3 9 -12 36
80-85 7 82.5 -2 4 -14 28
85-90 7 87.5 -1 1 -7 7
90-95 15 92.5 0 0 0 0
95-100 9 97.5 1 1 9 9
100-105 6 102.5 2 4 12 24
105-110 6 107.5 3 9 18 54
110-115 3 112.5 4 16 12 48
 

\sum{f_i} =N = 60

 

 

 

\sum f_iy_i

= 6

\sum f_iy_i ^2

=254

Mean,

\overline{y} = A + \frac{1}{N}\sum_{i=1}^{n}f_iy_i\times h =92.5 + \frac{6}{60}\times5 = 93

We know, Variance, \sigma^2 = \frac{1}{N^2}\left [N\sum f_iy_i^2 - (\sum f_iy_i)^2 \right ]\times h^2

\\ \implies \sigma^2 = \frac{1}{(60)^2}\left [60(254) - (6)^2 \right ] \\ = \frac{1}{(60)^2}\left [15240 - 36 \right ] \\ = \frac{15204}{144} = 105.583

We know,  Standard Deviation = \sigma = \sqrt{Variance}

\therefore \sigma = \sqrt{105.583} = 10.275

Hence, Mean = 93, Variance = 105.583 and Standard Deviation = 10.275

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