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  • Find the mean and standard deviation using short-cut method. (6)

6     Find the mean and standard deviation using short-cut method.

\small x_i 60 61 62 63 64 65 66 67 68
\small f_i 2 1 12 29 25 12 10 4 5

 

Answers (1)

best_answer

Let the assumed mean, A = 64 and h = 1

x_i f_i y_i = \frac{x_i-A}{h} y_i^2 f_iy_i f_iy_i^2
60    2 -4 16 -8 32
61 1 -3 9 -3 9
62 12 -2 4 -24 48
63 29 -1 1 -29 29
64 25 0 0 0 0
65 12 1 1 12 12
66 10 2 4 20 40
67 4 3 9 12 36
68 5 4 16 20 80
 

\sum{f_i}

=100

 

  

\sum f_iy_i

= 0

\sum f_iy_i ^2

=286

N = \sum_{i=1}^{9}{f_i} = 100 ; \sum_{i=1}^{9}{f_iy_i} = 0

Mean,

\overline{x} = A + \frac{1}{N}\sum_{i=1}^{n}f_iy_i\times h =64 + \frac{0}{100} = 64

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}{(100)^2}\left [100(286) - (0)^2 \right ] \\ = \frac{28600}{10000} = 2.86

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

\therefore \sigma = \sqrt{2.86} = 1.691

Hence, Mean = 64 and Standard Deviation = 1.691

Posted by

HARSH KANKARIA

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