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Q.

The meanand S.D of 20 items were found to be 10 and 2. If an item 12 was misread as 8 the correct mean and S.D are

Updated 5 months ago

Solution

Given mean = 10 = $\bar x$

Given standard Deviation = 2 = $\sigma$

No. of observations = 20 = n

Step (1)

${\sum_{i=1}^{20} x_i}\over n$ = $\bar x$

$\therefore$ ${\sum_{i=1}^{20} x_i}\over 20$ = 10  $\therefore$ ${\sum_{i=1}^{20} x_i}$ = 200

But observation 8 was wrong & is replaced by 12.

$\therefore$ ${\sum_{i=1}^{20} x_i}$ = $200-8 + 12 = 204$

$\therefore$ New mean = $204\over 20$ = 10.2

Step (2)

Given $\sigma$ = 2

Variance = $\sigma^2$ = $2^2$ = 4

$\therefore$ ${{\sum_{i=1}^{20} x_i^2}\over n} - \left({\sum_{i=1}^{20} x_i}\over n\right)^2$ = 4

${\sum_{i=1}^{20} x_i^2}\over 20$ $- 10^2 = 4$

$\therefore$ ${\sum_{i=1}^{20} x_i^2} = 104 \times 20 = 2080$

But observation 8 is replaced by 12 $\therefore$

${\sum_{i=1}^{20} x_i^2} = 2080 - 8^2 + 12^2 = 2160$

$\therefore$ New variance = ${{\sum_{i=1}^{20} x_i^2}\over 20} - \left({\sum {x_i}}\over 20\right)^2$

${2160 \over 20} - (10.2)^2$

= 3.96

$\therefore$ New S. D. = $\sqrt{3.96} = 1.99$

$\therefore$ New S. D. = 1.99

Updated 5 months ago