The mean and standard deviation of 20 observa | vedclass

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(A) Number of observations $(n) = 20$.Incorrect mean $(\bar{x}) = 10$.Incorrect standard deviation $(\sigma) = 2$.$\sum x_i = n 	imes \bar{x} = 20

Vedclass · Accepted Answer

$10.1, 2.02$

Vedclass · Answer

(A) Number of observations $(n) = 20$.Incorrect mean $(\bar{x}) = 10$.Incorrect standard deviation $(\sigma) = 2$.$\sum x_i = n 	imes \bar{x} = 20 	imes 10 = 200$.Correct sum of observations $= 200 - 8 = 192$.Correct mean $= \frac{192}{19} \approx 10.105$.Incorrect $\sum x_i^2 = n(\sigma^2 + \bar{x}^2) = 20(2^2 + 10^2) = 20(4 + 100) = 2080$.Correct $\sum x_i^2 = 2080 - 8^2 = 2080 - 64 = 2016$.Correct standard deviation $= \sqrt{\frac{\sum x_i^2}{n} - (	ext{mean})^2} = \sqrt{\frac{2016}{19} - (10.105)^2} = \sqrt{106.105 - 102.111} = \sqrt{3.994} \approx 2.00$.

Class:	$10-20$	$20-30$	$30-40$	$40-50$	$50-60$
Freq:	$\alpha$	$110$	$54$	$30$	$\beta$

Daily wage (Rs.)	$30$-$40$	$40$-$50$	$50$-$60$	$60$-$70$	$70$-$80$	$80$-$90$
No. of workers	$17$	$28$	$21$	$15$	$13$	$6$

The mean and standard deviation of $20$ observations are found to be $10$ and $2$ respectively. On rechecking,it was found that an observation $8$ was incorrect. Calculate the correct mean and standard deviation if the wrong item is omitted.

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