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 in each of the following cases:

(i) If wrong item is omitted.

(ii) If it is replaced by 12

Asked by Abhisek | 11 months ago |  127

Solution :-

(i) If wrong item is omitted,

From the question it is given that,

The number of observations i.e. n = 20

The incorrect mean = 20

The incorrect standard deviation = 2

$$\overline {X}= \dfrac{1}{n} \displaystyle\sum_{i=1}^{20} (X_1)$$

$$10= \dfrac{1}{20} \displaystyle\sum_{i=1}^{20} (X_1)$$

$$\displaystyle\sum_{i=1}^{20} (X_1)=200$$

That is, incorrect sum of observations = 200

Correct sum of observations = 200 – 8 = 192

Correct mean = $$\dfrac{Correct\;sum}{19}$$

$$\dfrac{192}{19}$$ = 10.1

Standard deviation σ=

$$(σ)= \sqrt{\dfrac{1}{n} \displaystyle\sum_{i=1}^{n} X_1-\dfrac{1}{n^2} (\displaystyle\sum_{i=1}^{n}X)^2}$$

$$2= \sqrt{\dfrac{1}{n} \displaystyle\sum_{i=1}^{n} (X_1^2-\overline X)^2}$$

$$4=\dfrac{1}{20} Incorrect \displaystyle\sum_{i=1}^{n}X_1^2 -100$$

$$Incorrect \displaystyle\sum_{i=1}^{n}X_1^2 =2080$$

Therefore,correct

$$correct \displaystyle\sum_{i=1}^{n}X_1^2= Incorrect \displaystyle\sum_{i=1}^{n}X_1^2 -(8)^2$$

= 2080 - 64

= 2016

Correct standard deviation =

$$\sqrt{\dfrac{correct \displaystyle\sum X_1^2}{n}-(correct\;mean)^2}$$

$$\sqrt{\dfrac{2016}{19}-(10.1)^2}$$

$$\sqrt{1061.1- 102.1}$$

$$\sqrt{4.09}$$

= 2.02

(ii) When 8 is replaced by 12,

Incorrect sum of observations = 200

Correct sum of observations = 200 – 8 + 12 = 204

Correct mean = $$\dfrac{Correct\;sum}{20}$$

$$\dfrac{204}{20}=10.2$$

Standard deviation $$(σ)= \sqrt{\dfrac{1}{n} \displaystyle\sum_{i=1}^{n} X_1-\dfrac{1}{n^2} (\displaystyle\sum_{i=1}^{n}X)^2}$$

$$2= \sqrt{\dfrac{1}{n} \displaystyle\sum_{i=1}^{n} (X_1^2-\overline X)^2}$$

$$4=\dfrac{1}{20} Incorrect \displaystyle\sum_{i=1}^{n}X_1^2 -100$$

$$Incorrect \displaystyle\sum_{i=1}^{n}X_1^2 =2080$$

Therefore,correct

$$correct \displaystyle\sum_{i=1}^{n}X_1^2= Incorrect \displaystyle\sum_{i=1}^{n}X_1^2 -(8)^2$$

= 2080 - 64

= 2160

Correct standard deviation=

$$\sqrt{\dfrac{correct \displaystyle\sum X_1^2}{n}-(correct\;mean)^2}$$

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

$$\sqrt{108-104.04}$$

$$\sqrt{3.96}$$

= 1.98

Answered by Pragya Singh | 11 months ago

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