Answer to Question #274102 in Statistics and Probability for Grace

Question

A data set of 20 observations has a mean 45 and standard deviation 25.4. The data set is reviewed and one observation which was incorrectly recorded is now corrected to 30. Determine the mean and standard deviation of the corrected data

Answer 1

Let observation $a$ was incorrect.

The incorrect sum of observations is $A_1=45(20)=900$

The correct sum of observations is $A_2=900-a+30=930-a$

The correct mean is $\bar{x}_{new}=(930-a)/20$

$$\sigma^2=\dfrac{1}{n}\sum_ix_i^2-\dfrac{1}{n^2}\bigg(\sum_ix_i\bigg)^2$$

The incorrect sum of squares is

$$B_1=20(25.4)^2+\dfrac{1}{20}(900)^2=53403.2$$

The correct sum of squares is

$$B_2=53403.2-a^2+(30)^2=54303.2-a^2$$

The correct variance is

$$\sigma_{new}^2=\dfrac{1}{20}B_2-\dfrac{1}{(20)^2}(A_2)^2$$

$$=\dfrac{1}{20}(54303.2-a^2)-\dfrac{1}{400}(930-a)^2$$

$$=552.91-0.0525a^2+4.65a$$

The correct standard deviation is

$$\sigma_{new}=\sqrt{552.91-0.0525a^2+4.65a}$$

The correct mean is

$$\bar{x}_{new}=(930-a)/20$$

The correct standard deviation is

$$\sigma_{new}=\sqrt{552.91-0.0525a^2+4.65a}$$