Question

The mean and standard deviation of 100 observations were calculated as 40 and 5.1 respectively by a student qho look by mistake 50 instead of 40 for one observation. What are the correct mean and standard deviation ?

Answer 1

Given thet number of observations (n) = 100

Incorrect mean $\left(\overline{x}\right)$ = 40,

Incorrect standard deviation $\left(\sigma \right)$ = 5.1

We know that,

$\overline{x}=\frac{1}{n}{\sum }_{i=1}^n{x}_i$

i.e., $40=\frac{1}{100}{\sum }_{i=1}^{100}{x}_{i}or{\sum }_{i=1}^{100}{x}_i=4000$

i.e., Incorrect sum of observations = 4000

Thus, the correct sum of observations = Incorrect sum - 50+40

= 4000-50+40 = 3990

Hence, correct mean = $\frac{correctsum}{100}=\frac{3990}{100}=39.9$

Also, standard deviation $\sigma$

$=\sqrt{\frac{1}{n}{\sum }_{i=1}^n{x}_i^2-\frac{1}{n^2}{\left({\sum }_{i=1}^n\right)}^2}$

$=\sqrt{\frac{1}{n}{\sum }_{i=1}^n{x}_i^2-(\overline{x}{)}^2}$

$5.1=\sqrt{\frac{1}{100}\times Incorrect{\sum }_{i=1}^n{x}_i^2-(40{)}^2}$

or 26.01 = $\frac{1}{100}\times Incorrect{\sum }_{i=1}^n{x}_i^2-1600$

Therefore, Incorrect ${\sum }_{i=1}^n{x}_i^2=100(26.01+1600)=162601$

Now, Correct ${\sum }_{i=1}^n{x}_i^2=Incorrect{\sum }_{i=1}^n{x}_i^2-(50{)}^2+(40{)}^2$

= 162601-2500+1600= 161701

Therefore , correct standard deviation

$=\sqrt{\frac{correct\sum x_i^2}{n}}-(Correctmean{)}^2$

$=\sqrt{\frac{161701}{100}-(39.9{)}^2}$

$=\sqrt{1617.01-1592.01}=\sqrt{25}=5$