Question

for a group of 100 candidates the mean and standard deviavation of the mark are
found to be 60 and 15 respectively. Later on it was to found that the scorces 45and72
were wrongly entered as 40 and 27. Find the correct mean and standard deviation​

Answer 1

Answer:

Mean = 60.5, SD = 14.61

Step-by-step explanation:

Answer 2

Correct Mean = 60.5

Correct Standard deviation = 14.61

Step-by-step explanation:

We are given that for a group of 100 candidates the mean and standard deviation of the mark are  found to be 60 and 15 respectively.

Later on it was to found that the scores 45 and 72  were wrongly entered as 40 and 27.

Firstly, the mean formula is given by;

                 Mean =  $\frac{\sum X}{n}$

where,  $\sum X$ = sum of all the observations in the data

                 n  = number of observations in the data

So, Incorrect  $\sum X$ = $60 \times 100$ = 6000

Now, Correct  $\sum X$ = 6000 - 40 - 27 + 45 + 72 = 6050

Correct n = 100 - 2 + 2 = 100

Hence, the correct mean  =  $\frac{6050}{100}$  = 60.5

Now, Variance formula is given by;

                Variance =  ${\frac{\sum X^{2}-n\bar X^{2} }{n-1} }$

Now, Incorrect  $\sum X^{2}$  =  $(15^{2} \times 99) + (100 \times 60^{2} )$ = 382275

Correct  $\sum X^{2}$  = 382275 - $40^{2}$ - $27^{2}$ + $45^{2}$ + $72^{2}$ = 387155

So, Correct variance =  ${\frac{387155-100 \times 60.5^{2} }{100-1} }$

                                  =  213.43

Hence, Correct standard deviation =  $\sqrt{Variance}$

                                                             =  $\sqrt{213.43}$  =  14.61