The mean and standard deviation of a group of 25 observations were calculated to be 20 and 5 respectively. It was later found that there was an error in recording one observation; it was recorded as 12 instead of 22. The standard deviation when the error is corrected is (approx.):

Concept:

Suppose a data consists of the n number of elements and the elements are x₁, x₂, x₃, …, x_n.

Then mean of the data is given by =

Standard deviation is the square root of the average of the square deviation of data points from the mean.

The standard deviation SD is given by,

SD =

Calculation:

The mean and standard deviation for a group of 25 observations is 20 and 5.

Hence, = 20

  = 20

  = 500

Since the standard deviation is 5, we can write,

SD = 5

  = 5

  = 25

  = 625

  = 625

 x_i² - 40x_i + 400 = 625

 x_i² - 40 × 500 + 400 × 25 = 625

 x_i² - 20000 + 10000 = 625

 x_i² = 625 + 10000 = 10625

It was later found that observation was recorded incorrectly as 12 instead of 22.

Hence the actual Sigma x = 500 - 12 + 22 = 510.

Actual mean = = 20.4 

Actual x_i² = 10625 - 12² + 22² = 10625 - 144 + 484 = 10625 + 340 = 10965

Now the new standard deviation will be given by the formula,

SD =

 SD =

 SD =​ 

​ SD = 

 SD = 

 SD = 

 SD =  = 4.737 ≈ 4.7

Hence the correct answer is option 3.