Mean and standard deviation of 100 observations were found to be 40 and 10 respectively. If at the time of calculation two observations were wrongly taken as 30 and 70 in place of 3 and 27 respectively, find the correct standard deviation.

Advertisement Remove all ads

#### Solution

Given:

Number of observations, *n* = 100

Mean,

\[\bar{x} \] = 40

Standard deviation,

\[\sigma\] = 10

We know that \[x = \frac{\sum_{} x_i}{100}\]

\[ \Rightarrow \frac{\sum_{} x_i}{100} = 40\]

\[ \Rightarrow \sum_{} x_i = 4000\]

\[ \Rightarrow \frac{\sum_{} x_i}{100} = 40\]

\[ \Rightarrow \sum_{} x_i = 4000\]

∴ Correct

\[ \sum_{} x_i = 4000 - \left( 30 + 70 \right) + \left( 3 + 27 \right) = 3930\]

Correct mean = \[\frac{\text{ Correct } \sum_{} x_i}{100} = \frac{3930}{100} = 39 . 3\]

Now,

Incorrect variance,

Incorrect variance,

\[\sigma^2 = \frac{\sum_{} x_i^2}{100} - \left( 40 \right)^2\]

\[\Rightarrow \frac{\sum_{} x_i^2}{100} = 100 + 1600 = 1700\]

\[ \Rightarrow \sum_{} x_i^2 = 170000\]

\[ \Rightarrow \sum_{} x_i^2 = 170000\]

Correct \[\sum_{} x_i^2 = 170000 - {30}^2 - {70}^2 + 3^2 + {27}^2 = 164939\]

∴ Correct standard deviation \[= \sqrt{\frac{164939}{100} - \left( 39 . 3 \right)^2}\]

\[ = \sqrt{1649 . 39 - 1544 . 49}\]

\[ = \sqrt{104 . 9}\]

\[ = 10 . 24\]

\[ = \sqrt{1649 . 39 - 1544 . 49}\]

\[ = \sqrt{104 . 9}\]

\[ = 10 . 24\]

Is there an error in this question or solution?

Advertisement Remove all ads

#### APPEARS IN

Advertisement Remove all ads

Advertisement Remove all ads