Question

Question 7 The mean and standard deviation of a group of 100 observations were found to be 20 and 3, respectively. Later on it was found that three observations were incorrect, which were recorded as 21, 21 and 18. Find the mean and standard deviation if the incorrect observations are omitted.

Class X1 - Maths -Statistics Page 380

Answer 1

Here, Given that
number of observation ( n ) = 100
$mean ( \overline{x}) = 20\\\sigma=3$
$\because\:\overline{x}=\frac{sum\:of\: observation}{total\: number\:of\: observation}\\\\\frac{\sum{x}}{100}=20\\\\\sum{x}=100\times\:20=2000$
now, here incorrect observation are 21 , 21 and 18 are omitted,
now, correct sum is
$\sum{x}=2000-21-21-18=2000-60=1940\\\\\sum{x}=1940$
Now, mean of 97 observations after correction,
$\overline{x}=\frac{1940}{97}=20\\\\\\again,\sigma=\sqrt{\frac{\sum{x}^2}{n}-(\overline{x})^2}$
so,

$3=\sigma=\sqrt{\frac{\sum{x}^2}{n}-(\overline{x})^2}\\\\9=\frac{\sum{x}^2}{n}-( \overline{x})^2 \\ \\ { \sum{x}}^{2} = ({20}^{2} + 9) \times 100 = 40900$

Now, correcting of ∑x² :
∑x² = 40900 - (21)² - (21)² - (18)²
= 40900 - 441 - 441 - 324
= 40900 - 1206
= 39694

Now, correct Standard deviation (SD ) of remaining 97 observations.
$\sigma=\sqrt{\frac{39694}{97}-(\frac{1940}{97})^2}\\\\=\sqrt{409.2-(20)^2}\\\\=\sqrt{409.2-400}\\\\=\sqrt{9.2}=3.03$