Question

The mean and standard deviation of the marks of 200 candidates were found to be 40 and 15 respectively. Later, it was discovered that a score of 40 was wrongly read as 50. The correct mean and standard deviation respectively are:

• A. 14.98, 39.95
• B. 39.95, 14.98
• C. 39.95, 224.5
• D. None the these

Answer 1

The correct option is B 39.95, 14.98

$Corrected\sum x=40\times 200-50+40=7990$
$\therefore Corrected\overline{x}=\frac{7990}{200}=39.95$
$\sum x^2=n[{\sigma }^2+{\overline{x}}^2]=200[{15}^2+{40}^2]=365000$
$Correct\sum x^2=365000-2500+1600=364100$
$\therefore Corrected\sigma =\sqrt{\frac{364100}{200}-(39.95{)}^2}$
$=\sqrt{(1820.5-1596)}=\sqrt{224.5}=\mathrm{14.98.}$