MCQ
The mean and variance of $10$ observations were calculated as $15$ and $15$ respectively by a student who took by mistake $25$ instead of $15$ for one observation. Then, the correct standard deviation is$.....$
- A$4$
- B$6$
- ✓$2$
- D$8$
$6^{2}=\frac{\sum x_{i}^{2}}{10}-(\bar{x})^{2}=15$
$\sum_{i=1}^{10} x_{i}=150$
$\sum_{i=1}^{9} x_{i}+25=150$
$\sum_{i=1}^{9} x_{i}=125$
$\sum_{i=1}^{9} x_{i}+15=140$
Actual mean $=\frac{140}{10}=14=\bar{x}_{\text {nev }}$
$\sum_{i=1}^{9} \frac{x_{i}^{2}+25^{2}-15^{2}}{10}=15$
$\sum_{i=1}^{9} x_{i}^{2}+625=2400$
$\sum_{i=1}^{9} x_{i}^{2}=1775$
$\sum_{i=1}^{9} x_{i}^{2}+15^{2}=2000=\left(\sum x_{i}^{2}\right)_{\text {acnaal }}$
$6_{\text {actual }}^{2}=\frac{\left(\sum x_{i}^{2}\right)_{\text {actual }}-\left(\bar{x}_{\text {new }}\right)^{2}}{10}$
$=\frac{2000}{10}-14^{2}$
$=200-196=4$
$(\text { S.D })_{\text {attul }}=6=2$
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