The mean and standard deviation of 15 observations were found to be 12 and 3 respectively.

Question

The mean and standard deviation of 15 observations were found to be 12 and 3 respectively. On rechecking it was found that an observation was read as 10 in place of 12. If  and $\sigma^2$ denote the mean and variance of the correct observations respectively, then $15(\mu+\mu^2+\sigma^2)$ is equal to \_\_\_

Answer: 2521

Accepted Answer

Original: $n=15$, mean $=12 \Rightarrow \sum x_i = 180$; S.D. $=3 \Rightarrow \sum x_i^2 = 15(9+144)=2295$.

Correcting 10 to 12: new sum $=180-10+12=182$, new sum of squares $=2295-100+144=2339$.

$$\mu = \frac{182}{15}, \qquad \sigma^2 = \frac{2339}{15}-\mu^2$$

$$15(\mu+\mu^2+\sigma^2) = 15\mu + 15\left(\frac{2339}{15}\right) = 15\left(\frac{182}{15}\right)+2339 = 182+2339 = 2521$$

Solutions · 1