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 $\mu$ and $\sigma^2$ denote the mean and variance of the correct observations respectively, then $15\left(\mu+\mu^2+\sigma^2\right)$ is equal to __________.

<p>Let the incorrect mean be $\mu^{\prime}$ and standard deviation be $\sigma^{\prime}$</p> <p>We have</p> <p>$$\mu^{\prime}=\frac{\Sigma \mathrm{x}_{\mathrm{i}}}{15}=12 \Rightarrow \Sigma \mathrm{x}_{\mathrm{i}}=180$$</p> <p>As per given information correct $\Sigma \mathrm{x}_{\mathrm{i}}=180-10+12$</p> <p>$\Rightarrow \mu(\text { correct mean})=\frac{182}{15}$</p> <p>Also</p> <p>$$\sigma^{\prime}=\sqrt{\frac{\Sigma \mathrm{x}_{\mathrm{i}}^2}{15}-144}=3 \Rightarrow \Sigma \mathrm{x}_{\mathrm{i}}^2=2295$$</p> <p>Correct $\Sigma \mathrm{x}_{\mathrm{i}}{ }^2=2295-100+144=2339$</p> <p>$$\sigma^2(\text { correct variance })=\frac{2339}{15}-\frac{182 \times 182}{15 \times 15} $$</p> <p>Required value</p> <p>$$\begin{aligned} & =15\left(\mu+\mu^2+\sigma^2\right) \\ & =15\left(\frac{182}{15}+\frac{182 \times 182}{15 \times 15}+\frac{2339}{15}-\frac{182 \times 182}{15 \times 15}\right) \\ & =15\left(\frac{182}{15}+\frac{2339}{15}\right) \\ & =2521 \end{aligned}$$</p>