The mean and standard deviation of 15 observa | vedclass

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(A) Let the incorrect mean be $\mu^{\prime}$ and standard deviation be $\sigma^{\prime}$.We have $\mu^{\prime} = \frac{\Sigma x_i}{15} = 12 \Rightarro

Vedclass · Accepted Answer

$2521$

Vedclass · Answer

(A) Let the incorrect mean be $\mu^{\prime}$ and standard deviation be $\sigma^{\prime}$.We have $\mu^{\prime} = \frac{\Sigma x_i}{15} = 12 \Rightarrow \Sigma x_i = 180$.As per the given information,the correct $\Sigma x_i = 180 - 10 + 12 = 182$.$\mu = \frac{182}{15}$.Also,$\sigma^{\prime} = \sqrt{\frac{\Sigma x_i^2}{15} - (12)^2} = 3$ $\Rightarrow \frac{\Sigma x_i^2}{15} - 144 = 9$ $\Rightarrow \Sigma x_i^2 = 15 	imes 153 = 2295$.Correct $\Sigma x_i^2 = 2295 - 10^2 + 12^2 = 2295 - 100 + 144 = 2339$.$\sigma^2 = \frac{\Sigma x_i^2}{15} - \mu^2 = \frac{2339}{15} - \left(\frac{182}{15}ight)^2$.We need to find $15(\mu + \mu^2 + \sigma^2)$.Substituting $\sigma^2 = \frac{\Sigma x_i^2}{15} - \mu^2$:$15(\mu + \mu^2 + \frac{\Sigma x_i^2}{15} - \mu^2) = 15(\mu + \frac{\Sigma x_i^2}{15}) = 15\mu + \Sigma x_i^2$.$= 15 	imes \frac{182}{15} + 2339 = 182 + 2339 = 2521$.

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