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 denote the mean and variance of the correct observations respectively, then

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Solution

Q.
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 $μ$ and $σ^{2}$ denote the mean and variance of the correct observations respectively, then $15 (μ + μ^{2} + σ^{2})$ is equal to _________ . [2024]

Ans.
(2521)

We have, mean = 12, standard deviation = 3

Let $x_{1}, x_{2}, \dots, x_{15}$ be 15 observations

$∴$ $n = 15$

$\frac{x_{1} + x_{2} + \dots + x_{14} + 10}{15} = 12$ $\Rightarrow x_{1} + x_{2} + \dots + x_{14} = 170$

Now, correct mean = $μ$

$\frac{x_{1} + x_{2} + \dots + x_{14} + 12}{15} = μ \Rightarrow \frac{170 + 12}{15} = μ \Rightarrow μ = \frac{182}{15}$

Standard deviation $= \frac{1}{15} \sqrt{15 \sum_{i = 1}^{15} x_{i}^{2} - {(\sum_{i = 1}^{15} x_{i})}^{2}}$

$\Rightarrow 3 = \frac{1}{15} \sqrt{15 \times \sum_{i = 1}^{15} x_{i}^{2} - {(180)}^{2}}$ $\Rightarrow 45 = \sqrt{15 \times \sum_{i = 1}^{15} x_{i}^{2} - 32400}$

$\Rightarrow 2025 = 15 \sum_{i = 1}^{15} x_{i}^{2} - 32400$

$\Rightarrow 2025 + 32400 = 15 (x_{1}^{2} + x_{2}^{2} + \dots + x_{14}^{2} + 100)$

$\Rightarrow 2295 = x_{1}^{2} + x_{2}^{2} + \dots + x_{14}^{2} + 100$

$\Rightarrow x_{1}^{2} + x_{2}^{2} + \dots + x_{14}^{2} = 2195$

Correct Variance, $σ^{2} = \frac{1}{n^{2}} (15 \times \sum_{i = 1}^{15} x_{i}^{2} - {(\sum_{i = 1}^{n} x_{i})}^{2})$

$= \frac{1}{{(15)}^{2}} (15 \times (x_{1}^{2} + x_{2}^{2} + \dots + x_{14}^{2} + 144) - {(182)}^{2})$

$= \frac{1}{225} (15 \times (2339) - 33124)$ $= \frac{1}{225} (35085 - 33124)$

$σ^{2} = \frac{1961}{225}$

$∴$ $15 (μ + μ^{2} + σ^{2}) = 15 (\frac{182}{15} + {(\frac{182}{15})}^{2} + \frac{1961}{225})$

$= 15 (\frac{2730 + 33124 + 1961}{225})$ $= \frac{37815}{15} = 2521$

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