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

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

Smart Achievers · Accepted Answer

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

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

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

$σ^{2} = \frac{1961}{225}$

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

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

Solution

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