The mean and standard deviation of 100 observations are 40 and 5.1, respectively. By mistake one observation is taken as 50 instead of 40. If the correct mean and the correct standard deviation are and respectively, then is equal to [2025]

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Solution

Q.
The mean and standard deviation of 100 observations are 40 and 5.1, respectively. By mistake one observation is taken as 50 instead of 40. If the correct mean and the correct standard deviation are $μ$ and $σ$ respectively, then $10 (μ + σ)$ is equal to [2025]

1 447

2 445

3 451

4 449

Ans.
(4)

We have, Mean = $(\bar{x}) = 40$

Mean $(\bar{x}) = \frac{\sum_{}^{} x_{i}^{}}{100}$

$\Rightarrow 40 \times 100 = \sum_{}^{} x_{i}^{} \Rightarrow \sum_{}^{} x_{i}^{} = 4000$

Correct mean, $μ = \frac{100 (40) - 50 + 40}{100}$

$= 40 - \frac{1}{10} = 39.9$

Also, S.D.( $σ$ ) = 5.1

$\Rightarrow {(5.1)}^{2} = \frac{\sum_{}^{} x_{i}^{2}}{100} - {(\bar{x})}^{2}$

$\Rightarrow \sum_{}^{} x_{i}^{2} = 100 \times {(40)}^{2} + 100 {(5.1)}^{2}$

$= 16 \times 10^{4} + {(5.1)}^{2} \times 100 = 162601$

Correct $σ^{2} = \frac{\sum_{}^{} x_{i}^{2} - 50^{2} + 40^{2}}{100} - {(μ)}^{2}$

$= 1617.01 - 1592.01 = 25 \Rightarrow σ = 5$

$∴ 10 (μ + σ) = 10 (39.9 + 5) = 10 \times 44.9 = 449$ .

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