Calculating Correct Mean ($\mu$)
We are given:
- Number of observations, $n = 100$
- Incorrect Mean, $\bar{x}_{inc} = 40$
- Incorrect standard deviation, $s_{inc} = 5.1$
- Incorrect observation recorded = 50
- Correct observation = 40
First, calculate the sum of the incorrect observations:
$ \text{Sum}_{inc} = n \times \bar{x}_{inc} = 100 \times 40 = 4000 $Now, adjust the sum to find the correct sum:
$ \text{Sum}_{corr} = \text{Sum}_{inc} - (\text{Incorrect Observation}) + (\text{Correct Observation}) $ $ \text{Sum}_{corr} = 4000 - 50 + 40 = 3990 $Calculate the correct mean ($\mu$):
$ \mu = \frac{\text{Sum}_{corr}}{n} = \frac{3990}{100} = 39.9 $Calculating Correct Standard Deviation ($\sigma$)
We use the formula for variance: $Var = \frac{\sum x^2}{n} - (\text{Mean})^2$.
First, find the incorrect sum of squares ($\sum x^2_{inc}$):
$ \text{Variance}_{inc} = (s_{inc})^2 = (5.1)^2 = 26.01 $ $ \sum x^2_{inc} = n \times (\text{Variance}_{inc} + (\bar{x}_{inc})^2) $ $ \sum x^2_{inc} = 100 \times (26.01 + (40)^2) = 100 \times (26.01 + 1600) = 100 \times 1626.01 = 162601 $Adjust the sum of squares to find the correct sum of squares:
$ \sum x^2_{corr} = \sum x^2_{inc} - (\text{Incorrect Observation})^2 + (\text{Correct Observation})^2 $ $ \sum x^2_{corr} = 162601 - (50)^2 + (40)^2 = 162601 - 2500 + 1600 = 161701 $Now, calculate the correct variance:
$ \text{Variance}_{corr} = \frac{\sum x^2_{corr}}{n} - \mu^2 $ $ \text{Variance}_{corr} = \frac{161701}{100} - (39.9)^2 = 1617.01 - 1592.01 = 25 $Calculate the correct standard deviation ($\sigma$):
$ \sigma = \sqrt{\text{Variance}_{corr}} = \sqrt{25} = 5 $Final Calculation: $10(\mu + \sigma)$
Substitute the values of $\mu$ and $\sigma$:
$ 10(\mu + \sigma) = 10(39.9 + 5) = 10(44.9) = 449 $Thus, the value of $10(\mu + \sigma)$ is 449.
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