Question

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 $\mu$ and $\sigma$ respectively, then $10(\mu + \sigma)$ is equal to

• A. 447
• B. 445
• C. 449
• D. 451

Answer 1

Answer: (C)

449

Answer 2

Explanation of the solution:

1. Calculate the incorrect sum of observations: Given incorrect mean ($\bar{x}_{inc}$) = 40 and number of observations (n) = 100. $\sum x_{inc} = n \times \bar{x}_{inc} = 100 \times 40 = 4000$.

2. Calculate the correct sum of observations: The observation 50 was taken instead of 40. $\sum x_{corr} = \sum x_{inc} - (\text{incorrect observation}) + (\text{correct observation})$ $\sum x_{corr} = 4000 - 50 + 40 = 3990$.

3. Calculate the correct mean ($\mu$): $\mu = \frac{\sum x_{corr}}{n} = \frac{3990}{100} = 39.9$.

4. Calculate the incorrect sum of squares: Given incorrect standard deviation ($\sigma_{inc}$) = 5.1. The formula for variance is $\sigma^2 = \frac{\sum x^2}{n} - (\bar{x})^2$. So, $\sum x^2 = n(\sigma^2 + \bar{x}^2)$. $\sum x_{inc}^2 = 100 \times ((5.1)^2 + (40)^2)$ $\sum x_{inc}^2 = 100 \times (26.01 + 1600)$ $\sum x_{inc}^2 = 100 \times 1626.01 = 162601$.

5. Calculate the correct sum of squares: $\sum x_{corr}^2 = \sum x_{inc}^2 - (\text{incorrect observation})^2 + (\text{correct observation})^2$ $\sum x_{corr}^2 = 162601 - (50)^2 + (40)^2$ $\sum x_{corr}^2 = 162601 - 2500 + 1600$ $\sum x_{corr}^2 = 162601 - 900 = 161701$.

6. Calculate the correct standard deviation ($\sigma$): $\sigma^2 = \frac{\sum x_{corr}^2}{n} - (\mu)^2$ $\sigma^2 = \frac{161701}{100} - (39.9)^2$ $\sigma^2 = 1617.01 - 1592.01$ $\sigma^2 = 25$ $\sigma = \sqrt{25} = 5$.

7. Calculate $10(\mu + \sigma)$: $10(\mu + \sigma) = 10(39.9 + 5)$ $10(\mu + \sigma) = 10(44.9)$ $10(\mu + \sigma) = 449$.