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. \( 445 \)
• B. \( 451 \)
• C. \( 447 \)
• D. \( 449 \)

Hint:

When dealing with problems involving the correction of mistaken observations, carefully compute the sum, mean, and standard deviation based on the correct data.

Answer 1

The Correct Option is D

We are given: - The mean of 100 observations is 40.
- The standard deviation of 100 observations is 5.1.
- One observation is mistakenly taken as 50 instead of 40.
Let the correct mean and standard deviation be \( \mu \) and \( \sigma \), respectively.
Step 1: Calculation of Correct Mean \( \mu \)
The incorrect mean is given by: \[ \text{Incorrect mean} = \frac{\sum x}{100} = 40 \] Since the incorrect observation is taken as 50 instead of 40, the incorrect sum is: \[ \sum x = 100 \times 40 = 4000 \] Now, the correct sum is: \[ \text{Correct sum} = 4000 - 50 + 40 = 3990 \]
Thus, the correct mean is: \[ \mu = \frac{3990}{100} = 39.9 \]
Step 2: Calculation of Correct Standard Deviation \( \sigma \)
The incorrect standard deviation is 5.1. The formula for standard deviation is: \[ \sigma = \sqrt{\frac{1}{N} \sum (x_i - \mu)^2} \] After applying the correction to the mistaken observation, we find: \[ 10(\mu + \sigma) = 449 \]
Thus, the correct answer is 449.

Answer 2

Actual mean: \[ \mu = \frac{100(40) - 50 + 40}{100} \] \[ \mu = 40 - \frac{1}{10} = 39.9 \] Incorrect variance: \[ (5.1)^2 = \frac{\sum x_i^2}{100} - (\bar{x})^2 \] \[ \sum x_i^2 = 100 \times (40^2) + 100 \times (5.1)^2 \] \[ \sum x_i^2 = 16 \times 10^4 + (5.1)^2 \times 100 = 162601 \] \[ \sigma^2 = \frac{\sum x_i^2 - 50^2 + 40^2}{100} - (\mu)^2 \] \[ \sigma^2 = 1617.01 - (39.9)^2 = 25 \] \[ \sigma = 5 \] \[ 10(\mu + \sigma) = 10(39.9 + 5) \] \[ 10 \times 44.9 = 449 \] \[ \boxed{10(\mu + \sigma) = 449} \]