Question

The mean and standard deviation of 20 observations were calculated as 10 and 2.5 respectively. It was found that by mistake one data value was taken as 25 instead of 35. If \(\alpha\) and \(\sqrt{\beta}\) are the mean and standard deviation respectively for correct data, then \((\alpha, \beta)\) is :

• A. \((11, 25)\)
• B. \((11, 26)\)
• C. \((10.5, 26)\)
• D. \((10.5, 25)\)

Hint:

In statistics correction problems, remember that \(\sum x_i^2\) is always corrected using the difference of the squares of the wrong and correct values. Also, the variance \(\beta\) is the square of the standard deviation.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
We correct the mean by adjusting the sum of observations and correct the standard deviation by adjusting the sum of the squares of the observations.
Step 2: Key Formula or Approach:
1. \(\bar{x} = \frac{\sum x_i}{n}\). 2. \(\sigma^2 = \frac{\sum x_i^2}{n} - (\bar{x})^2\).
Step 3: Detailed Explanation:
Incorrect values: \(n = 20, \bar{x} = 10, \sigma = 2.5 \implies \sigma^2 = 6.25\). Incorrect \(\sum x_i = 20 \cdot 10 = 200\). Correct \(\sum x_i = 200 - 25 + 35 = 210 \implies \text{Correct mean } \alpha = \frac{210}{20} = 10.5\). Incorrect \(\sum x_i^2\): \[ \sigma^2 = \frac{\sum x_i^2}{20} - 100 = 6.25 \implies \frac{\sum x_i^2}{20} = 106.25 \implies \sum x_i^2 = 2125 \] Correct \(\sum x_i^2 = 2125 - 25^2 + 35^2 = 2125 - 625 + 1225 = 2725\). Correct variance \(\beta\): \[ \beta = \frac{2725}{20} - (10.5)^2 = 136.25 - 110.25 = 26 \] So, \((\alpha, \beta) = (10.5, 26)\).
Step 4: Final Answer:
The pair \((\alpha, \beta)\) is \((10.5, 26)\).