Question

A student scores the following marks in five tests: 45, 54, 41, 57, 43. His score is not known for the sixth test. If the mean score is 48 in the six tests, then the standard deviation of the marks in six tests is:

• A. \( \frac{100}{3} \)
• B. \( \frac{10}{3} \)
• C. \( \frac{10}{\sqrt{3}} \)
• D. \( \frac{100}{\sqrt{3}} \)

Hint:

To calculate the standard deviation, first find the variance using \( \frac{1}{n} \sum_{i=1}^{n} (x_i - \mu)^2 \), then take the square root of the variance.

Answer 1

The Correct Option is C

Step 1: Sum the five provided marks: 45, 54, 41, 57, and 43. The sum is:
\[
45 + 54 + 41 + 57 + 43 = 240.
\]

Step 2: Determine the sixth test score using the mean. The mean score for six tests is 48. The total sum of all six marks is:
\[
\text{Sum of six marks} = 48 \times 6 = 288.
\]
The sixth test score is therefore:
\[
x_6 = 288 - 240 = 48.
\]

Step 3: Calculate the variance. The six marks are 45, 54, 41, 57, 43, and 48, with a mean \( \mu = 48 \). The variance is computed as:
\[
\text{Variance} = \frac{1}{6} \sum_{i=1}^{6} (x_i - \mu)^2
\]
Calculate \( (x_i - 48)^2 \) for each mark:
- \( (45 - 48)^2 = 9 \)
- \( (54 - 48)^2 = 36 \)
- \( (41 - 48)^2 = 49 \)
- \( (57 - 48)^2 = 81 \)
- \( (43 - 48)^2 = 25 \)
- \( (48 - 48)^2 = 0 \)

The variance is:
\[
\text{Variance} = \frac{9 + 36 + 49 + 81 + 25 + 0}{6} = \frac{200}{6} = \frac{100}{3}.
\]

Step 4: Calculate the standard deviation. The standard deviation is the square root of the variance:
\[
\text{Standard Deviation} = \sqrt{\frac{100}{3}} = \frac{10}{\sqrt{3}}.
\]