Question

The probability distribution function of a random variable \( X \) is shown in the figure. From this distribution, random samples with sample size \( n = 68 \) are taken. If \( \bar{X} \) is the sample mean, the standard deviation of the probability distribution of \( \bar{X} \), i.e. \( \sigma_{\bar{X}} \), is \(\underline{\hspace{1cm}}\) (rounded off to 3 decimal places). 

Hint:

For sampling distributions, the standard deviation of the sample mean decreases as sample size increases: \( \sigma_{\bar{X}} = \frac{\sigma}{\sqrt{n}} \).

Answer 1

Correct Answer: 0.069

From the figure, the distribution of \( X \) is a uniform distribution from 1 to 3.
The variance \( \sigma^2 \) of a uniform distribution \( U(a, b) \) is given by:
\[ \sigma^2 = \frac{(b - a)^2}{12} \]
For the given distribution \( a = 1 \) and \( b = 3 \):
\[ \sigma^2 = \frac{(3 - 1)^2}{12} = \frac{4}{12} = \frac{1}{3} \approx 0.3333 \]
The standard deviation \( \sigma = \sqrt{0.3333} \approx 0.5774 \).
For the sampling distribution of the sample mean \( \bar{X} \), the standard deviation is:
\[ \sigma_{\bar{X}} = \frac{\sigma}{\sqrt{n}} \]
Substitute \( \sigma = 0.5774 \) and \( n = 68 \):
\[ \sigma_{\bar{X}} = \frac{0.5774}{\sqrt{68}} \approx 0.069 \]
Rounded to three decimal places: \[ \sigma_{\bar{X}} \approx 0.071 \]