Questions: The scores on an examination in biology are approximately normally distributed with mean 500 and an unknown standard deviation. The following is a random sample of scores from this examination. 403, 411, 455, 494, 525 Find a 95% confidence interval for the population standard deviation. Then give its lower limit and upper limit, Carry your intermediate computations to at least three decimal places. Round your answers to at least two decimal places.

Solution Steps

The sample mean \( \mu \) is calculated as follows: \[ \mu = \frac{\sum x_i}{n} = \frac{2288}{5} = 457.6 \]

The sample variance \( s^2 \) is given by: \[ s^2 = \frac{\sum (x_i - \mu)^2}{n-1} = 2756.8 \]

The sample standard deviation \( s \) is then: \[ s = \sqrt{2756.8} = 52.505 \]

The confidence interval for the variance of a single population with known population mean is calculated using the formula: \[ \left(\frac{(n - 1)s^2}{\chi^2_{\alpha/2}}, \frac{(n - 1)s^2}{\chi^2_{1 - \alpha/2}}\right) \]

Substituting the values: \[ = \left(\frac{(5 - 1) \times 2756.8}{\chi^2_{\alpha/2}}, \frac{(5 - 1) \times 2756.8}{\chi^2_{1 - \alpha/2}}\right) \] \[ = \left(\frac{4 \times 2756.8}{\chi^2_{\alpha/2}}, \frac{4 \times 2756.8}{\chi^2_{1 - \alpha/2}}\right) \]

Assuming \( \chi^2_{\alpha/2} \) and \( \chi^2_{1 - \alpha/2} \) are known, we find: \[ CI = (989.58, 22763.79) \]

The confidence interval for the standard deviation is derived from the variance confidence interval: \[ \text{Lower limit} = \sqrt{989.58} \quad \text{and} \quad \text{Upper limit} = \sqrt{22763.79} \]

Calculating these gives: \[ \text{Lower limit} = 31.46 \quad \text{and} \quad \text{Upper limit} = 150.88 \]

Final Answer