Question

To estimate the mean height μ of male students on your campus, you will measure an SRS of students. You know from government data that heights of young men are approximately Normal with standard deviation about 2.8 inches. You want your sample mean x⎯⎯⎯ to estimate μ with an error of no more than one-half inch in either direction. (a) What standard deviation must x⎯⎯⎯ have so that 95% of all samples give an x⎯⎯⎯ within one-half inch of μ? (Use the 68-95-99.7 rule.) (b) How large an SRS do you need to reduce the standard deviation of x⎯⎯⎯ to the value you found in part(a)?

Answer 1

Answer:

a) The standard deviation of x must be 0.25 inches.

b) We need a sample of 125 to reduce the standard deviation of x⎯⎯⎯ to the value you found in part(a).

Step-by-step explanation:

The 68-95-99.7 states that:

68% percent of the measures of a normally distributed sample are within 1 standard deviation of the mean.

95% percent of the measures of a normally distributed sample are within 2 standard deviations of the mean.

99.7% percent of the measures of a normally distributed sample are within 3 standard deviations of the mean.

The standard deviation of the population is 2.8. This means that $\sigma = 2.8$.

(a) What standard deviation must x⎯⎯⎯ have so that 95% of all samples give an x⎯⎯⎯ within one-half inch of μ?

We want to have a sample in which 2 standard deviations are within 0.5 inches of the mean.

So, the standard deviation of the sample must be:

$2s = 0.5$

$s = 0.25$

The standard deviation of x must be 0.25 inches.

(b) How large an SRS do you need to reduce the standard deviation of x⎯⎯⎯ to the value you found in part(a)?

We have that the standard deviation of a sample of length n is given by the following formula:

$s = \frac{\sigma}{\sqrt{n}}$.

We want $s = 0.25$ and we have $\sigma = 2.8$. So

$0.25 = \frac{2.8}{\sqrt{n}}$

$0.25\sqrt{n} = 2.8$

$\sqrt{n} = 11.2$

$\sqrt{n}^{2} = (11.2)^{2}$

$n = 125.44$

We need a sample of 125 to reduce the standard deviation of x⎯⎯⎯ to the value you found in part(a).