Question

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

Answer 1

Answer:

a) $\sigma = 0.167$

b) We need a sample of at least 282 young men.

Step-by-step explanation:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$, the zscore of a measure X is given by

$Z = \frac{X - \mu}{\sigma}$

This Zscore is how many standard deviations the value of the measure X is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

(a) What standard deviation must x have so that 99.7% of allsamples give an x within one-half inch of μ?

To solve this problem, we use the 68-95-99.7 rule. This rule states that:

68% of the measures are within 1 standard deviation of the mean.

95% of the measures are within 2 standard deviations of the mean.

99.7% of the measures are within 3 standard deviations of the mean.

In this problem, we want 99.7% of all samples give X within one-half inch of $\mu$. So $X - \mu = 0.5$ must have $Z = 3$ and $X - \mu = -0.5$ must have $Z = -3$.

So

$Z = \frac{X - \mu}{\sigma}$

$3 = \frac{0.5}{\sigma}$

$3\sigma = 0.5$

$\sigma = \frac{0.5}{3}$

$\sigma = 0.167$

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

You know from government data that heights of young men are approximately Normal with standard deviation about 2.8 inches. This means that $\sigma = 2.8$

The standard deviation of a sample of n young man is given by the following formula

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

We want to have $s = 0.167$

$0.167 = \frac{2.8}{\sqrt{n}}$

$0.167\sqrt{n} = 2.8$

$\sqrt{n} = \frac{2.8}{0.167}$

$\sqrt{n} = 16.77$

$\sqrt{n}^{2} = 16.77^{2}$

$n = 281.23$

We need a sample of at least 282 young men.