Question

Suppose the heights of 18-year-old men are approximately normally distributed, with mean 69 inches and standard deviation 2 inches.

(a) What is the probability that an 18-year-old man selected at random is between 68 and 70 inches tall?
(b) If a random sample of sixteen 18-year-old men is selected, what is the probability that the mean height x is between 68 and 70 inches?

Answer 1

Answer:

a) 38.3% probability that an 18-year-old man selected at random is between 68 and 70 inches tall.

b) 95.44% probability that the mean height x is between 68 and 70 inches.

Step-by-step explanation:

To solve this question, it is important to know the normal probability distribution and the Central Limit Theorem.

Normal probability distribution

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:

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

The Z-score measures how many standard deviations the measure 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.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a random variable X, with mean $\mu$ and standard deviation $\sigma$, a large sample size can be approximated to a normal distribution with mean $\mu$ and standard deviation $\frac{\sigma}{\sqrt{n}}$.

In this problem, we have that:

$\mu = 69, \sigma = 2$

(a) What is the probability that an 18-year-old man selected at random is between 68 and 70 inches tall?

This is the pvalue of Z when X = 70 subtracted by the pvalue of Z when X = 68. So

X = 70

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

$Z = \frac{70 - 69}{2}$

$Z = 0.5$

$Z = 0.5$ has a pvalue of 0.6915.

X = 68

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

$Z = \frac{68 - 69}{2}$

$Z = -0.5$

$Z = -0.5$ has a pvalue of 0.3085.

So there is a 0.6915 - 0.3085 = 0.383 = 38.3% probability that an 18-year-old man selected at random is between 68 and 70 inches tall.

(b) If a random sample of sixteen 18-year-old men is selected, what is the probability that the mean height x is between 68 and 70 inches?

Now we use the Central Limit Theorem, with $n = 16, s = \frac{2}{\sqrt{16}} = 0.5$

The probability is also the pvalue of Z when X = 70 subtracted by the pvalue of Z when X = 68, but with s as the standard deviation. So

X = 70

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

$Z = \frac{70 - 69}{0.5}$

$Z = 2$

$Z = 2$ has a pvalue of 0.9772.

X = 68

$Z = \frac{X - \mu}{0.5}$

$Z = \frac{68 - 69}{0.5}$

$Z = -2$

$Z = -2$ has a pvalue of 0.0228.

So there is a 0.9772 - 0.0228 = 0.9544 = 95.44% probability that the mean height x is between 68 and 70 inches.