Question

In the United States, the mean and standard deviation of adult women’s heights are 65 inches (5 feet 5 inches) and 3.5 inches, respectively. Suppose the American adult women’s heights have a normal distribution. a. If a woman is selected at random in the United States, fnd the probability that she is taller than 5 feet 8 inches. b. Find the 72nd percentile of the distribution of heights of American women. c. If 100 women are selected at random in the United States, fnd an approximate probability that exactly 20 of them are taller than 5 feet 8 inches.

Answer 1

Answer:

a) There is a 19.49% probability that she is taller than 5 feet 8 inches.

b) The 72nd percentile of the distribution of heights of American women is 67.0475 inches.

c) There is a 9.85% probability that exactly 20 of them are taller than 5 feet 8 inches.

Step-by-step explanation:

Questions a and b are solved using the zscore formula

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

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

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 This p-value is the probability that the value of the measure is greater than X.

Question c is solved using the binomial probability distribution

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinatios of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

In this problem, we have that:

In the United States, the mean and standard deviation of adult women’s heights are 65 inches (5 feet 5 inches) and 3.5 inches. So $\mu = 65, \sigma = 3.5$

a) If a woman is selected at random in the United States, fnd the probability that she is taller than 5 feet 8 inches.

Each feet has 12 inches. So 5 feet 8 inches is 12*5 + 8 = 68 inches.

So this probability is 1 subtracted by the pvalue of $X = 68$.

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

$Z = \frac{68 - 65}{3.5}$

$Z = 0.86$

$Z = 0.86$ has a pvalue of 0.8051.

So, there is a 1-0.8051 = 0.1949 = 19.49% probability that she is taller than 5 feet 8 inches.

b. Find the 72nd percentile of the distribution of heights of American women.

This is the value of X when Z has a pvalue of 0.72. This is between Z = 0.58 and Z = 59, so we use $Z = 0.585$

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

$0.585 = \frac{X - 65}{3.5}$

$X - 65 = 0.585*3.5$

$X = 67.0475$

The 72nd percentile of the distribution of heights of American women is 67.0475 inches.

c. If 100 women are selected at random in the United States, fnd an approximate probability that exactly 20 of them are taller than 5 feet 8 inches.

For each women, there are only two possible outcomes. Either she is taller than 5 feet 8 inches, or she is smaller. This is why we use the binomial probability distribution to solve this problem.

There are 100 women, so $n = 100$.

From a), we found that the probabilith that a women is taller than 5 feet 8 inches is 0.1949. So $p = 0.1949$.

We want P(X = 20).

Using a binomial probability calculator

$P(X = 20) = C_{100,20}.(0.1949)^{20}.(0.8051)^{80} = 0.0985$

There is a 9.85% probability that exactly 20 of them are taller than 5 feet 8 inches.