Question

In a survey of US men, the heights in the 20-29 age group were normally distributed, with a mean of 69.4 inches and a standard deviation of 2.9 inches. Find the probability that a randomly selected study participant has a height that is
Less than 66 inches,


Between 66 and 72 inches,


More than 72 inches.


Are any of the above events unusual? Why or why not?

Answer 1

Using the normal distribution, it is found that:

• There is a 0.121 = 12.1% probability that a randomly selected study participant has a height that is less than 66 inches.

• There is a 0.695 = 69.5% probability that a randomly selected study participant has a height that is between 66 and 72 inches.

• There is a 0.184 = 18.4% probability that a randomly selected study participant has a height that is more than 72 inches.

• Since for all events, |Z| < 2, none are unusual.

Normal Probability Distribution

In a normal distribution with mean $\mu$ and standard deviation $\sigma$, the z-score of a measure X is given by:

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

• It 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, which is the percentile of X.

• If for a measure X, |Z| > 2, the measure is considered either unusually low or unusually high.

In this problem:

• The mean is of 69.4 inches, hence $\mu = 69.4$.

• The standard deviation is of 2.9 inches, hence $\sigma = 2.9$.

The probability that a randomly selected study participant has a height that is less than 66 inches is the p-value of Z when X = 66, hence:

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

$Z = \frac{66 - 69.4}{2.9}$

$Z = -1.17$

$Z = -1.17$ has a p-value of 0.121.

There is a 0.121 = 12.1% probability that a randomly selected study participant has a height that is less than 66 inches.

The probability that a randomly selected study participant has a height that is between 66 and 72 inches is the p-value of Z when X = 72 subtracted by the p-value of Z when X = 66, hence:

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

$Z = \frac{72 - 69.4}{2.9}$

$Z = 0.9$

$Z = 0.9$ has a p-value of 0.816.

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

$Z = \frac{66 - 69.4}{2.9}$

$Z = -1.17$

$Z = -1.17$ has a p-value of 0.121.

0.816 - 0.121 = 0.695.

There is a 0.695 = 69.5% probability that a randomly selected study participant has a height that is between 66 and 72 inches.

The probability that a randomly selected study participant has a height that is more than 72 inches is 1 subtracted by the p-value of Z when X = 72, hence:

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

$Z = \frac{72 - 69.4}{2.9}$

$Z = 0.9$

$Z = 0.9$ has a p-value of 0.816.

1 - 0.816 = 0.184.

There is a 0.184 = 18.4% probability that a randomly selected study participant has a height that is more than 72 inches.

Since for all events, |Z| < 2, none are unusual.

You can learn more about the normal distribution at https://brainly.com/question/24663213