Question

Heights On the basis of Data Set 1 “Body Data” in Appendix B, assume that heights of men are normally distributed, with a mean of 68.6 in. and a standard deviation of 2.8 in.

a. The U.S. Coast Guard requires that men must have a height between 60 in. and 80 in. Findthe percentage of men who satisfy that height requirement.

b. Find the probability that 4 randomly selected men have heights with a mean greater than 70 in.

Short answer

a. The percentage of men who satisfy that height requirement to be between 60 (inches) to 80 (inches) would be approximately 99.89%.

b. The probability that 4 randomly selected men have heve heights with a mean greater than 70 (inches). would be approximately 0.1587.

Step-by-step solution

Given information

The given data depicts the height of the men.

The data is normally distributed with mean of 68.6 (inches). and a standard deviation of 2.8 (inches).

Describe the random variable

Define X as the random variable for heights of men.

\(\begin{aligned}{c}X \sim N\left( {\mu ,{\sigma ^2}} \right)\\ \sim N\left( {68.6,{{2.8}^2}} \right)\end{aligned}\)

The z-score is the standard value such that it is the quotient of difference between raw score and mean with the standard deviation measure.

Formula for computing Z-score:

\(Z = \frac{{X - \mu }}{\sigma }\)

Compute the z-score

a.

The Z-score associated both of these given limit values 60(inches) and 80(inches)are computed as follows,

The Z-score for the lower limit (60):

\(\begin{aligned}{c}{z_1} = \frac{{60 - 68.6}}{{2.8}}\\ = - 3.071\end{aligned}\)

The Z-score for the upper limit:

\(\begin{aligned}{c}{z_2} = \frac{{80 - 68.6}}{{2.8}}\\ = 4.071\end{aligned}\)

Compute the probabilities

From the standard normal-ditribution table,the left tailed probability of a z-score is equal to the area under the curve towards the left.

The area under the curve to the left of -3.071 is 0.00107 and the area under the curve to the left of 4.071 is 0.99998.

Therefore, the probability that a random man who satisfy that height requirement would be as follows.

\(\begin{aligned}{c}{\rm{Area}}\;{\rm{between}}\; - 3.071\;{\rm{and}}\;4.071 = {\rm{Area}}\;{\rm{to}}\;{\rm{left}}\;{\rm{of}}\;4.071 - {\rm{Area}}\;{\rm{to}}\;{\rm{left}}\;{\rm{of}} - 3.071\\ = 0.99998 - 0.00107\\ = 0.99891\end{aligned}\)

Thus, approximately 99.89% of men satify the height requirement.

State the distribution of sample means

b.

Define\(\bar X\)be the random variable of the sampling distribution of samples mean of size 4.

As the population is normally distributed, the sampling distribution of sample means would be normal.

\(\begin{aligned}{c}\bar X \sim N\left( {{\mu _{\bar X}} = \mu ,{\sigma _{\bar X}}^2 = {{\left( {\frac{\sigma }{{\sqrt n }}} \right)}^2}} \right)\\ \sim N\left( {68.6,{\sigma _{\bar X}}^2 = {{\left( {\frac{{2.8}}{{\sqrt 4 }}} \right)}^2}} \right)\\ \sim N\left( {68.6,{{1.4}^2}} \right)\end{aligned}\)

Compute the probabily for the sample mean

The probability that 4 randomly selected men have mean heights greater than 70 (inches).

The z-score corresponding to 70 is,

\(\begin{aligned}{c}{z_3} = \frac{{\bar x - {\mu _{\bar x}}}}{{{\sigma _{\bar x}}}}\\ = \frac{{70 - 68.6}}{{1.4}}\\ = 1\end{aligned}\)

Using the standard normal table, the area to the left of z-score 1 is 0.8413.

The probability that mean height is greater than 70 inches is,

\(\begin{aligned}{c}{\rm{Area}}\;{\rm{to}}\;{\rm{right}}\;{\rm{of}}\;1 = 1 - {\rm{Area}}\;{\rm{to}}\;{\rm{left}}\;{\rm{of}}\;1\\ = 1 - 0.8413\\ = 0.1587\end{aligned}\)

Therefore, the probability that 4 randomly selected men have mean heights greater than 70 (inches) would be approximately 0.1587.