Question

The heights of the $430$National Basketball Association players were listed on team rosters at the start of the $2005–2006$season. The heights of basketball players have an approximately normal distribution with mean, $\mu =79$inches and a standard deviation, $\sigma =3.89$inches. For each of the following heights, calculate the z-score and interpret it using complete sentences

a. $77$inches

b. $85$inches

c. If an NBA player reported his height had a z-score of $3.5$, would you believe him? Explain your answer

Short answer

1. The z-score $77$inch height is $-0.5141$that is negative So, the$77$inch is shorter that the average height of players.
2. The z-score of $85$is $1.5424$that is positive. So the height$85$ is taller than the height of the all players.
3. The obtained height is $92.615$ inches and it is more than the mean height of $79$ inches. So the claim of the players can not believe.

Step-by-step solution

Given information (Part a)

Given in the question that

Mean$=79$

Standard deviation$=3.89$

$X~N(79,3.89)$

Solution (Part a)

We need to calculate the z-score of $x=77$

$z=\frac{x-\mu }{\sigma }$

$=\frac{77-79}{3.89}$

$=-0.5141$

Given information (Part b)

Given in the question that

Mean$=79$

Standard deviation$=3.89$

$X~N(79,3.89)$

Solution (part b)

Here we need to find the z-score of $x=85$

$z=\frac{x-\mu }{\sigma }$

$=\frac{85-79}{3.89}$

$=1.5424$

Given information (Part c)

Given in the question that

Mean$=79$

Standard deviation $=3.89$

$X~N(79,3.89)$

Solution (Part c)

Here the z-score is given as $3.5$.So, we need to find the height of the players.

$z=\frac{x-\mu }{\sigma }$

$3.5=\frac{x-79}{3.89}$

$x=3.5\times 3.89+79$

$=92.615$