Question

NBA players have heights that are normally distributed with μ =79 in. and σ =3.5 in.If one player is selected, find the probability that he is:a.) shorter than 75 inb.) between 73 and 83 inC.) What height would cut off the tallest 7% of players?Now, 16 players are selected.d.) Find the probability that their average height is above 80 in.

Answer 1

Given

$μ=79in.\text{ }and\text{ }σ=3.5in.$

Explanation

Part A: shorter than 75 in

First we will find the z-score

$\begin{gathered} z_{75}=\frac{x-\mu}{\sigma}=\frac{75-79}{3.5}=-1.14286 \\ \end{gathered}$

We can then get the P-value from Z-Table:

Answer:

$P\left(x<75\right)=0.12655$

Part B: Between 73 and 83 in

$\begin{gathered} When\text{ x =73} \\ Zscore=\frac{x-μ}{\sigma}=\frac{73-79}{3.5}=-1.71429 \\ When\text{ x =83} \\ Zscore=\frac{x-μ}{\sigma}=\frac{83-79}{3.5}=1.14286 \\ \end{gathered}$

We can then get the P-value from Z-Table for the range 73 to 83

$P\left(-1.71429

Answer:

[tex]P(73Part C

To solve this, we will find the z score for p(x>z)=7% . Using the z-score calculator.

[tex]\begin{gathered} For\text{ }p(x>z)=0.07 \\ z=1.476 \end{gathered}$

Then we will find the value of x using the z score formula;

$\begin{gathered} z=\frac{x-\mu}{\sigma} \\ 1.476=\frac{x-79}{3.5} \\ 1.476\times3.5=x-79 \\ switch\text{ sides} \\ x-79=5.166 \\ x=79+5.166 \\ x=84.166 \end{gathered}$

Answer: Approximately 84

Part D

Since a sample of 16 students is now selected, the formula to use will be changed to accommodate the sample.

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

Therefore, the z score becomes

$z=\frac{80-79}{\frac{3.5}{\sqrt{16}}}=\frac{1}{\frac{3.5}{4}}=\frac{4}{3.5}=1.14285$

Using the z-score calculator

$p(x>80)=p(x>1.14285)=0.12655$

Answer: 0.12655