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.`

Step-by-step explanation

Part A: shorter than 75 in

First we will find the z-score

`\begin{gathered} z_(75)=(x-\mu)/(\sigma)=(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=(x-μ)/(\sigma)=(73-79)/(3.5)=-1.71429 \\ When\text{ x =83} \\ Zscore=(x-μ)/(\sigma)=(83-79)/(3.5)=1.14286 \\ \end{gathered}`

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

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

`\begin{gathered} z=(x-\mu)/(\sigma) \\ 1.476=(x-79)/(3.5) \\ 1.476*3.5=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=(x-\mu)/((\sigma)/(√(n)))`

Therefore, the z score becomes

`z=(80-79)/((3.5)/(√(16)))=(1)/((3.5)/(4))=(4)/(3.5)=1.14285`

Using the z-score calculator

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

Answer: 0.12655