Question

The average adult male has a height of 70 inches, with a standard deviation of 3 inches. What percentage of adult males have a height between 65 and 75 inches?

Answer 1

The rule of the z-score is

`z=(x-\mu)/(\sigma)`

μ is the mean

σ is the standard deviation

Since the average of the height of an adult male is 70 inches, then

`\mu=70`

Since the standard deviation is 3 inches, then

`\sigma=3`

We need to find the percentage of the height of the adult males between 65 and 75 inches, then

`x_1=65,x_2=75`

We will find the z-score of them

`\begin{gathered} z_1=(65-70)/(3) \\ z_1=-(5)/(3) \\ z_1=-1.6667 \end{gathered}`
`\begin{gathered} z_2=(75-70)/(3) \\ z_2=(5)/(3) \\ z_2=1.6667 \end{gathered}`

We will use the given table to find their areas

`\begin{gathered} P(x>65)=0.04746 \\ P(x<75)=0.95254 \end{gathered}`

We will subtract both values to find the answer

[tex]\begin{gathered} P(65We will change it to percent, then[tex]\begin{gathered} P(65The answer is 90.5%