Question

The average woman’s height is 65 inches with a standard deviation of 3.5 inches.What proportion of women are shorter than 71 inches tall?(enter answer as a percent rounded to the nearest hundredth as needed)

Answer 1

Step 1:

In this question, we are given the following:

The average woman’s height is 65 inches with a standard deviation of 3.5 inches.

What proportion of women is shorter than 71 inches tall?

(enter the answer as a percent rounded to the nearest hundredth as needed)

Step 2:

The details of the solution are as follows:

Using the formulae, we have that:

`Z=\frac{x_i\text{ -}\mu}{\sigma}`
`\begin{gathered} where\text{ Z = z- score , x}_i\text{ \lparen Given score \rparen = 71 inches} \\ \mu\text{= Mean \lparen Average\rparen = 65 inches} \\ \sigma\text{ = standard deviation = 3. 5 inches} \end{gathered}`

Step 3:

Computing the values, we have that:

`Z\text{ =}\frac{71\text{ - 65}}{3.\text{ 5}}\text{ = }\frac{6}{3.\text{ 5}}\text{ =1. 7142 \lparen 4 decimal places\rparen}`

Now, we need to use the z-score calculator, to find the:

`Probability\text{ \lparen x< 1.7142 \rparen= ?}`

CONCLUSION:

The percentage of the women that are shorter than 71 inches tall is:

`95.\text{ 68 \% \lparen to the nearest hundredth \rparen}`