Question

In a survey of women in the United States (ages 20-29) the mean height was 64.2 inches with a standard deviation of 2.9 inches.(a) What height represents the 95th percentile?(B) What height represents the first quartile

Answer 1

Given:

Mean = 64.2

Standard deviation = 2.9 inches

Let's solve for the following:

• (a). What height represents the 95th percentile?

At the 95th percentile, the z-score is 1.645

Apply the z-score formula:

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

Where:

z = 1.645

μ = 64.2

σ = 2.9

Let's solve for x:

Rewrite the equation for x:

$\begin{gathered} x=z\sigma+\mu \\ \\ x=1.645\times2.9+64.2 \\ \\ x=4.7705+64.2 \\ \\ x=68.97 \end{gathered}$

Therefore, the height that represents the 95th percentile is 68.97 inches.

• (b). , What height represents the first quartile?

We have:

The z-value for the height of the first quartile is:

P(Z < z) = 0.25

Using the standard normal distribution table, we have:

NORMSINV(0.25) = -0.67

z = -0.67

Now, apply the z-score formula:

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

Where:

z = -0.67

u = 64.2

σ = 2.9

Let's solve for x:

$\begin{gathered} x=z\sigma+\mu \\ \\ x=-0.67\times2.9+64.2 \\ \\ x=62.26 \end{gathered}$

Therefore, the height that represents the first quartile is 62.26 inches.

ANSWER:

• (a). 68.97 inches

,

• (b). 62.26 inches