Question

In a survey men in a certain country (ages 20-29), the mean height was 62.8 inches with a standard deviation of 2.8 inches, what height represents the 99th percentile?

Answer 1

Answer:

the height that represents the 99th percentile is 69.324  inches

Step-by-step explanation:

Given that :

the mean height =  62.8 inches

standard deviation = 2.8  inches

For 99th percentile;

Let X be the random variable;

SO, P(Z≤ z)  = 0.99

From the standard normal z tables

P(Z )= 2.33

The standard z score formula is :

$z = \dfrac{X- \mu}{\sigma}$

$2.33 = \dfrac{X- 62.8}{2.8}$

2.33 × 2.8 = X - 62.8

6.524 = X - 62.8

6.524 +62.8 =  X

69.324  = X

X = 69.324

Therefore; the height that represents the 99th percentile is 69.324  inches