Question

For the women age 18-24 in HANES2, the average height was about 64.3 inches; the SD was about 2.6 inches. Express each height in standard units. (Decimal form; two places after decimal.) 66 inches:

Answer 1

Answer:

For a height of 66 inches, Z = 0.65.

Step-by-step explanation:

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$, the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

The average height was about 64.3 inches; the SD was about 2.6 inches.

This means that $\mu = 64.3, \sigma = 2.6$

66 inches:

The z-score for a height of 66 inches is:

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{66 - 64.3}{2.6}$

$Z = 0.65$

For a height of 66 inches, Z = 0.65.