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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:

User Shenise
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Answer:

For a height of 66 inches, Z = 0.65.

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 = (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 = (X - \mu)/(\sigma)


Z = (66 - 64.3)/(2.6)


Z = 0.65

For a height of 66 inches, Z = 0.65.

User Zet
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