The heights of women aged 20–29 in the United States are approximately Normal with mean 64.1 inches and standard deviation 3.7 inches. Men the same age have mean height 69.4 inches, with standard deviation 3.1 inches.

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Using the normal distribution, it is found that the z-score of the height of a women 5.5 feet tall is 0.51.

What is the missing information?

The problem asks for the z-score of the height of a women 5.5 feet tall.

Normal Probability Distribution

The z-score of a measure X of a normally distributed variable with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex] is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

  • The z-score measures how many standard deviations the measure is above or below the mean.
  • Looking at the z-score table, the p-value associated with this z-score is found, which is the percentile of X.

Considering that each feet has 12 inches, the parameters are given as follows:

[tex]X = 5.5 \times 12 = 66, \mu = 64.1, \sigma = 3.7[/tex]

Hence the z-score is:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

Z = (66 - 64.1)/3.7

Z = 0.51.

The z-score of the height of a women 5.5 feet tall is 0.51.

More can be learned about the normal distribution at https://brainly.com/question/15181104

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