Question

suppose the heights (in inches) of adult females (ages 20 - 29) in the united states are normally distributed with a mean of 64.1 inches and a standard deviation of 2.75 inches.

1. find the percent of women who are less than 61.6 inches tall.

Explanation:

Step1: Calculate the z - score

The formula for the z - score is \(z=\frac{x-\mu}{\sigma}\), where \(x = 61.6\), \(\mu=64.1\), and \(\sigma = 2.75\).

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Step2: Find the probability using the standard normal table

We want to find \(P(X\lt61.6)\), which is equivalent to \(P(Z\lt - 0.91)\) in the standard normal distribution (\(X\) is the height variable and \(Z\) is the standard normal variable).
Looking up the value of \(z=-0.91\) in the standard normal table (the cumulative - distribution function for the standard normal distribution \(\varPhi(z)\)), we find that \(P(Z\lt - 0.91)=0.1814\)

Answer:

Approximately \(18.14\%\) of women are less than \(61.6\) inches tall.