The stature of men is normally distributed, with a mean of 69.0 inches and a standard deviation of 2.8 inches. The height of women is normally distributed, with a mean of 63.6 inches and a standard deviation of 2.5 inches. Modeling academy standards require women to be models taller than 66 inches (or 5 feet 6 inches). What percentage of women meet this requirement?

Question
Modeling
asked 2021-02-11
The stature of men is normally distributed, with a mean of 69.0 inches and a standard deviation of 2.8 inches. The height of women is normally distributed, with a mean of 63.6 inches and a standard deviation of 2.5 inches. Modeling academy standards require women to be models taller than 66 inches (or 5 feet 6 inches). What percentage of women meet this requirement?

Answers (1)

2021-02-12
Step 1 From the given information, the mean of women is 63.6 inches and standard deviation is 2.5 inches. Let X be the height of the women follows normal distribution with \(\mu=63.6\ and\ \sigma=2.5.\) Modeling academy standards require women to be models taller than 66 inches Step 2 The percentage of women meet this requirement is, Percentage \(=P(X\ >\ 66)^{*}\ 100\%\)
\(=P\left(\frac{X\ -\ \mu}{\sigma}\ >\ (66\ -\ \mu)\sigma\right)^{*}\ 100\%\)
\(=P\left(z\ >\ \frac{66\ -\ 63.6}{2.5}\right)^{*}\ 100\%\)
\(=P(z\ >\ 0.96)^{*}\ 100\%\)
\(=[1\ -\ P(z\ \leq\ 0.96)]^{*}\ 100\%\)
\(=0.1685^{*}\ 100\%\) from the excel function, \(=1\ −\ NORM.DIST(0.96,\ 0,\ 1,\ TRUE)\)
\(=16.85\%\)
0

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