Tall Clubs The social organization Tall Clubs International has a requirement that women must be at least 70 in. tall. Assume that women have normally distributed heights with a mean of 63.7 in. and a standard deviation of 2.9 in. (based on Data Set 1 “Body Data” in Appendix B).

a. Find the percentage of women who satisfy the height requirement.

b. If the height requirement is to be changed so that the tallest 2.5% of women are eligible, what is the new height requirement?

The height of women is normally distributed with mean 63.7 in. and standard deviation is 2.9 in.

The height requirement is at least 70 in.

Let X be the random variable for heights of women.

\(X \sim N\left( {\mu = 63.7,{\sigma ^2} = {{2.9}^2}} \right)\)

a. The probability that the height of a randomly selected woman is at least 70 inches expressed as \(P\left( {X \ge 70} \right) = 1 - P\left( {X < 70} \right)\) .

The z-score corresponding to 70 is computed as,

\(\begin{array}{c}{\bf{z = }}\frac{{{\bf{x - \mu }}}}{{\bf{\sigma }}}\\ = \frac{{70 - 63.7}}{{2.9}}\\ = 2.17\end{array}\)

Refer to the standard normal table for the cumulative probability for the z-score 2.17.

The value corresponds to row 2.1 and column 0.07 represents\(P\left( {Z < 2.17} \right) = 0.9850\).

The probability that the height of women is at least 70 in.,

\(\begin{array}{c}P\left( {X \ge 70} \right) = 1 - 0.9850\\ = 0.015\end{array}\)

Now, the percentage of height of women is given by,

\(0.015 \times 100 = 1.50\% \)

Therefore, the percentage of women that meet the height requirement of least 70 inch is is 1.50%.

b.

Let x denote the minimum height of the tallest 2.5% women and z be the corresponding z-score.

Thus,

\(\begin{array}{c}P\left( {X > x} \right) = 0.025\\P\left( {Z > z} \right) = 0.025\\P\left( {Z < z} \right) = 1 - 0.025\\ = 0.975\end{array}\)

Refer to the standard normal table for the cumulative area of 0.975, which occurs corresponding to row 1.9 and column 0.06, which implies z-score is 1.96.

The required height of women is given by,

\[\begin{array}{c}x = \left( {1.96 \times 2.9} \right) + 63.7\\x = 69.4\end{array}\]

Therefore, the required height of women is 69.4 in.