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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.