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Forward grip reaches of women are distributed normally with a mean equal to 686 mm and standard deviation equal to 34 mm.

a.

It is given that 95% can reach the dashboard.

Let\({z_p}\)denote the z-score of the minimum value of the forward grip reach.

This implies the following expression:

\(\begin{aligned}{c}P\left( {z \ge {z_p}} \right) = 0.95\\1 - P\left( {z < {z_p}} \right) = 0.95\\P\left( {z < {z_p}} \right) = 0.05\end{aligned}\)

Thus, the value of the z-score for an area under the curve equal to 0.05 is equal to -1.645.

It is known that:

\(\begin{aligned}{l}\mu = 686\;mm\\\sigma = 34\;mm\end{aligned}\)

Let X denote the shortest forward grip reach.

To compute the value of the shortest forward grip reach, the following calculations are done:

\(\begin{aligned}{c}z = \frac{{x - \mu }}{\sigma }\\ - 1.645 = \frac{{x - 686}}{{34}}\\x = \left( { - 1.645} \right)34 + 686\\ = 630.07\\ \approx 630\end{aligned}\)

Therefore, if 95% of women can reach the car dashboard, the value of the shortest forward grip reach is equal to 630mm.

b.

Let X denote the forward grip reach value.

The car dashboard can be reached by women with a grip reach greater than 650 mm.

Therefore, women with forward grip reach less than 650 mm cannot reach the dashboard.

The proportion of women who cannot reach the dashboard is computed below:

\(\begin{aligned}{c}P\left( {X < 650} \right) = P\left( {\frac{{X - \mu }}{\sigma } < \frac{{650 - \mu }}{\sigma }} \right)\\ = P\left( {Z < \frac{{650 - 686}}{{34}}} \right)\\ = P\left( {Z < - 1.06} \right)\\ = 1 - P\left( {Z < 1.06} \right)\end{aligned}\)

\(\begin{aligned}{c} = 1 - 0.8554\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\left( {\therefore {\rm{From}}\;{\rm{Standard}}\;{\rm{normal}}\;{\rm{table}}} \right)\\ = 0.1446\\ = 14.46\% \end{aligned}\)

Therefore, the percentage of women who cannot reach the dashboard is equal to 14.46%.

The percentage is too high, indicating that a significant portion of women will be left out.

c.

If the forward grip reach value denoted by X follows normal distribution with mean\(\left( \mu \right)\)and standard deviation\(\left( \sigma \right)\), then the mean forward grip reach value follows normal distribution with mean\(\left( \mu \right)\)and standard deviation\(\left( {\frac{\sigma }{{\sqrt n }}} \right)\).

Here, n is given to be equal to 16.

The probability of selecting 16 women with a mean forward grip reach value greater than 680 mm is computed below:

\(\begin{aligned}{c}P\left( {\bar X > 680} \right) = P\left( {\frac{{\bar X > \mu }}{{\frac{\sigma }{{\sqrt n }}}} > \frac{{680 - \mu }}{{\frac{\sigma }{{\sqrt n }}}}} \right)\\ = P\left( {Z > \frac{{680 - 686}}{{\frac{{34}}{{\sqrt {16} }}}}} \right)\\ = P\left( {Z > - 0.71} \right)\end{aligned}\)

Simplifying this further, the following value is obtained using the standard normal table:

\(\begin{aligned}{c}P\left( {Z > - 0.71} \right) = P\left( {Z < 0.71} \right)\\ = 0.7611\end{aligned}\)

Therefore, the probability of selecting 16 women with a mean forward grip reach value greater than 680 mm is equal to 0.7611.

Only individual women occupy the driver's seat and not a group of 16 women; hence this conclusion has no effect on the design.