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It is given that the population of the standing heights of women is normally distributed with a mean value equal to 59.7 inches and a standard deviation equal to 2.5 inches. Women with a standing height less than 54 inches find the height of the eye recognition security system to be uncomfortable.

Here, the population mean value is equal to $渭=59.7$.

The population standard deviation is equal to $蟽=2.5$.

The sample value given is equal to x=54 inches.

The following formula is used to convert a given sample value (x=54) to a z-score:

$$\begin{gathered} z=\frac{x-渭}{蟽} \\ =\frac{54-59.7}{2.5} \\ =-2.28 \end{gathered}$$

By referring to the standard normal table, the required probability value can be computed using the value of the z-score.

a.

The probability of getting a standing height less than 54 inches is computed below.

$$\begin{gathered} P\left(x<54\right)=P\left(z<-2.28\right) \\ =0.0113 \end{gathered}$$

By converting the probability value to a percentage, the following value is obtained:

$$\begin{gathered} \text{Percentage}=0.0113脳100\% \\ =1.13\% \end{gathered}$$

Therefore, the percentage of women who will find the height of the eye recognition system uncomfortable is equal to 1.13%.

b.

Let X denote the standing height of men.

Now, it is given that the positioning of the eye recognition system suits the shortest 95% of women.

Thus, the value that separates the bottom 95%of the standing eye height from the highest5% has the following expression:

$P\left(Z<z\right)=0.95$

Now, the corresponding z-score for the left-tailed probability value equal to 0.95 is seen from the table and is approximately equal to 1.645.

Thus, $P\left(z<1.645\right)=0.95$.

The value of the standing eye height corresponding to the z-score of 1.645 is computed below.

$$\begin{gathered} z=\frac{x-渭}{蟽} \\ x=渭+z蟽 \\ =59.7+\left(1.645\right)\left(2.5\right) \\ =63.8 \end{gathered}$$

Therefore, the standing height of women that separates the lowest 95% of standing eye heights from the top 5% is equal to 63.8 inches.