Question

Navy PilotsThe US Navy requires that fighter pilots have heights between 62 inches and78 inches.(a) Find the percentage of women meeting the height requirement.(b) Find the percentage of men meeting the height requirement.(c) If the height requirements are changed to exclude the tallest 10% of men and the shortest10% of women, what are the new height requirements?

Answer 1

The first part of the question is missing and it says;

Use these parameters: Men's heights are normally distributed with mean 68.6 in. and standard deviation 2.8 in. Women's heights are normally distributed with mean 63.7 in. and standard deviation 2.9 in.

Answer:

A) Percentage of women meeting the height requirement = 72.24%

B) Percentage of men meeting the height requirement = 0.875%

C) Corresponding women's height =67.42 inches while corresponding men's height = 72.19 inches

Step-by-step explanation:

From the question,

For men;

Mean μ = 68.6 in

Standard deviation σ = 2.8 in

For women;

Mean μ = 63.7 in

Standard deviation σ = 2.9 in

Now let's calculate the standardized scores;

The formula is z = (x - μ)/σ

A) For women;

Z = (62 - 63.7)/2.9 = - 0.59

Z = (78 - 63.7)/2.9 = 4.93

The original question cam be framed as;

P(62 < X < 78).

So thus, the probability of only women will take the form of;

P(-0.59 < Z < 4.93) = P(Z<4.93) - P(Z > - 0.59)

From the normal probability table attached, when we interpolate, we'll arrive at P(Z<4.93) = 0.9999996

And P(Z > - 0.59) = 0.277595

Thus;

P(Z<4.93) - P(Z > - 0.59) =0.9999996 - 0.277595 = 0.7224

So, percentage of women meeting the height requirement is 72.24%.

B) For men;

Z = (62 - 68.6)/2.8 = -2.36

Z = (78 - 68.6)/2.8 = 3.36

Thus, the probability of only men will take the form of;

P(-2.36 < Z < 3.36) = P(Z<3.36) - P(Z > - 2.36)

From the normal probability table attached, when we interpolate, we'll arrive at P(Z<3.36) = 0.99961

And P(Z > -2.36) = 0.99086

Thus;

P(Z<3.36) - P(Z > -2.36) 0.99961 - 0.99086 = 0.00875

So, percentage of women meeting the height requirement is 72.24%.

B)For women;

Z = (62 - 63.7)/2.9 = - 0.59

Z = (78 - 63.7)/2.9 = 4.93

The original question cam be framed as;

P(62 < X < 78).

So thus, the probability of only women will take the form of;

P(-0.59 < Z < 4.93) = P(Z<4.93) - P(Z > - 0.59)

From the normal probability table attached, when we interpolate, we'll arrive at P(Z<4.93) = 0.9999996

And P(Z > - 0.59) = 0.277595

Thus;

P(Z<4.93) - P(Z > - 0.59) =0.9999996 - 0.277595 = 0.00875

So, percentage of women meeting the height requirement is 0.875%

C) Since the height requirements are changed to exclude the tallest 10% of men and the shortest10% of women.

For women;

Let's find the z-value with a right-tail of 10%. From the second table i attached ;

invNorm(0.90) = 1.2816

Thus, the corresponding women's height:: x = (1.2816 x 2.9) + 63.7= 67.42 inches

For men;

We have seen that,

invNorm(0.90) = 1.2816

Thus ;

Thus, the corresponding men's height:: x = (1.2816 x 2.8) + 68.6 = 72.19 inches