Assume that women's heights are normally distributed with a mean=63.6 in. and standard deviation = 2.5 in.

a) To be eligible for the US Marine Corps, a woman must have a height between 58 in. and 73 in. Find the percentage of women who satisfy that requirement.

b)If the requirement is changed to exclude the shortest 1% and exclude the tallest 1%, find the heights that are acceptable.

a) To find the percentage of women who satisfy the height requirement for the US Marine Corps, we need to use the standard normal distribution table.

First, we need to convert the given heights to z-scores using the formula:
z = (x - μ) / σ

For the lower height requirement:
z1 = (58 - 63.6) / 2.5 = -2.24

For the upper height requirement:
z2 = (73 - 63.6) / 2.5 = 3.76

Next, we look up the corresponding cumulative probability for each z-score in the standard normal table.

The cumulative probability for z1 (lower height requirement) is approximately 0.0122.

The cumulative probability for z2 (upper height requirement) is approximately 0.9998.

To find the percentage of women who satisfy the requirement, we subtract the cumulative probabilities:
Percentage = (0.9998 - 0.0122) * 100
Percentage ≈ 98.76%

Therefore, approximately 98.76% of women satisfy the height requirement for the US Marine Corps.

b) If the requirement is changed to exclude the shortest 1% and exclude the tallest 1%, we need to find the heights that correspond to these cutoff probabilities.

The cutoff probability for excluding the shortest 1% is 0.01, so we need to find the z-score that corresponds to this probability. Looking up the z-score for a cumulative probability of 0.01 in the standard normal table, we find z = -2.33.

To find the height corresponding to this z-score, we use the formula:
x1 = μ + (z1 * σ)
x1 = 63.6 + (-2.33 * 2.5) ≈ 58.67 inches

The cutoff probability for excluding the tallest 1% is also 0.01, so we need to find the z-score that corresponds to this probability. Looking up the z-score for a cumulative probability of 0.99 in the standard normal table (complementary probability), we find z = 2.33 (positive instead of negative).

To find the height corresponding to this z-score, we use the same formula:
x2 = μ + (z2 * σ)
x2 = 63.6 + (2.33 * 2.5) ≈ 68.68 inches

Therefore, the acceptable heights are approximately 58.67 inches to 68.68 inches for the changed requirement.

To solve these problems, we will use the properties of the normal distribution to find the appropriate probabilities and heights.

a) To find the percentage of women who satisfy the height requirement of being between 58 in. and 73 in., we need to find the area under the normal distribution curve between these two heights.

Step 1: Standardize the heights
We need to convert the given heights to standard scores (also known as z-scores) to use the standard normal table.

The z-score formula is:
z = (x - μ) / σ

where:
z is the standard score,
x is the given value,
μ is the mean of the distribution, and
σ is the standard deviation of the distribution.

For the lower height limit of 58 in., the z-score will be:
z1 = (58 - 63.6) / 2.5

For the upper height limit of 73 in., the z-score will be:
z2 = (73 - 63.6) / 2.5

Step 2: Use the standard normal table
Consulting the standard normal table, we can find the probabilities associated with these z-scores.

The standard normal table provides the percentage of observations that fall below a particular z-score. Since we want to find the percentage between two z-scores, we need to subtract the lower probability from the higher probability.

So, the percentage of women who satisfy the height requirement is:
Percentage = (Upper probability) - (Lower probability)

b) If the requirement is changed to exclude the shortest 1% and tallest 1%, we need to find the corresponding z-scores that represent these percentiles.

Step 1: Find the z-scores
The standard normal table also provides the z-score corresponding to a particular percentile.

For the shortest 1% height, we need to find the z-score that represents the lower 1% tail. This z-score is found by looking up the cumulative probability in the standard normal table.

The same process applies to finding the z-score for the tallest 1% height, but in this case, we look up the probability in the upper tail of the distribution.

Step 2: Convert z-scores back to heights
Once we have the z-scores, we can convert them back to heights using the formula:

x = μ + (z * σ)

where:
x is the height,
μ is the mean,
σ is the standard deviation, and
z is the appropriate z-score.

Combining the information obtained from steps 1 and 2 will give you the heights that are acceptable for the changed requirement.

Use z-scores for both a) and b).

Formula:

z = (x - mean)/sd

a) Find both z-scores.
First z: x = 58, mean = 63.6, sd = 2.5
Second z: x = 73, mean = 63.6, sd = 2.5

Once you have both z-scores, check a z-table for the probability between the two scores. Convert to a percentage.

b) Find the two z-scores using a z-table for the shortest 1% and the tallest 1%. Once you have both z-scores, find x for both using the formula above, then go from there.

I hope this will help get you started.