Cholesterol levels for a group of women aged 30-39 follow an approximately normal distribution with mean 190.14 milligrams per deciliter (mg/dl). Medical guidelines state that women with cholesterol levels above 240 mg/dl are considered to have high cholesterol and about 9.3% of women fall into this category.

1. What is the Z-score that corresponds to the top 9.3% (or the 90.7-th percentile) of the standard normal distribution? Round your answer to three decimal places.

2. Find the standard deviation of the distribution in the situation stated above. Round your answer to 1 decimal place.

1. You need to find the corresponding Z-value that corresponds to the top 9.3% of the distribution, i.e. is the value of the standard normal distribution that has above it 0.093 of the distribution and below it is 0.907, symbolically:

P(Z≥z₀)= 0.093

-*or*-

P(Z≤z₀)= 0.907

Since the Z-table shows accumulative probabilities P(Z<Z₁₋α) I'll work with the second expression:

P(Z≤z₀)= 0.907

Now all you have to do is look for the given probability in the body of the table and reach the margins to obtain the corresponding Z value. The first column gives you the integer and first decimal value and the first row gives you the second decimal value:

z₀= 1.323

2.

Using the Z value from 1., the mean Cholesterol level (μ= 190.14 mg/dl) and the Medical guideline that indicates that 9.3% of the women have levels above 240 mg/dl you can clear the standard deviation of the distribution from the Z-formula:

Z= (X- μ)/δ ~N(0;1)

Z= (X- μ)/δ

Z*δ= X- μ

δ=(X- μ)/Z

δ=(240-190.14)/1.323

δ= 37.687 ≅ 37.7 mg/dl

I hope it helps!