Math Problem Statement

Assume the cholesterol levels of women can be described by a Normal model with a mean of 188 mg/dL and a standard deviation of 29 mg/dL. Complete parts a through e. Part b) What percent of women do you expect to have cholesterol levels over 210 mg/dL? Part c) What percent of women do you expect to have cholesterol levels between 150 mg/dL and 170 mg/dL?

Solution

Let's go through each part of this question based on the provided image.

The problem describes the cholesterol levels of women as following a normal distribution with:

  • Mean (μ\mu) = 188 mg/dL
  • Standard deviation (σ\sigma) = 29 mg/dL

Part a)

The correct answer for drawing and labeling the Normal model is marked as C in the image. This figure represents the mean (188 mg/dL) in the center and marks each standard deviation interval (68%, 95%, and 99.7%) symmetrically around the mean.

Part b)

For this part, we are asked:

What percent of women do you expect to have cholesterol levels over 210 mg/dL?

To solve this:

  1. Find the z-score for 210 mg/dL: z=21018829=22290.76z = \frac{210 - 188}{29} = \frac{22}{29} \approx 0.76

  2. Use a z-table or calculator to find the probability for z=0.76z = 0.76.

    • The probability for z0.76z \leq 0.76 is approximately 0.7764.
    • Since we want the percentage above 210 mg/dL, we calculate 10.7764=0.22361 - 0.7764 = 0.2236, or 22.36%.

So, the answer is 22.36%.

Part c)

What percent of women do you expect to have cholesterol levels between 150 mg/dL and 170 mg/dL?

To solve this:

  1. Calculate the z-score for 150 mg/dL: z=15018829=38291.31z = \frac{150 - 188}{29} = \frac{-38}{29} \approx -1.31

  2. Calculate the z-score for 170 mg/dL: z=17018829=18290.62z = \frac{170 - 188}{29} = \frac{-18}{29} \approx -0.62

  3. Use a z-table or calculator to find the probabilities:

    • For z=1.31z = -1.31, the probability is approximately 0.0951.
    • For z=0.62z = -0.62, the probability is approximately 0.2676.
  4. Find the percentage of women between 150 mg/dL and 170 mg/dL: 0.26760.0951=0.17250.2676 - 0.0951 = 0.1725 So, the answer is approximately 17.25%.

If you need further details on any calculation or have more questions, feel free to ask!


Related Questions to Consider:

  1. What is the probability of cholesterol levels being below 150 mg/dL?
  2. How would the percentages change if the standard deviation was different?
  3. What cholesterol level corresponds to the top 5% of women in this distribution?
  4. How would you interpret a z-score of 1.5 in this context?
  5. How can the Empirical Rule be applied to estimate probabilities without calculating exact values?

Tip:

Always remember that in a normal distribution, about 68% of values fall within one standard deviation of the mean, 95% within two, and 99.7% within three. This can help you estimate probabilities quickly.

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Math Problem Analysis

Mathematical Concepts

Statistics
Normal Distribution
Z-scores
Probability

Formulas

Z-score formula: z = (X - μ) / σ
Percentage above a given value using: 1 - P(Z <= z)
Percentage between two values: P(Z <= z2) - P(Z <= z1)

Theorems

Empirical Rule
Properties of Normal Distribution

Suitable Grade Level

College Level