91Ó°ÊÓ

Compared to their stay-at-home peers, women employed outside the home have higher levels of high-density lipoproteins (HDL), the "good" cholesterol associated with lower risk for heart attacks. A study of cholesterol levels in 2000 women, aged \(25-64\), living in Augsburg, Germany, was conducted by Ursula Haertel, Ulrigh Keil, and colleagues \(^{\star}\) at the GSF-Medis Institut in Munich. Of these 2000 women, the \(48 \%\) who worked outside the home had HDL levels that were between 2.5 and 3.6 milligrams per deciliter (mg/dL) higher than the HDL levels of their stay-at-home counterparts. Suppose that the difference in HDL levels is normally distributed, with mean 0 (indicating no difference between the two groups of women) and standard deviation \(1.2 \mathrm{mg} / \mathrm{dL}\). If you were to select an employed woman and a stay-at-home counterpart at random, what is the probability that the difference in their HDL levels would be between 1.2 and \(2.4 ?\)

Step by step solution

01

Understand the Problem

We need to find the probability of the difference in HDL levels between employed women and stay-at-home women, which is normally distributed with a mean of 0 and a standard deviation of 1.2 mg/dL, being between 1.2 and 2.4 mg/dL.

02

Standardize the Range

To calculate probabilities for a normal distribution, we first convert the HDL difference range (1.2 to 2.4 mg/dL) to standard normal variable (z-scores). Use the formula for z-score: \( z = \frac{(X - \mu)}{\sigma} \). Given \( \mu = 0 \) and \( \sigma = 1.2 \), compute the z-scores for 1.2 and 2.4.

03

Calculate Z-scores

Convert the range limits to z-scores: - Lower limit: \( z = \frac{(1.2 - 0)}{1.2} = 1.0 \)- Upper limit: \( z = \frac{(2.4 - 0)}{1.2} = 2.0 \). These z-scores help in looking up probabilities in the standard normal distribution table.

04

Find Probabilities from Standard Normal Table

Look up the cumulative probabilities for \( z = 1.0 \) and \( z = 2.0 \):- Probability for \( z = 1.0 \) is approximately 0.8413- Probability for \( z = 2.0 \) is approximately 0.9772.

05

Calculate the Desired Probability

Subtract the cumulative probability at \( z = 1.0 \) from the cumulative probability at \( z = 2.0 \): \[ P(1.2 < X < 2.4) = P(Z < 2.0) - P(Z < 1.0) = 0.9772 - 0.8413 = 0.1359 \].

06

Final Result

The probability that the difference in HDL levels is between 1.2 and 2.4 mg/dL is 0.1359, or approximately 13.59%.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Z-score

A Z-score tells us how many standard deviations away a data point is from the mean of a distribution. It helps in standardizing different data points on the same scale. By converting raw scores into Z-scores, we can easily determine their position in the normal distribution. For this exercise, we're given that the mean difference in HDL levels between employed and stay-at-home women is 0, with a standard deviation of 1.2 mg/dL.
To compute a Z-score, use the formula: \( z = \frac{(X - \mu)}{\sigma} \).

• Lower limit: For an HDL difference of 1.2 mg/dL, the Z-score calculation is \( z = \frac{(1.2 - 0)}{1.2} = 1.0 \).
• Upper limit: For an HDL difference of 2.4 mg/dL, you calculate \( z = \frac{(2.4 - 0)}{1.2} = 2.0 \).

These Z-scores allow us to use the standard normal distribution table to find probabilities. They provide a straightforward way to interpret where the values lie in the distribution.

Cumulative Probability

Cumulative probability refers to the likelihood that a random variable is less than or equal to a specific value. In the context of a normal distribution, it's the integral of the probability density function (PDF) from negative infinity up to a specific value. Finding the cumulative probability involves looking up a value in the standard normal distribution table using a Z-score.
For this exercise, we use cumulative probability to find the likelihood that the HDL level difference falls within a specified range.

• The cumulative probability for Z = 1.0 is indicated as 0.8413, which represents the probability that a value is less than 1.0 standard deviations from the mean.
• For Z = 2.0, the cumulative probability is approximately 0.9772, indicating a higher likelihood of being less than 2.0 standard deviations.

To find the probability of the HDL level difference being between 1.2 mg/dL and 2.4 mg/dL, we subtract the cumulative probability at Z = 1.0 from the cumulative probability at Z = 2.0, yielding the desired probability—approximately 13.59%.

Standard Normal Table

The standard normal table, also known as the Z-table, is a handy tool that provides the cumulative probability associated with each Z-score in a standard normal distribution. The table essentially maps Z-scores to cumulative probabilities, which indicate the probability of a standard normal variable falling below the given Z-score.

• In our example problem, after calculating the Z-scores (1.0 and 2.0) for the specified HDL difference range, we refer to the standard normal table.
• At Z = 1.0, the table provides the cumulative probability of 0.8413, while for Z = 2.0, it gives 0.9772.

Using these cumulative probabilities, we determine the specific probability of the HDL level difference falling between the calculated Z-scores. This table is crucial for understanding the spread of data in a normal distribution, enabling efficient probability calculation without needing complex mathematical functions every time.