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Chapter 10: Problem 35

Cholesterol \((6.2)\) The level of cholesterol in the blood for all men aged 20 to 34 follows a Normal distribution with mean 188 milligrams per deciliter (mg/dl) and standard deviation 41 \(\mathrm{mg} / \mathrm{dl}\) . For 14 -year- old boys, blood cholesterol levels follow a Normal distribution with mean 170 \(\mathrm{mg} / \mathrm{dl}\) and standard deviation 30 \(\mathrm{mg} / \mathrm{d}\) . (a) Let \(M=\) the cholesterol level of a randomly selected 20 - to 34 -ycar-old man and \(B=\) the cholesterol level of a randomly selected 14 -year-old boy. Describe the shape, center, and spread of the distribution of \(M-B\) . (b) Find the probability that a randomly selected 14 -year-old boy has higher cholesterol than a ran- domly selected man aged 20 to \(34 .\) Show your work.

Short Answer

Expert verified
(a) Normal, mean 18, SD 50.8; (b) Probability is approximately 0.361.

Step by step solution

01

Understand the Problem

We need to find the shape, center, and spread of the distribution of the random variable \( M - B \), where \( M \) is the cholesterol level for men aged 20 to 34, and \( B \) is the cholesterol level for boys aged 14.
02

Describe the Distribution Shape

Since both \( M \) and \( B \) follow Normal distributions, the difference \( M - B \) also follows a Normal distribution. This is a property of the Normal distribution: the difference or sum of two independent Normal random variables is also Normal.
03

Find the Distribution Center

The mean of \( M - B \) can be calculated using the means of \( M \) and \( B \). \( M \) has mean 188 mg/dl, and \( B \) has mean 170 mg/dl.\[E(M - B) = E(M) - E(B) = 188 - 170 = 18 \text{ mg/dl}\]
04

Find the Spread (Standard Deviation)

The standard deviation of \( M - B \) can be found using the variances. The variances are the squares of the standard deviations. Since \( M \) and \( B \) are independent:\[\text{Var}(M) = 41^2, \quad \text{Var}(B) = 30^2\]The variance of \( M - B \) is:\[\text{Var}(M - B) = 41^2 + 30^2 = 1681 + 900 = 2581\]Thus, the standard deviation is:\[\text{SD}(M - B) = \sqrt{2581} \approx 50.8 \text{ mg/dl}\]
05

Find the Probability that B > M

We want to find \( P(B > M) \), which is equivalent to \( P(M - B < 0) \). For \( M - B \) which is normally distributed with \( \mu = 18 \) and \( \sigma \approx 50.8 \), we need to find the standard normal probability:\[Z = \frac{0 - 18}{50.8} \approx -0.354\]Using standard normal tables or a calculator, find \( P(Z < -0.354) \).
06

Compute the Probability

Using standard normal distribution tables or a calculator, \( P(Z < -0.354) \approx 0.361 \). This is the probability that a randomly selected 14-year-old boy has higher cholesterol than a randomly selected man aged 20 to 34.

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Key Concepts

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

Probability
Probability is a branch of mathematics that helps us assess the likelihood of certain events happening. Think of it like forecasting the weather; probability gives us a way to predict future outcomes based on existing data. When dealing with normal distributions, such as cholesterol levels, probability helps to determine how likely it is for one individual or another to have a particular cholesterol level.

In the exercise, we calculated the probability of a 14-year-old boy having a higher cholesterol level than a man aged 20 to 34.
Here's how you can approach such problems:
  • Understand what you're comparing鈥攊n this case, cholesterol levels of different age groups.
  • Work with the statistics given, such as means and standard deviations.
  • Use the properties of the normal distribution to facilitate your probability calculations, such as finding Z-scores.
By understanding probability, you can make educated guesses on various outcomes based on numerical data.
Random Variables
Random variables are a fundamental concept in statistics and probability. They are essentially variables whose possible values are numerical outcomes of a random phenomenon. For example, the cholesterol level in an individual can vary randomly depending on numerous factors.

In our exercise, both the cholesterol levels of men aged 20 to 34 and boys aged 14 are random variables. Each of these variables follows its own normal distribution:
  • For men aged 20 to 34, the random variable representing cholesterol levels follows a normal distribution with a mean of 188 and a standard deviation of 41 mg/dl.
  • For the boys aged 14, the cholesterol levels are another random variable, with a mean of 170 and a standard deviation of 30 mg/dl.
  • When looking at the difference between these two, the resulting random variable ( M-B) is also normally distributed, showcasing the beautiful properties of normal distributions.
Understanding random variables is crucial as they allow us to model real-world phenomena and analyze their behavior using statistical methods.
Standard Deviation
Standard deviation is a statistical measure that provides insight into the spread or variability of a set of numbers. In simpler terms, it tells us how much the individual data points deviate from the average (mean) of the data set. This is crucial when interpreting normal distributions, as it helps define the width of the bell curve.

In our cholesterol exercise, the standard deviation played a vital role in determining the spread for each group:
  • The standard deviation for men aged 20 to 34 was given as 41 mg/dl, indicating a wider range of cholesterol levels around the mean of 188 mg/dl.
  • For the 14-year-old boys, the standard deviation was smaller at 30 mg/dl, indicating that their cholesterol levels are less spread out around the mean of 170 mg/dl.
  • When we looked at the combined standard deviation of the distribution of the difference ( M-B), we calculated it to be approximately 50.8 mg/dl. This provides a measure of how much the differences in cholesterol levels can spread around the average difference of 18 mg/dl.
By understanding standard deviation, you get a better grasp of how variable the data can be and how typical or atypical specific outcomes are within the context of a probability distribution.

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Most popular questions from this chapter

Young adults living at home A surprising number of young adults (ages 19 to 25) still live in their parents鈥� homes. A random sample by the National Institutes of Health included 2253 men and 2629 women in this age group.\(^{11}\) The survey found that 986 of the men and 923 of the women lived with their parents. (a) Construct and interpret a 99% confidence interval for the difference in population proportions (men minus women). (b) Does your interval from part (a) give convincing evidence of a difference between the population proportions? Explain.

Paying for college College financial aid offices expect students to use summer earnings to help pay for college. But how large are these earnings? One large university studied this question by asking a random sample of 1296 students who had summer jobs how much they earned. The financial aid office separated the responses into two groups based on gender. Here are the data in summary form:\(^{33}\) $$\begin{array}{llll}{\text { Group }} & {n} & {\overline{x}} & {s_{x}} \\\ \hline \text { Males } & {675} & {\$ 1884.52} & {\$ 13688.37} \\ {\text { Females }} & {621} & {\$ 1360.39} & {\$ 1037.46}\end{array}$$ (a) How can you tell from the summary statistics that the distribution of earnings in each group is strongly skewed to the right? A graph of the data reveals no outliers. The use of two-sample t procedures is still justified. Why? (b) Construct and interpret a 90% confidence interval for the difference between the mean summer earnings of male and female students at this university. (c) Interpret the 90% confidence level in the context of this study.

Paired or unpaired? In each of the following settings, decide whether you should use paired \(t\) procedures or two-sample t procedures to perform inference. Explain your choice.\(^{43}\) (a) To compare the average weight gain of pigs fed two different rations, nine pairs of pigs were used. The pigs in each pair were littermates. A coin toss was used to decide which pig in each pair got Ration A and which got Ration B. (b) A random sample of college professors is taken. We wish to compare the average salaries of male and female teachers. (c) To test the effects of a new fertilizer, 100 plots are treated with the new fertilizer, and 100 plots are treated with another fertilizer. A computer鈥檚 random number generator is used to determine which plots get which fertilizer.

Exercises 23 through 26 involve the following setting. Some women would like to have children but cannot do so for medical reasons. One option for these women is a procedure called in vitro fertilization (IVF), which involves injecting a fertilized egg into the woman鈥檚 uterus. Acupuncture and pregnancy A study reported in the medical journal Fertility and Sterility sought to determine whether the ancient Chinese art of acupuncture could help infertile women become pregnant.\(^{18}\) One hundred sixty healthy women who planned to have IVF were recruited for the study. Half of the subjects (80) were randomly assigned to receive acupuncture 25 minutes before embryo transfer and again 25 minutes after the transfer. The remaining 80 women were assigned to a control group and instructed to lie still for 25 minutes after the embryo transfer. Results are shown in the table below. $$\begin{array}{ll}&{\text { Acupuncture group }} & {\text { Control group }} \\\ \text { Pregnant } & \quad\quad\quad\quad {34} & \quad\quad\quad {21} \\\ \text { Not Pregnant } & \quad\quad\quad\quad {46} & \quad\quad\quad {59} \\\ \text { Total } & \quad\quad\quad\quad {80} & \quad\quad\quad {80}\end{array}$$ Is the pregnancy rate significantly higher for women who received acupuncture? To find out, researchers perform a test of \(H_{0} : p_{1}=p_{2}\) versus \(H_{a} : p_{1}>p_{2},\) where \(p_{1}\) and \(p_{2}\) are the actual pregnancy rates for women like those in the study who do and don't receive acupuncture, respectively. (a) Name the appropriate test and check that the conditions for carrying out this test are met. (b) The appropriate test from part (a) yields a P-value of 0.0152. Interpret this P-value in context. (c) What conclusion should researchers draw at the \(\alpha=0.05\) significance level? Explain. (d) What flaw in the design of the experiment prevents us from drawing a cause-and-effect conclusion? Explain.

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