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Chapter 6: Problem 6

Assume that females have pulserates that are normally distributed with a mean of 74.0 beats per minute and a standard deviation of 12.5 beats per minute (based on Data Set 1 鈥淏ody Data鈥� in Appendix B). a. If 1 adult female is randomly selected, find the probability that her pulse rate is greater than 70 beats per minute. b. If 25 adult females are randomly selected, find the probability that they have pulse rates with a mean greater than 70 beats per minute. c. Why can the normal distribution be used in part (b), even though the sample size does not exceed \(30 ?\)

Short Answer

Expert verified
Part a: 0.6255, Part b: 0.9452, Part c: Central Limit Theorem allows the use of normal distribution.

Step by step solution

01

Convert to Z-score (Part a)

To find the probability, convert the given value (70 bpm) to a Z-score. Use the Z-score formula: \[ Z = \frac{X - \mu}{\sigma} \] where \(X = 70\), \(\mu = 74.0\) (mean pulse rate), \(\sigma = 12.5\) (standard deviation).
02

Calculate Z-score (Part a)

Substitute the values into the Z-score formula: \[ Z = \frac{70 - 74.0}{12.5} = \frac{-4.0}{12.5} = -0.32 \]
03

Find the Probability (Part a)

Use the Z-score to find the corresponding probability from the standard normal distribution table. For \(Z = -0.32\), the cumulative probability is approximately 0.3745. Since we want the probability that the pulse rate is greater than 70, find the complement: \[ P(X > 70) = 1 - P(X \leq 70) = 1 - 0.3745 = 0.6255 \]
04

Calculate the Standard Error (Part b)

For a sample size of 25, calculate the standard error of the mean: \[ \text{SE} = \frac{\sigma}{\sqrt{n}} \] Here \(\sigma = 12.5\), \(n = 25\): \[ \text{SE} = \frac{12.5}{\sqrt{25}} = \frac{12.5}{5} = 2.5 \]
05

Convert to Z-score (Part b)

Convert the mean pulse rate (70 bpm) to a Z-score for the sample mean: \[ Z = \frac{\bar{X} - \mu}{SE} \] where \(\bar{X} = 70\), \(\mu = 74.0\) (population mean), \(SE = 2.5\). Substitute the values: \[ Z = \frac{70 - 74.0}{2.5} = \frac{-4.0}{2.5} = -1.6 \]
06

Find the Probability (Part b)

Use the Z-score to find the corresponding probability from the standard normal distribution table. For \(Z = -1.6\), the cumulative probability is approximately 0.0548. Since we want the probability that the mean pulse rate is greater than 70, find the complement: \[ P(\bar{X} > 70) = 1 - P(\bar{X} \leq 70) = 1 - 0.0548 = 0.9452 \]
07

Explanation for Using Normal Distribution (Part c)

The normal distribution can be used in part (b) because of the Central Limit Theorem, which states that the sampling distribution of the sample mean will be approximately normally distributed if the sample size is sufficiently large (typically \(n \geq 30 \)). However, for smaller sample sizes, like \(n = 25\), the distribution can still be considered approximately normal if the underlying population is normally distributed, as given in this problem.

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Key Concepts

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

Z-score calculation
To understand Z-scores, let's start with the basics. A Z-score tells us how many standard deviations a data point is from the mean. Here is the formula to calculate it:
en Z = \( \frac{X - \mu}{\sigma} \)Where:
  • \(X\) is the value we are examining (e.g., 70 beats per minute in our problem).
  • \(\mu\) (mu) is the population mean (74 beats per minute).
  • \(\sigma\) (sigma) is the population standard deviation (12.5 beats per minute).
For instance, if we calculate the Z-score for a pulse rate of 70 beats per minute, we find it is \(-0.32\), meaning the pulse rate is \(0.32\) standard deviations below the mean. Understanding Z-scores helps us determine probabilities from the standard normal distribution.
Central Limit Theorem
The Central Limit Theorem (CLT) is a fundamental principle in statistics. It states that the sampling distribution of the sample mean will approximate a normal distribution, provided the sample size is sufficiently large (commonly \(n \geq 30\)). This principle allows us to work with normal distribution properties, even if the population distribution itself is not normal.
However, in our problem, the sample size is more minor than 30 (i.e., 25). The CLT still holds because we are told the population distribution is normal. Hence, even with smaller sample sizes, the sampling distribution of the sample mean is normally distributed.
Probability distribution
Probability distribution tells us the likelihood of all possible outcomes. For normal distribution, the probabilities are spread in a characteristic bell curve.
In our scenario, calculating probabilities using Z-scores and the standard normal distribution table gives us insight into the probability of a specific pulse rate. For instance, for a single female with a pulse rate higher than 70 beats per minute, the Z-score calculation helps find this probability by converting it into a standardized form. Then, we look up this Z-score in standard normal distribution tables to find the corresponding probability.
Sample mean
The sample mean is the average value calculated from a sample, not the entire population. It often approximates the population mean as the sample size grows. We use the sample mean to make inferences about the population.
In our problem, we consider a sample of 25 females and calculate the sample mean of pulse rates. We use the sample mean and standard error to compute the Z-score, which tells us how the sample mean compares to the population mean, thereby helping us determine the probability of observing such a sample mean in our context.
Standard error
The standard error (SE) measures the dispersion of sample means around the population mean. It tells us how much variability to expect in the sample mean due to random sampling.
The standard error formula is:
SE = \( \frac{\sigma}{\sqrt{n}} \)Where:
  • \(\sigma\) is the population standard deviation, and
  • \(n\) is the sample size.
In our problem, the SE for a sample size of 25 females is calculated as 2.5 beats per minute. This smaller value compared to the population standard deviation (12.5) highlights that sample means tend to cluster more tightly around the population mean.

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Most popular questions from this chapter

In a study of babies born with very low birth weights, 275 children were given IQ tests at age 8 , and their scores approximated a normal distribution with \(\mu=95.5\) and \(\sigma=16.0\) (based on data from "Neurobehavioral Outcomes of School-age Children Born Extremely Low Birth Weight or Very Preterm," by Anderson et al., Journal of the American Medical Association, Vol. 289, No. 24 ). Fifty of those children are to be randomly selected without replacement for a follow-up study. a. When considering the distribution of the mean IQ scores for samples of 50 children, should \(\sigma_{-}\) be corrected by using the finite population correction factor? Why or why not? What is the value of \(\sigma_{\bar{x}}\) ? b. Find the probability that the mean IQ score of the follow-up sample is between 95 and 105 .

There are about 4200 college presidents in the United States, and they have annual incomes with a distribution that is skewed instead of being normal. Many different samples of 40 college presidents are randomly selected, and the mean annual income is computed for each sample. a. What is the approximate shape of the distribution of the sample means (uniform, normal, skewed, other)? b. What value do the sample means target? That is, what is the mean of all such sample means?

Common tests such as the SAT, ACT, Law School Admission test (LSAT), and Medical College Admission Test (MCAT) use multiple choice test questions, each with possible answers of \(a, b, c, d, e\), and each question has only one correct answer. We want to find the probability of getting at least 25 correct answers for someone who makes random guesses for answers to a block of 100 questions. If we plan to use the methods of this section with a normal distribution used to approximate a binomial distribution, are the necessary requirements satisfied? Explain.

Data Set 4 "Births" in Appendix B includes birth weights of 400 babies. If we compute the values of sample statistics from that sample, which of the following statistics are unbiased estimators of the corresponding population parameters: sample mean; sample median; sample range; sample variance; sample standard deviation; sample proportion?

Do the following: If the requirements of \(n p \geq 5\) and \(n q \geq 5\) are both satisfied, estimate the indicated probability by using the normal distribution as an approximation to the binomial distribution; if \(n p<5\) or \(n q<5\), then state that the normal approximation should not be used. With \(n=20\) births and \(p=0.512\) for a boy, find \(P(\) fewer than 8 boys).
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