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Chapter 6: Problem 7

Assuming that the heights of college women are normally distributed with mean 65 inches and standard deviation \(2.5\) inches (based on information from Statistical Abstract of the United States, 112 th Edition), answer the following questions. Hint: Use Problems 5 and 6 and Figure \(6-3\). (a) What percentage of women are taller than 65 inches? (b) What percentage of women are shorter than 65 inches? (c) What percentage of women are between \(62.5\) inches and \(67.5\) inches? (d) What percentage of women are between 60 inches and 70 inches?

Short Answer

Expert verified
(a) 50%, (b) 50%, (c) 68.27%, (d) 95.45%.

Step by step solution

01

Understand the Normal Distribution

The problem states that heights are normally distributed with a mean \( \mu = 65 \) inches and standard deviation \( \sigma = 2.5 \) inches. The normal distribution is a continuous probability distribution that is symmetric around the mean.
02

Calculate Percentage for (a) Taller than 65 inches

Since 65 inches is the mean, half of the distribution is above 65 inches. Using properties of the normal distribution, 50% of the heights will be above 65 inches.
03

Calculate Percentage for (b) Shorter than 65 inches

Similarly, half of the distribution is below 65 inches. Thus, 50% of women are shorter than 65 inches.
04

Calculate Percentage for (c) Between 62.5 inches and 67.5 inches

First, find the z-scores for 62.5 and 67.5 inches:\[ z = \frac{x - \mu}{\sigma} \]\[ z_{62.5} = \frac{62.5 - 65}{2.5} = -1 \] \[ z_{67.5} = \frac{67.5 - 65}{2.5} = 1 \]Using the standard normal distribution table, find the area between these z-scores, which is approximately 68.27%.
05

Calculate Percentage for (d) Between 60 inches and 70 inches

Find the z-scores for 60 and 70 inches:\[ z_{60} = \frac{60 - 65}{2.5} = -2 \]\[ z_{70} = \frac{70 - 65}{2.5} = 2 \]Using the standard normal distribution table, the area between these z-scores is approximately 95.45%.

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Key Concepts

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

College Women Heights
When studying the normal distribution of college women's heights, the average or mean height is a central concept. In this context, the mean height is 65 inches. This number represents the average height of college women and serves as the peak or center of the normal distribution curve. Normal distribution, often depicted as a symmetric bell shape, shows how different possible outcomes are distributed around the mean. For heights that follow this pattern, most individuals will cluster closely around the 65 inches mark, with fewer individuals being much taller or much shorter. Understanding heights as a dataset following a normal distribution allows us to expect that approximately half of the population will fall below this average, and the other half will be taller. It makes predicting and analyzing the data more straightforward since we have these symmetrical properties.
Standard Deviation
Standard deviation is a key measure to understand when dealing with normally distributed data. It quantifies the amount of variation or dispersion in a set of data values. In the current example, the standard deviation of college women's heights is 2.5 inches. This means that most of the data points (in this case, heights) lie within a certain range around the mean. For instance, within one standard deviation (2.5 inches) from the mean, you can find a significant portion of the data. In tallies percentages, roughly 68% of the data points are expected to be within 7.5 inches and 67.5 inches. A smaller standard deviation would suggest that the data points are closely packed around the mean, whereas a larger standard deviation indicates more widespread data. Thus, understanding the standard deviation helps in analyzing the spread of college women's heights around the average.
Probability Distribution
Probability distribution is an essential concept that helps determine the likelihood of any particular outcome. In the case of normally distributed college women's heights, probability distribution is used to find what portion of women fall within specific height ranges. Using a normal distribution, you can determine percentages like what part of the population is shorter or taller than a certain height. For example, since the mean is 65 inches, you already know that 50% of the college women are taller than this height, and 50% are shorter. This symmetry is a property of the normal distribution. To find more specific probabilities, such as the percentage of women with heights between 62.5 inches and 67.5 inches, you would calculate the z-scores for these values. Using a standard normal distribution table, you can then determine that approximately 68% of the women fall within this range. Understanding probability distribution is crucial for making statistically significant predictions about the dataset.

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Most popular questions from this chapter

Let \(z\) be a random variable with a standard normal distribution. Find the indicated probability, and shade the corresponding area under the standard normal curve. $$ P(z \geq 2.17) $$

Let \(z\) be a random variable with a standard normal distribution. Find the indicated probability, and shade the corresponding area under the standard normal curve. $$ P(-0.73 \leq z \leq 3.12) $$

Distribution Suppose we have a binomial experiment in which success is defined to be a particular quality or attribute that interests us. (a) Suppose \(n=33\) and \(p=0.21\). Can we approximate the \(\hat{p}\) distribution by a normal distribution? Why? What are the values of \(\mu_{\hat{p}}\) and \(\sigma_{\hat{p}} ?\) (b) Suppose \(n=25\) and \(p=0.15 .\) Can we safely approximate the \(\hat{p}\) distribution by a normal distribution? Why or why not? (c) Suppose \(n=48\) and \(p=0.15 .\) Can we approximate the \(\hat{p}\) distribution by a normal distribution? Why? What are the values of \(\mu_{\hat{p}}\) and \(\sigma_{\hat{p}} ?\)

Sketch the areas under the standard normal curve over the indicated intervals, and find the specified areas. Between \(z=-2.42\) and \(z=-1.77\)

Let \(x\) be a random variable that represents the level of glucose in the blood (milligrams per deciliter of blood) after a 12 -hour fast. Assume that for people under 50 years old, \(x\) has a distribution that is approximately normal, with mean \(\mu=85\) and estimated standard deviation \(\sigma=25\) (based on information from Diagnostic Tests with Nursing Applications, edited by S. Loeb, Springhouse). A test result \(x<40\) is an indication of severe excess insulin, and medication is usually prescribed. (a) What is the probability that, on a single test, \(x<40\) ? (b) Suppose a doctor uses the average \(\bar{x}\) for two tests taken about a week apart. What can we say about the probability distribution of \(\bar{x}\) ? Hint: See Theorem 6.1. What is the probability that \(\bar{x}<40\) ? (c) Repeat part (b) for \(n=3\) tests taken a week apart. (d) Repeat part (b) for \(n=5\) tests taken a week apart. (e) Compare your answers to parts (a), (b), (c), and (d). Did the probabilities decrease as \(n\) increased? Explain what this might imply if you were a doctor or a nurse. If a patient had a test result of \(\bar{x}<40\) based on five tests, explain why either you are looking at an extremely rare event or (more likely) the person has a case of excess insulin.
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