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The heights of 18 -year-old men are approximately normally distributed, with mean 68 inches and standard deviation 3 inches (based on information from Statistical Abstract of the United States, 112 th Edition). (a) What is the probability that an 18 -year-old man selected at random is between 67 and 69 inches tall? (b) If a random sample of nine 18 -year-old men is selected, what is the probability that the mean height \(\bar{x}\) is between 67 and 69 inches? (c) Compare your answers to parts (a) and (b). Is the probability in part (b) much higher? Why would you expect this?

Short Answer

Expert verified
Part (b) has a higher probability (68.26%) than part (a) (62.93%). This is because the mean of a sample is more constrained and less variable.

Step by step solution

01

Understanding Normal Distribution

To solve parts (a) and (b), we need to understand that the heights are normally distributed with mean \( \mu = 68 \) inches and standard deviation \( \sigma = 3 \) inches.
02

Define Z-Score Formula

The Z-score formula is \[ Z = \frac{X - \mu}{\sigma} \], where \( X \) is the value of the random variable, \( \mu \) is the mean, and \( \sigma \) is the standard deviation. We will use this formula to find probabilities.
03

Calculate Z-Scores for Part (a)

For part (a), calculate the Z-scores for 67 inches and 69 inches. \[\text{For } X = 67: Z = \frac{67 - 68}{3} = -\frac{1}{3} \approx -0.333, \\text{For } X = 69: Z = \frac{69 - 68}{3} = \frac{1}{3} \approx 0.333\]
04

Find Probabilities for Part (a)

Use the standard normal distribution table (or calculator) to find the probabilities corresponding to Z-scores \(-0.333\) and \(0.333\). Find the probability that \( P(-0.333 < Z < 0.333) \). This is approximately \(0.6293\), meaning a 62.93% chance.
05

Calculate Standard Error for Part (b)

When considering a sample, we use the standard error, \( SE = \frac{\sigma}{\sqrt{n}} \), where \( n \) is the sample size. For a sample size of 9, \( SE = \frac{3}{\sqrt{9}} = 1 \).
06

Calculate Z-Scores for Part (b)

Now calculate Z-scores for the sample mean between 67 and 69 inches with the standard error. \[\text{For } \bar{x} = 67: Z = \frac{67 - 68}{1} = -1, \\text{For } \bar{x} = 69: Z = \frac{69 - 68}{1} = 1.\]
07

Find Probabilities for Part (b)

Use the standard normal distribution table to find \( P(-1 < Z < 1) \). This probability is approximately \(0.6826\), meaning a 68.26% chance.
08

Compare Results of (a) and (b)

The probability for part (b) is higher (68.26%) compared to part (a) (62.93%). This is expected because the standard error of the mean reduces the variability, making the distribution of sample means narrower.

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Key Concepts

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

Z-score
To understand probability in the context of a normal distribution, we make use of the Z-score. The Z-score is a way to standardize a value within a data set to see how it compares to the mean. By converting a score into a Z-score, you are finding out how many standard deviations an original score is from the mean.The formula to calculate the Z-score is: \[ Z = \frac{X - \mu}{\sigma} \] where:
  • \(X\) is the original data point,
  • \(\mu\) is the mean of the data set,
  • \(\sigma\) is the standard deviation.
Using the Z-score, we can translate an individual data point within a normal distribution to the standard normal distribution, which has a mean of 0 and a standard deviation of 1. This makes it easier to calculate the probability of a data point occurring within a certain range. For instance, for the heights of 18-year-old men, you would convert the specific heights (like 67 or 69 inches) into Z-scores to find where they fall in a standard normal distribution.
standard deviation
Standard deviation is a measure of the spread, or variance, of a set of data. Within a normal distribution, the standard deviation tells us how much the individual data points differ from the mean of the data set. The formula for calculating standard deviation in a population is: \[ \sigma = \sqrt{\frac{\sum (X - \mu)^2}{N}} \] where:
  • \(\sigma\) is the standard deviation,
  • \(X\) represents each value in the data set,
  • \(\mu\) is the mean of the data set,
  • \(N\) is the number of observations.
Standard deviation is key in determining how spread out the data points are in a normally distributed data set. In this exercise, the standard deviation of the heights of 18-year-old men is 3 inches. This means most individuals' heights will fall within 3 inches above or below the mean height of 68 inches. It is important in calculating the probability as it defines the width of the normal distribution curve.
probability calculation
After determining the Z-scores, we then calculate the probability that a value falls within a certain range by considering these scores. This is done using the standard normal distribution table or a calculator that can interpret Z-scores:- **Interpreting Z-scores:** For the heights problem, once you've found the Z-scores for the range (67 and 69 inches converted to -0.333 and 0.333), use the table to find the probability of values falling between these scores.- **Calculating probability:** For a single value, you would find the area under the normal curve between your calculated Z-scores. Usually, these standardized tables give you the cumulative aggregate from the mean to the Z-score, allowing you to ascertain the probability of a value falling within your range.When dealing with a sample mean, like in part (b) of the exercise, you must consider the standard error (SE), which uses the formula:\[ SE = \frac{\sigma}{\sqrt{n}} \] where \(n\) is the sample size.The probability for a group, calculated using the standard error, usually results in higher values due to reduced variance in sample means. For example, finding \(P(-1 < Z < 1)\) doesn't only depend on individual height, but rather on the distribution of the sample mean, making the range tighter and, therefore, the probability higher.

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