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Chapter 7: Problem 28

For the most recent year available, the mean annual cost to attend a private university in the United States was \(\$ 42,224 .\) Assume the distribution of annual costs follows the normal probability distribution and the standard deviation is \(\$ 4,500 .\) Ninety-five percent of all students at private universities pay less than what amount?

Short Answer

Expert verified
95% of students pay less than $49,627 at most.

Step by step solution

01

Identify the Known Values

We are provided with the mean annual cost, which is \( \mu = 42,224 \) dollars, and the standard deviation, \( \sigma = 4,500 \) dollars. We need to determine the amount below which 95% of the costs fall, corresponding to the 95th percentile of the distribution.
02

Understand the Normal Distribution

In a normal distribution, the area under the curve represents probability. The 95th percentile indicates the point where the cumulative probability reaches 0.95. In a standard normal distribution, this corresponds to a z-score of about 1.645 (for a one-tailed test).
03

Calculate the 95th Percentile

The z-score formula relates the z-score to the amount \( X \) using the equation: \[ X = \mu + z \cdot \sigma \] Insert the known values: \( \mu = 42,224 \), \( \sigma = 4,500 \), and \( z = 1.645 \).
04

Solve the Equation

Plug the values into the equation: \[ X = 42,224 + 1.645 \times 4,500 \] This calculation results in: \[ X = 42,224 + 7,402.5 = 49,626.5 \] dollars.

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Key Concepts

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

Understanding the Mean
The mean, often referred to as the average, is a measure of central tendency. It provides a single value that represents the center point of a dataset. In the context of a normal distribution, the mean is depicted right at the center of the bell curve. For the annual cost to attend a private university, the mean is given as \( \\(42,224 \).
  • This means that if you sum up all the costs of attending these universities and divide by the number of universities, you'd get \( \\)42,224 \).
  • The mean is a key component when working with distributions because it helps locate where most data points congregate.
With a normal distribution, data spreads evenly around the mean, which simplifies the prediction of future data based on past data. This predictability is one of the reasons why understanding the mean is so crucial.
Exploring Standard Deviation
Standard deviation is a measure of how spread out the numbers in a data set are. In a normal distribution, it tells us how much values deviate from the mean.
  • A small standard deviation means that most of the data points are close to the mean, leading to a steep, narrow bell curve.
  • A larger standard deviation indicates that data points are more spread out, creating a flatter, wider bell.
For the private university cost data, the standard deviation is \(\\(4,500\). This tells us that the costs typically vary by about \(\\)4,500\) from the mean of \(\$42,224\).Understanding standard deviation helps in comprehending the variability or consistency of data鈥攅ssential when analyzing data distributions.
Defining Percentile
A percentile is a measure that indicates the relative standing of a value within a dataset. When discussing percentiles, we look at how a particular value compares to the others in the dataset.
  • The term '95th percentile' refers to the point below which 95% of the data falls. For the university cost problem, we are interested in finding the cost point where only 5% of the values are greater.
  • Percentiles are vital in understanding what proportion of the data falls below a particular value and are used in many fields, such as finance, education, and health statistics.
Working with percentiles allows us to describe data points in relation to the entire dataset, providing insight into how extreme or typical certain values are.
Understanding Z-Score
The z-score is a statistical measurement that describes a value's position relative to the mean of a group of values. It tells us how many standard deviations away a point is from the mean.
  • In the context of normal distribution, a z-score helps find the percentile rank of a value.
  • For instance, a z-score of 0 means the value is exactly at the mean, while a positive z-score indicates it is above the mean.
  • A z-score of 1.645, as used in this problem, is significant because it corresponds to the 95th percentile in a normal distribution.
Calculating the z-score helps us translate between the standard normal distribution and the actual dataset, making it crucial for estimating probabilities and determining percentile rankings.

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Most popular questions from this chapter

Fast Service Truck Lines uses the Ford Super Duty F-750 exclusively. Management made a study of the maintenance costs and determined the number of miles traveled during the year followed the normal distribution. The mean of the distribution was 60,000 miles and the standard deviation was 2,000 miles. a. What percent of the Ford Super Duty F-750s logged 65,200 miles or more? b. What percent of the trucks logged more than 57,060 but less than 58,280 miles? c. What percent of the Fords traveled 62,000 miles or less during the year? d. Is it reasonable to conclude that any of the trucks were driven more than 70,000 miles? Explain.

The funds dispensed at the ATM machine located near the checkout line at the Kroger's in Union, Kentucky, follows a normal probability distribution with a mean of \(\$ 4,200\) per day and a standard deviation of \(\$ 720\) per day. The machine is programmed to notify the nearby bank if the amount dispensed is very low (less than \(\$ 2,500\) ) or very high (more than \(\$ 6,000\) ). a. What percent of the days will the bank be notified because the amount dispensed is very low? b. What percent of the time will the bank be notified because the amount dispensed is high? c. What percent of the time will the bank not be notified regarding the amount of funds dispersed?

A recent Gallup study (http://www.gallup.com/poll/175286/hour-workweek- actually-Ionger-seven- hours.aspx?g_source=polls+work+hours\(\&\)g_medium=search\(\&\)g_campaign=tiles) found the typical American works an average of 46.7 hour per week. The study did not report the shape of the distribution of hours worked or the standard deviation. It did, however, indicate that \(40 \%\) of the workers worked less than 40 hours a week and that 18 percent worked more than 60 hours. a. If we assume that the distribution of hours worked is normally distributed, and knowing \(40 \%\) of the workers worked less than 40 hours, find the standard deviation of the distribution. b. If we assume that the distribution of hours worked is normally distributed and \(18 \%\) of the workers worked more than 60 hours, find the standard deviation of the distribution. c. Compare the standard deviations computed in parts \(a\) and \(b\). Is the assumption that the distribution of hours worked is approximately normal reasonable? Why?

In establishing warranties on HDTVs, the manufacturer wants to set the limits so that few will need repair at the manufacturer's expense. On the other hand, the warranty period must be long enough to make the purchase attractive to the buyer. For a new HDTV, the mean number of months until repairs are needed is 36.84 with a standard deviation of 3.34 months. Where should the warranty limits be set so that only \(10 \%\) of the HDTVs need repairs at the manufacturer's expense?

WNAE, an all-news AM station, finds that the distribution of the lengths of time listeners are tuned to the station follows the normal distribution. The mean of the distribution is 15.0 minutes and the standard deviation is 3.5 minutes. What is the probability that a particular listener will tune in for: a. More than 20 minutes? b. 20 minutes or less? c. Between 10 and 12 minutes?
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