/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 37 A recent report in USA Today ind... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A recent report in USA Today indicated a typical family of four spends \(\$ 490\) per month on food. Assume the distribution of food expenditures for a family of four follows the normal distribution, with a mean of \(\$ 490\) and a standard deviation of \(\$ 90\). a. What percent of the families spend more than \(\$ 30\) but less than \(\$ 490\) per month on food? b. What percent of the families spend less than \(\$ 430\) per month on food? c. What percent spend between \(\$ 430\) and \(\$ 600\) per month on food? d. What percent spend between \(\$ 500\) and \(\$ 600\) per month on food?

Short Answer

Expert verified
a) 50%, b) 25.25%, c) 63.72%, d) 34.56%

Step by step solution

01

Understanding the Problem

We need to determine the percentages of families falling within specific spending ranges on food, given a normal distribution with mean $490 and standard deviation $90. We'll use the properties of the normal distribution to solve each sub-question.
02

Part a: Calculating P(30 < X < 490)

Firstly, we convert the spending values into z-scores. For \(30, the z-score is \( z_{30} = \frac{30 - 490}{90} \). For \)490, \( z_{490} = \frac{490 - 490}{90} \).\[ z_{30} = -5.1111 \quad \text{and} \quad z_{490} = 0 \]Next, we find the cumulative probability for these z-scores from the standard normal distribution table. \( P(Z = 0) = 0.5 \) and \( P(Z = -5.1111) \approx 0 \).Thus, \( P(30 < X < 490) = P(Z = 0) - P(Z = -5.1111) = 0.5 - 0 \).\[ P(30 < X < 490) = 0.5 = 50\% \].
03

Part b: Calculating P(X < 430)

Convert $430 into a z-score: \( z_{430} = \frac{430 - 490}{90} = -0.6667 \).Check the standard normal distribution table or use a calculator to find \( P(Z < -0.6667) \approx 0.2525 \).\[ P(X < 430) = 25.25\% \].
04

Part c: Calculating P(430 < X < 600)

Calculate z-scores for \(430 and \)600:\( z_{430} = \frac{430 - 490}{90} = -0.6667 \) and \( z_{600} = \frac{600 - 490}{90} = 1.2222 \).From the z-tables: \( P(Z < 1.2222) \approx 0.8897 \) and \( P(Z < -0.6667) \approx 0.2525 \).Find the difference: \( P(430 < X < 600) = 0.8897 - 0.2525 = 0.6372 \).\[ P(430 < X < 600) = 63.72\% \].
05

Part d: Calculating P(500 < X < 600)

Find z-scores for \(500 and \)600:\( z_{500} = \frac{500 - 490}{90} = 0.1111 \) and \( z_{600} = 1.2222 \).Using z-tables: \( P(Z < 1.2222) \approx 0.8897 \) and \( P(Z < 0.1111) \approx 0.5441 \).Calculate difference: \( P(500 < X < 600) = 0.8897 - 0.5441 = 0.3456 \).\[ P(500 < X < 600) = 34.56\% \].

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Z-Scores
Z-scores are a fundamental concept in statistics, especially when dealing with normal distributions. A z-score measures how many standard deviations an element is from the mean. To calculate a z-score, you subtract the mean from the value of interest and divide it by the standard deviation. The formula is:\[ z = \frac{X - \mu}{\sigma} \]where:
  • \( X \) is the value of interest,
  • \( \mu \) is the mean of the distribution,
  • and \( \sigma \) is the standard deviation.
Z-scores help to standardize different data sets, making them comparable. They also enable us to use standard normal distribution tables to find cumulative probabilities. Understanding how to convert values to z-scores is crucial as it allows you to work with the standard normal distribution, which can greatly simplify the process of finding probabilities.
Cumulative Probability
Cumulative probability is the probability that a random variable is less than or equal to a particular value. In the context of normal distributions, cumulative probabilities are found using z-scores and can be looked up in the standard normal distribution table.For example, if you have a z-score of \( 1.2222 \), the cumulative probability tells you the likelihood that a random variable is less than or equal to that z-score. This is helpful in determining the probability of falling within a certain range in a distribution.

Using Cumulative Probability

The cumulative probability allows you to determine the likelihood of various outcomes:
  • You can find the probability of being below a particular threshold by looking up the z-score in the standard normal table.
  • To find the probability that a value is between two thresholds, calculate the cumulative probability for each z-score and subtract the smaller from the larger.
This concept is invaluable when dealing with questions asking about the probability of certain ranges in a distribution.
Standard Deviation
The standard deviation is a measure of the amount of variation or dispersion in a set of values. It tells us how much the individual data points differ from the mean. A small standard deviation indicates that the data points tend to be close to the mean, while a large standard deviation indicates that the data points are more spread out.

Importance of Standard Deviation

Understanding standard deviation is crucial when analyzing data:
  • It helps in understanding the distribution of data.
  • It is particularly important in the calculation of z-scores, as it allows for standardization.
  • In a normal distribution, it defines the width of the bell curve, influencing how probabilities are distributed around the mean.
Comprehending the standard deviation aids in grasping the variability of a dataset and is a keystone in any statistical analysis.
Mean
The mean, often referred to as the average, is the sum of all values in a dataset divided by the number of values. It represents the central tendency of a distribution and is a key measure of where the center of the data lies.

Calculating the Mean

To calculate the mean for a set of values \( X_1, X_2, \ldots, X_n \), you would use:\[ \mu = \frac{\sum_{i=1}^n X_i}{n} \]where:
  • \( \mu \) is the mean,
  • \( X \) represents each value,
  • and \( n \) is the number of values.
The mean forms the basis for more complex statistical concepts such as variance and standard deviation. In normal distributions, the mean is at the center of the bell curve, which reflects the typical or "average" scenario for the dataset.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A recent study of the hourly wages of maintenance crew members for major airlines showed that the mean hourly salary was \(\$ 20.50,\) with a standard deviation of \(\$ 3.50 .\) Assume the distribution of hourly wages follows the normal probability distribution. If we select a crew member at random, what is the probability the crew member earns: a. Between \(\$ 20.50\) and \(\$ 24.00\) per hour? b. More than \(\$ 24.00\) per hour? c. Less than \(\$ 19.00\) per hour?

The net sales and the number of employees for aluminum fabricators with similar characteristics are organized into frequency distributions. Both are normally distributed. For the net sales, the mean is \(\$ 180\) million and the standard deviation is \(\$ 25\) million. For the number of employees, the mean is 1,500 and the standard deviation is \(120 .\) Clarion Fabricators had sales of \(\$ 170\) million and 1,850 employees. a. Convert Clarion's sales and number of employees to \(z\) values. b. Locate the two \(z\) values. c. Compare Clarion's sales and number of employees with those of the other fabricators.

The mean of a normal probability distribution is \(60 ;\) the standard deviation is \(5 .\) a. About what percent of the observations lie between 55 and \(65 ?\) b. About what percent of the observations lie between 50 and \(70 ?\) c. About what percent of the observations lie between 45 and \(75 ?\)

The mean of a normal probability distribution is 400 pounds. The standard deviation is 10 pounds. a. What is the area between 415 pounds and the mean of 400 pounds? b. What is the area between the mean and 395 pounds? c. What is the probability of selecting a value at random and discovering that it has a value of less than 395 pounds?

Most four-year automobile leases allow up to 60,000 miles. If the lessee goes beyond this amount, a penalty of 20 cents per mile is added to the lease cost. Suppose the distribution of miles driven on four-year leases follows the normal distribution. The mean is 52,000 miles and the standard deviation is 5,000 miles. a. What percent of the leases will yield a penalty because of excess mileage? b. If the automobile company wanted to change the terms of the lease so that 25 percent of the leases went over the limit, where should the new upper limit be set? c. One definition of a low-mileage car is one that is 4 years old and has been driven less than 45,000 miles. What percent of the cars returned are considered lowmileaqe?

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.