/*! 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 32 The average monthly mortgage pay... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

The average monthly mortgage payment for all homeowners in a city is $$\$ 2850.$$ Suppose that the distribution of monthly mortgages paid by homeowners in this city follow an approximate normal distribution with a mean of $$\$ 2850$$ and a standard deviation of $$\$ 420.$$ Find the probability that the monthly mortgage paid by a randomly selected homeowner from this city is a. less than $$\$ 1200$$ b. between $$\$ 2300$$ and $$\$ 3140$$ c. more than $$\$ 3600$$ d. between $$\$ 3200$$ and $$\$ 3700$$

Short Answer

Expert verified
a. The probability that the mortgage is less than $1200 is approximately 0. b. The probability that the mortgage is between $2300 and $3140 is approximately 0.4125. c. The probability that the mortgage is more than $3600 is approximately 0.0359. d. The probability that the mortgage is between $3200 and $3700 is approximately 0.0968.

Step by step solution

01

Understand The Problem

The problem is related to normal distribution. The given values are the mean (\$2850) and standard deviation (\$420) of the distribution. We need to find probabilities of different ranges.
02

Calculate The Z-scores

The general formula to calculate a z-score from a data point in a distribution is: Z = (X - μ) / σ, where X is a data point, μ is the mean and σ is the standard deviation. Using this formula, the z-scores for all given values can be calculated.
03

Calculate Probabilities

After obtaining the Z-scores, use the Standard Normal Distribution table to find the associated probabilities. The standard normal distribution table provides the probability that a normally distributed random variable Z is less than or equal to z. Subtract for ranges.
04

Solve For All Cases

Use these steps to solve for all payment ranges requested in the problem.

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.

Understanding Z-Scores
The z-score is an essential concept when working with normal distributions, like in our exercise about mortgage payments. Z-score helps us understand how far away a particular data point is from the mean, in terms of standard deviations. For instance, if the mean monthly mortgage is \(2850, and you're interested in a payment of \)1200, you can use the z-score formula:
  • \( Z = \frac{X - \mu}{\sigma} \)
Here, \(X\) is the data point in question, \(\mu\) is the mean, and \(\sigma \) is the standard deviation. Calculating a z-score is helpful because it translates values into a common scale: the number of standard deviations below or above the mean.
The z-score allows us to compare data points from different normal distributions or to identify how uncommon or common a particular data point is in the context of its specific distribution.
Importance of Standard Deviation
Standard deviation is a critical measure when analyzing normal distribution data. It helps us understand the amount of variation or spread in a set of values. In our exercise, the standard deviation of the monthly mortgage payments is \( \\(420 \). This indicates that most homeowners' payments won't deviate too far from the average of \( \\)2850 \).
The closer the data points are to the mean, the smaller the standard deviation, indicating that most people pay around the average amount. Conversely, a larger standard deviation would imply more significant differences among individuals' payments.A practical understanding of standard deviation helps with interpreting a normal distribution curve on a graph:
  • Approximately 68% of data falls within one standard deviation from the mean.
  • About 95% is within two standard deviations.
  • 99.7% lies within three standard deviations.
This helps you predict the likelihood of seeing a particular data point in the distribution.
Significance of Mean Value
In the context of normal distribution and our exercise, the mean value is the sum of all data points divided by the number of data points, often referred to as the average. For these mortgage payments, the mean value is \( \$2850 \).
The mean serves as the "central" point of our normal distribution curve, with half of the values lying above it and the other half below it. Understanding the mean is essential as it represents the typical value of the data set, providing a baseline for comparison.
In a nicely symmetrical normal distribution, the mean also coincides with the peak of the curve. For data analysis, the mean is critical in calculating z-scores and understanding what is "ordinary" versus "extraordinary" within the given data set.
Proficiency in Probability Calculation
Calculating probabilities within a normal distribution is a valuable skill, especially when determining the likelihood of certain events occurring. Once you have the z-scores for particular data points, as we saw with the mortgage payment ranges provided, they can be used to look up probabilities on the Standard Normal Distribution table.
This table gives the probability that a standard normal variable \( Z \) is less than or equal to a specific z-score:
  • To find the probability that a value is less than a certain number, locate the z-score in the table.
  • To determine probability between two values, find their z-scores and subtract their respective probabilities.
Being skilled in probability calculation allows for making informed predictions and decisions based on statistical data to handle queries like those in the exercise effectively.

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

The management of a supermarket wants to adopt a new promotional policy of giving a free gift to every customer who spends more than a certain amount per visit at this supermarket. The expectation of the management is that after this promotional policy is advertised, the expenditures for all customers at this supermarket will be normally distributed with a mean of $$\$ 95$$ and a standard deviation of $$\$ 20.$$ If the management decides to give free gifts to all those customers who spend more than $$\$ 130$$ at this supermarket during a visit, what percentage of the customers are expected to receive free gifts?

What are the parameters of the normal distribution?

Stress on the job is a major concern of a large number of people who go into managerial positions. It is estimated that \(80 \%\) of the managers of all companies suffer from job-related stress. a. What is the probability that in a sample of 200 managers of companies, exactly 150 suffer from job-related stress? b. Find the probability that in a sample of 200 managers of companies, at least 170 suffer from job-related stress. c. What is the probability that in a sample of 200 managers of companies, 165 or fewer suffer from job-related stress? d. Find the probability that in a sample of 200 managers of companies, 164 to 172 suffer from job-related stress.

Determine the value of \(z\) so that the area under the standard normal curve a. in the right tail is \(.0500\) b. in the left tail is \(.0250\) c. in the left tail is \(.0100\) d. in the right tail is \(.0050\)

The Bank of Connecticut issues Visa and MasterCard credit cards. It is estimated that the balances on all Visa credit cards issued by the Bank of Connecticut have a mean of $$\$ 845$$ and a standard deviation of $$\$ 270.$$ Assume that the balances on all these Visa cards follow a normal distribution. a. What is the probability that a randomly selected Visa card issued by this bank has a balance between $$\$ 1000$$ and $$\$ 1440? $$ b. What percentage of the Visa cards issued by this bank have a balance of $$\$ 730$$ or more?

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.