/*! 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 61 The U.S. Department of Education... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

The U.S. Department of Education reports that about \(50 \%\) of all college students use a student loan to help cover college expenses (National Center for Educational Studies, January 2006 ). A sample of students who graduated with student loan debt is shown here. The data, in thousands of dollars, show typical amounts of debt upon graduation. $$\begin{array}{lllllllll} 10.1 & 14.8 & 5.0 & 10.2 & 12.4 & 12.2 & 2.0 & 11.5 & 17.8 & 4.0 \end{array}$$ a. For those students who use a student loan, what is the mean loan debt upon graduation? b. What is the variance? Standard deviation?

Short Answer

Expert verified
Mean debt is $10,000. Variance is 22.178, and standard deviation is approximately 4.71.

Step by step solution

01

Calculate the Mean

The mean or average value is calculated by summing all the loan amounts and then dividing by the number of data points. Add all student debts: 10.1 + 14.8 + 5.0 + 10.2 + 12.4 + 12.2 + 2.0 + 11.5 + 17.8 + 4.0 = 100.0. There are 10 data points. So, the mean is: \( \text{Mean} = \frac{100.0}{10} = 10.0 \) thousand dollars.
02

Calculate each Data Point's Deviation from the Mean

First, subtract the mean (10.0) from each data point to find the deviation: 0.1, 4.8, -5.0, 0.2, 2.4, 2.2, -8.0, 1.5, 7.8, -6.0.
03

Square the Deviations

Square each deviation: 0.01, 23.04, 25.0, 0.04, 5.76, 4.84, 64.0, 2.25, 60.84, 36.0.
04

Compute the Variance

Sum the squared deviations and divide by the number of data points to find the variance. Sum: 0.01 + 23.04 + 25.0 + 0.04 + 5.76 + 4.84 + 64.0 + 2.25 + 60.84 + 36.0 = 221.78. There are 10 data points, so the variance is: \( \text{Variance} = \frac{221.78}{10} = 22.178 \).
05

Calculate the Standard Deviation

The standard deviation is the square root of the variance. Therefore, the standard deviation is: \( \text{Standard Deviation} = \sqrt{22.178} \approx 4.71 \) thousand dollars.

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.

Mean Calculation
The mean, often called the average, is a central concept in statistical analysis that helps us understand the typical value of a dataset. In the context of student loans, calculating the mean debt amount allows us to gauge the average borrowing among students.
Calculating the mean is straightforward. Start by adding all the debt figures together - in this case, the amounts are 10.1, 14.8, 5.0, 10.2, 12.4, 12.2, 2.0, 11.5, 17.8, and 4.0 thousand dollars.
This sum totals 100.0 thousand dollars. The next step is to divide this total by the number of entries (data points), which is 10 in this dataset.

So, the mean is given by:
  • Total Sum of Loan Amounts: 100.0
  • Number of Students: 10
  • Mean Loan Amount: \( \text{Mean} = \frac{100.0}{10} = 10.0 \) thousand dollars
Understanding the mean provides a basic summary of the data, offering a quick sense of what a typical student might owe.
Variance Calculation
Variance is a statistical measure that provides insight into the degree of spread in a dataset. It helps us understand how much the individual loan amounts differ from the mean value. Higher variance indicates larger differences between data points and the mean.

To calculate variance, follow these steps:
  • Firstly, determine the deviation of each data point from the mean. This is done by subtracting the mean (10.0) from each loan amount.
  • Then, square each of these deviations to avoid negative values which would cancel each other out.
  • Finally, sum all the squared deviations and divide by the number of data points to get the variance.
Let's see this calculation:
Deviations from mean: 0.1, 4.8, -5.0, 0.2, 2.4, 2.2, -8.0, 1.5, 7.8, -6.0
Squared deviations: 0.01, 23.04, 25.0, 0.04, 5.76, 4.84, 64.0, 2.25, 60.84, 36.0
The sum of squared deviations is 221.78. Divide this by 10 (the number of data points) to calculate the variance:
\[ \text{Variance} = \frac{221.78}{10} = 22.178 \]

Thus, the variance of 22.178 tells us the typical squared deviation from the mean, providing a measure of data spread.
Standard Deviation
The standard deviation is a crucial statistical tool that indicates how spread out the data is around the mean. It differs from variance as it is expressed in the same units as the data, making it easier to interpret.

To find the standard deviation, take the square root of the variance. Let's carry out this calculation step by step:
  • First, we have the variance which is 22.178 (calculated previously).
  • Next, compute the square root of this variance.
This calculation is done as follows:
\[ \text{Standard Deviation} = \sqrt{22.178} \approx 4.71 \] thousand dollars.

The standard deviation of around 4.71 tells us that, on average, the student loan debts deviate by this amount from the mean of 10.0 thousand dollars. It provides a meaningful insight into how consistently (or inconsistently) students deviate from the average debt through their borrowing patterns. This makes it a truly informative measure for comparing variations.

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

According to an annual consumer spending survey, the average monthly Bank of America Visa credit card charge was \(\$ 1838\) (U.S. Airways Attaché Magazine, December 2003 ). A sample of monthly credit card charges provides the following data. $$\begin{array}{llrrr} 236 & 1710 & 1351 & 825 & 7450 \\ 316 & 4135 & 1333 & 1584 & 387 \\ 991 & 3396 & 170 & 1428 & 1688 \end{array}$$ a. Compute the mean and median. b. Compute the first and third quartiles. c. Compute the range and interquartile range. d. Compute the variance and standard deviation. e. The skewness measure for these data is \(2.12 .\) Comment on the shape of this distribution. Is it the shape you would expect? Why or why not? f. Do the data contain outliers?

A bowler's scores for six games were \(182,168,184,190,170,\) and \(174 .\) Using these data as a sample, compute the following descriptive statistics: a. Range b. Variance c. Standard deviation d. Coefficient of variation

The following frequency distribution shows the price per share of the 30 companies in the Dow Jones Industrial Average (Barron s, February 2,2009 ). $$\begin{array}{cc} \text { Price per } & \text { Number of } \\ \text { Share } & \text { Companies } \\ \$ 0-9 & 4 \\ \$ 10-19 & 5 \\ \$ 20-29 & 7 \\ \$ 30-39 & 3 \\ \$ 40-49 & 4 \\ \$ 50-59 & 4 \\ \$ 60-69 & 0 \\ \$ 70-79 & 2 \\ \$ 80-89 & 0 \\ \$ 90-99 & 1 \end{array}$$ a. Compute the mean price per share and the standard deviation of the price per share for the Dow Jones Industrial Average companies. b. On January \(16,2006,\) the mean price per share was \(\$ 45.83\) and the standard deviation was \(\$ 18.14 .\) Comment on the changes in the price per share over the three-year period.

Consider a sample with data values of \(27,25,20,15,30,34,28,\) and \(25 .\) Compute the 20 th, \(25 \mathrm{th}, 65 \mathrm{th},\) and 75 th percentiles.

Consider the following data and corresponding weights. $$\begin{array}{cc} \boldsymbol{x}_{\boldsymbol{i}} & \text { Weight }\left(\boldsymbol{w}_{\boldsymbol{i}}\right) \\ 3.2 & 6 \\ 2.0 & 3 \\ 2.5 & 2 \\ 5.0 & 8 \end{array}$$ a. Compute the weighted mean. b. Compute the sample mean of the four data values without weighting. Note the difference in the results provided by the two computations.

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.