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Chapter 3: Problem 61

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}{llllllll} 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
The mean loan debt is 9.9k, variance is 22.188k虏, and standard deviation is 4.71k.

Step by step solution

01

Calculate the Mean

To find the mean loan debt, sum all the loan amounts and then divide by the number of students. The data consists of the amounts: 10.1, 14.8, 5.0, 10.2, 12.4, 12.2, 2.0, 11.5, 17.8, 4.0. \[\text{Mean} = \frac{10.1 + 14.8 + 5.0 + 10.2 + 12.4 + 12.2 + 2.0 + 11.5 + 17.8 + 4.0}{10} = \frac{99.0}{10} = 9.9\] So, the mean loan debt is 9.9 thousand dollars.
02

Calculate Each Deviation from the Mean

Subtract the mean (9.9) from each loan amount to find the deviations. They are: 10.1-9.9=0.2, 14.8-9.9=4.9, 5.0-9.9=-4.9, 10.2-9.9=0.3, 12.4-9.9=2.5, 12.2-9.9=2.3, 2.0-9.9=-7.9, 11.5-9.9=1.6, 17.8-9.9=7.9, 4.0-9.9=-5.9.
03

Square Each Deviation

Square each of the deviations found in the previous step: \(0.2^2 = 0.04\), \(4.9^2 = 24.01\), \((-4.9)^2 = 24.01\), \(0.3^2 = 0.09\), \(2.5^2 = 6.25\), \(2.3^2 = 5.29\), \((-7.9)^2 = 62.41\), \(1.6^2 = 2.56\), \(7.9^2 = 62.41\), and \((-5.9)^2 = 34.81\).
04

Calculate the Variance

Sum all the squared deviations and then divide by the number of observations to find the variance. \[\text{Variance} = \frac{0.04 + 24.01 + 24.01 + 0.09 + 6.25 + 5.29 + 62.41 + 2.56 + 62.41 + 34.81}{10} = \frac{221.88}{10} = 22.188\] So, the variance is 22.188 thousand dollars squared.
05

Calculate the Standard Deviation

Take the square root of the variance to find the standard deviation. \[\text{{Standard Deviation}} = \sqrt{22.188} \approx 4.71\] So, the standard deviation is approximately 4.71 thousand dollars.

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Key Concepts

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

Understanding Mean Calculation
The concept of mean is crucial in descriptive statistics as it gives the average value of a dataset. When you have a set of values, like the loan debts of students in the problem, the mean tells you what value you might expect if the debt were evenly distributed among all students. To calculate the mean, you add up all the values and then divide by the number of values you have. It's like sharing equally. For the student loans data:
  • Add up all the loan amounts: 10.1 + 14.8 + 5.0 + 10.2 + 12.4 + 12.2 + 2.0 + 11.5 + 17.8 + 4.0.
  • This gives a total of 99.0 thousand dollars.
  • Since there are 10 students, you divide 99.0 by 10.
This calculation gives a mean of 9.9 thousand dollars. The mean is a simple yet powerful way to summarize data and allows you to compare different sets of data easily.
Grasping Variance Calculation
Variance is an important statistical concept that tells you how much the values in a dataset differ from the mean. A high variance means the values are widely spread out, while a low variance indicates they are closer to the mean. To calculate variance, follow these steps:
  • First, find the difference between each data point and the mean (these are called deviations).
  • Next, square each of these deviations to eliminate negative values and place emphasis on larger deviations.
  • Finally, sum up all the squared deviations and divide by the number of data points.
For our student loans, after computing the squared deviations for each loan amount, we summed them to get 221.88. Dividing by 10 gives a variance of 22.188. Variance is crucial as it provides context to the mean, indicating whether the loan amounts are generally uniform or vary significantly.
Demystifying Standard Deviation Calculation
Standard deviation is a key statistical measure that builds on variance by providing a more interpretable form. It tells you, on average, how far each data point is from the mean. This is highly useful as it has the same unit as the original data, making it easier to understand. To calculate standard deviation, simply take the square root of the variance.
  • For the student loans data, the variance we calculated was 22.188.
  • The square root of 22.188 is approximately 4.71.
This means that on average, the amount of debt deviates from the mean by about 4.71 thousand dollars. Understanding standard deviation helps you see the distribution of data, and whether most values fall close to the mean or are spread out over a wide range.

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Most popular questions from this chapter

The following times were recorded by the quarter-mile and mile runners of a university track team (times are in minutes). \\[\begin{array}{lrrrrr} \text {Quarter-Mile Times:} & .92 & .98 & 1.04 & .90 & .99 \\ \text {Mile Times:} & 4.52 & 4.35 & 4.60 & 4.70 & 4.50 \end{array}\\] After viewing this sample of running times, one of the coaches commented that the quartermilers turned in the more consistent times. Use the standard deviation and the coefficient of variation to summarize the variability in the data. Does the use of the coefficient of variation indicate that the coach's statement should be qualified?

Scores turned in by an amateur golfer at the Bonita Fairways Golf Course in Bonita Springs, Florida, during 2005 and 2006 are as follows: 2005 Season \(74 \quad 78 \quad 79 \quad 77 \quad 75 \quad 73 \quad 75 \quad 77\) 2006 Season \(71 \quad 70 \quad 75 \quad 77 \quad 85 \quad 80 \quad 71 \quad 79\) a. Use the mean and standard deviation to evaluate the golfer's performance over the two-year period. b. What is the primary difference in performance between 2005 and \(2006 ?\) What improvement, if any, can be seen in the 2006 scores?

Consider a sample with data values of \(10,20,21,17,16,\) and \(12 .\) Compute the mean and median.

A forecasting technique referred to as moving averages uses the average or mean of the most recent \(n\) periods to forecast the next value for time series data. With a three-period moving average, the most recent three periods of data are used in the forecast computation. Consider a product with the following demand for the first three months of the current year: January ( 800 units), February (750 units), and March (900 units). a. What is the three-month moving average forecast for April? b. \(\quad\) A variation of this forecasting technique is called weighted moving averages. The weighting allows the more recent time series data to receive more weight or more importance in the computation of the forecast. For example, a weighted three-month moving average might give a weight of 3 to data one month old, a weight of 2 to data two months old, and a weight of 1 to data three months old. Use the data given to provide a three-month weighted moving average forecast for April.

Consider the following data and corresponding weights. $$\begin{array}{cc} x_{i} & \text { Weight }\left(w_{i}\right) \\ 3.2 & 6 \\ 2.0 & 3 \\ 2.5 & 2 \\ 5.0 & 8 \end{array}$$ 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.
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