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Chapter 3: Problem 74

A sample of 2000 observations has a mean of 74 and a standard deviation of 12 . Using Chebyshev's theorem, find at least what percentage of the observations fall in the intervals \(\bar{x} \pm 2 s, \bar{x} \pm 2.5 s\), and \(\bar{x} \pm 3 s\). Note that here \(\bar{x} \pm 2 s\) represents the interval \(\bar{x}-2 s\) to \(\bar{x}+2 s\), and so on.

Short Answer

Expert verified
At least 75% of the observations fall within the interval 74卤24, at least 84% of the observations fall within the interval 74卤30, and at least 89% of the observations fall within the interval 74卤36.

Step by step solution

01

Identify values

Directly from the exercise, we have the mean \(\bar{x} = 74\), the standard deviation \(s = 12\), and the values of k given as \(k = 2\), \(k = 2.5\), and \(k = 3\).
02

Apply Chebyshev's theorem for k = 2

Putting \(k = 2\) in Chebyshev's theorem, we get \(1 - (1/k^2) = 1 - 1/4 = 0.75\) or 75% of the samples fall within an interval of \(\bar{x} \pm 2s = 74 \pm 2*12\).
03

Apply Chebyshev's theorem for k = 2.5

Putting \(k = 2.5\) in Chebyshev's theorem, we get \(1 - (1/k^2) = 1 - 1/6.25 = 0.84\) or 84% of the samples fall within an interval of \(\bar{x} \pm 2.5s = 74 \pm 2.5*12\).
04

Apply Chebyshev's theorem for k = 3

Putting \(k = 3\) in Chebyshev's theorem, we get \(1 - (1/k^2) = 1 - 1/9 = 0.89\) or 89% of the samples fall within an interval of \(\bar{x} \pm 3s = 74 \pm 3*12\).

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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 is a central concept in statistics often referred to as the "average." It is calculated by summing all the observations and dividing by the total number of observations. In our example, a sample of 2000 observations has a mean of 74. This mean provides a central value around which data points are distributed.

Think of the mean as a balancing point. It gives us an idea of where most values cluster. When analyzing data, knowing the mean helps us further explore how other values vary. This is where measures like standard deviation come into play.
Demystifying Standard Deviation
Standard deviation measures how spread out the numbers in a dataset are. A low standard deviation means the numbers are close to the mean, while a high standard deviation indicates a wide range of values. In this context, the standard deviation is 12.

With a mean of 74 and a standard deviation of 12,
  • Numbers close to 74 are common.
  • Values much higher or lower than 74 are further from the average.
Understanding this spread helps us apply Chebyshev's theorem, which uses standard deviation to estimate how many observations fall within a certain distance from the mean.
Interval Estimation with Chebyshev's Theorem
Interval estimation allows us to understand the proportion of data within specific ranges. Using Chebyshev's theorem, we can determine what percentage of observations lie within certain standard deviations from the mean.

Chebyshev's theorem states that for any dataset, at least \(1 - \frac{1}{k^2}\) of the observations fall within \(k\) standard deviations from the mean:
  • When \(k = 2\), at least 75% of observations fall between \(74 \pm 24\).
  • When \(k = 2.5\), 84% fall between \(74 \pm 30\).
  • When \(k = 3\), 89% are within \(74 \pm 36\).
This principle is powerful because it applies to any shaped distribution, providing a way to gauge data dispersion without knowing the precise distribution type.

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Most popular questions from this chapter

The following data give the numbers of car thefts that occurred in a city in the past 12 days \(\begin{array}{llllllll}6 & 3 & 7 & 1 & 14 & 3 & 8 & 7\end{array}\) \(\begin{array}{lll}2 & 6 & 9\end{array}\) 5 Calculate the range, variance, and standard deviation.

A brochure from the department of public safety in a northern state recommends that motorists should carry 12 items (flashlights, blankets, and so forth) in their vehicles for emergency use while driving in winter. The following data give the number of items out of these 12 that were carried in their vehicles by 15 randomly selected motorists. \(\begin{array}{llllllllllllllll}5 & 3 & 7 & 8 & 0 & 1 & 0 & 5 & 1 & 21 & 07 & 6 & 7 & 1 & 19\end{array}\) Find the mean, median, and mode for these data. Are the values of these summary measures population parameters or sample statistics? Explain.

The following data give the speeds of 13 cars (in mph) measured by radar, traveling on I-84. \(\begin{array}{lllllll}73 & 75 & 69 & 68 & 78 & 69 & 74 \\\ 76 & 72 & 79 & 68 & 77 & 71 & \end{array}\) a. Find the values of the three quartiles and the interquartile range. b. Calculate the (approximate) value of the 35 th percentile. c. Compute the percentile rank of 71 .

Suppose that there are 150 freshmen engineering majors at a college and each of them will take the same five courses next semester. Four of these courses will be taught in small sections of 25 students each, whereas the fifth course will be taught in one section containing all 150 freshmen. To accommodate all 150 students, there must be six sections of each of the four courses taught in 25 -student sections. Thus, there are 24 classes of 25 students each and one class of 150 students. a. Find the mean size of these 25 classes. b. Find the mean class size from a student's point of view, noting that each student has five classes containing \(25,25,25,25\), and 150 students, respectively. Are the means in parts a and \(\mathrm{b}\) equal? If not, why not?

In some applications, certain values in a data set may be considered more important than others. For example, to determine students' grades in a course, an instructor may assign a weight to the final exam that is twice as much as that to each of the other exams. In such cases, it is more appropriate to use the weighted mean. In general, for a sequence of \(n\) data values \(x_{1}, x_{2}, \ldots, x_{n}\) that are assigned weights \(w_{1}\), \(w_{2}, \ldots, w_{n}\), respectively, the weighted mean is found by the formula $$ \text { Weighted mean }=\frac{\sum x w}{\sum w} $$ where \(\Sigma x w\) is obtained by multiplying each data value by its weight and then adding the products. Suppose an instructor gives two exams and a final, assigning the final exam a weight twice that of each of the other exams. Find the weighted mean for a student who scores 73 and 67 on the first two exams and 85 on the final. (Hint: Here, \(x_{1}=73, x_{2}=67, x_{3}=85, w_{1}=w_{2}=1\), and \(w_{3}=2 .\) )
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