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Chapter 3: Problem 141

The distribution of the lengths of fish in a certain lake is not known, but it is definitely not bell shaped. It is estimated that the mean length is 6 inches with a standard deviation of 2 inches. a. At least what proportion of fish in the lake are between 3 inches and 9 inches long? b. What is the smallest interval that will contain the lengths of at least \(84 \%\) of the fish? c. Find an interval so that fewer than \(36 \%\) of the fish have lengths outside this interval.

Short Answer

Expert verified
a. At least 55.56% of the fish are between 3 and 9 inches long. b. The smallest interval that contains at least 84% of the fish is roughly between 1.53 and 10.47 inches. c. An interval that contains at least 64% of the fish, leaving less than 36% of fish lengths outside, is approximately between 2.876 and 9.124 inches.

Step by step solution

01

Applying Chebyshev's theorem for part a

For part a, we need to find the proportion within 1.5 standard deviations of the mean: \( k = \frac{|9-6|}{2} = 1.5 \). By applying Chebyshev's theorem which states that at least \( 1 - \frac{1}{k^2} \) of data fall within \( k \) standard deviations of the mean, we get \( 1 - \frac{1}{1.5^2} = 0.5556 \) or approximately \( 55.56 \% \).
02

Applying Chebyshev's theorem for part b

For part b, we need to find the smallest interval that will contain the lengths of at least 84 \% of the fish. Setting \( 1 - \frac{1}{k^2} \) to be equal to or greater than 0.84, we need to solve for \( k \). Through rearranging the equation and taking the square root, we get \( k = \sqrt{\frac{1}{1-0.84}} \approx 2.236 \). Thus, the smallest interval is \( \mu - k\sigma, \mu + k\sigma \) which is \( 6 - 2.236*2, 6 + 2.236*2 \) or approximately \( 1.53, 10.47 \) inches.
03

Applying Chebyshev's theorem for part c

For part c, we need to find an interval so that fewer than 36 \% of the fish have lengths outside this interval. This means that the interval should contain more than 64 \% of the data, hence set \( 1 - \frac{1}{k^2} \) to be equal to or greater than 0.64. Solving for \( k \), we get \( k = \sqrt{\frac{1}{1-0.64}} \approx 1.562 \). Consequently, the required interval is \( \mu - k\sigma, \mu + k\sigma \) which is \( 6 - 1.562*2, 6 + 1.562*2 \) or approximately \( 2.876, 9.124 \) inches.

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Most popular questions from this chapter

One disadvantage of the standard deviation as a measure of dispersion is that it is a measure of absolute variability and not of relative variability. Sometimes we may need to compare the variability of two different data sets that have different units of measurement. The coefficient of variation is one such measure. The coefficient of variation, denoted by CV, expresses standard deviation as a percentage of the mean and is computed as follows: For population data: \(\mathrm{CV}=\frac{\sigma}{\mu} \times 100 \%\) For sample data: \(\quad \mathrm{CV}=\frac{s}{\bar{x}} \times 100 \%\) The yearly salaries of all employees who work for a company have a mean of \(\$ 62,350\) and a standard deviation of \(\$ 6820\). The years of experience for the same employees have a mean of 15 years and a standard deviation of 2 years. Is the relative variation in the salaries larger or smaller than that in years of experience for these employees?

Briefly explain Chebyshev's theorem and its applications.

The mean time taken to learn the basics of a word processor by all students is 200 minutes with a standard deviation of 20 minutes. a. Using Chebyshev's theorem, find at least what percentage of students will learn the basics of this word processor in i. 160 to 240 minutes ii. 140 to 260 minutes \({ }^{*}\) b. Using Chebyshev's theorem, find the interval that contains the time taken by at least \(75 \%\) of all students to learn this word processor.

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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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