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Consider a sample with a mean of 30 and a standard deviation of \(5 .\) Use Chebyshev's theorem to determine the percentage of the data within each of the following ranges. a. 20 to 40 b. 15 to 45 c. 22 to 38 d. 18 to 42 e. 12 to 48

Short Answer

Expert verified
a) 75%, b) 88.89%, c) 60.94%, d) 82.64%, e) 92.28%.

Step by step solution

01

Identify the Mean and Standard Deviation

The sample has a mean, \( \mu = 30 \), and a standard deviation, \( \sigma = 5 \). We'll use these values along with Chebyshev's theorem to determine the percentage of data within given ranges.
02

Apply Chebyshev's Theorem Formula

Chebyshev's theorem states that at least \( 1 - \frac{1}{k^2} \) of the data from a sample will fall within \( k \) standard deviations from the mean. To apply this theorem, we need to determine \( k \) in each case based on the given range.
03

Calculate \( k \) for Range 20 to 40

The range 20 to 40 is \(10\) units below and above the mean. Calculate \( k \) using \( k = \frac{\text{Range Extremes} - \mu}{\sigma} \). Here, \( k = \frac{40 - 30}{5} = \frac{10}{5} = 2 \).
04

Determine Percentage for Range 20 to 40

Using Chebyshev's theorem, calculate the percentage with \( k = 2 \): \( 1 - \frac{1}{2^2} = 1 - \frac{1}{4} = 0.75 \). Thus, at least 75% of the data falls within this range.
05

Calculate \( k \) for Range 15 to 45

The range 15 to 45 is \(15\) units below and above the mean. \( k = \frac{45 - 30}{5} = \frac{15}{5} = 3 \).
06

Determine Percentage for Range 15 to 45

With \( k = 3 \), the theorem gives: \( 1 - \frac{1}{3^2} = 1 - \frac{1}{9} \approx 0.8889 \). Hence, at least 88.89% of the data are within this range.
07

Calculate \( k \) for Range 22 to 38

The range 22 to 38 is \(8\) units below and above the mean. \( k = \frac{38 - 30}{5} = \frac{8}{5} = 1.6 \).
08

Determine Percentage for Range 22 to 38

With \( k = 1.6 \): \( 1 - \frac{1}{(1.6)^2} = 1 - \frac{1}{2.56} \approx 0.6094 \). At least 60.94% of the data falls within this range.
09

Calculate \( k \) for Range 18 to 42

The range 18 to 42 is \(12\) units below and above the mean. \( k = \frac{42 - 30}{5} = \frac{12}{5} = 2.4 \).
10

Determine Percentage for Range 18 to 42

With \( k = 2.4 \): \( 1 - \frac{1}{(2.4)^2} = 1 - \frac{1}{5.76} \approx 0.8264 \). At least 82.64% of the data are within this range.
11

Calculate \( k \) for Range 12 to 48

The range 12 to 48 is \(18\) units below and above the mean. \( k = \frac{48 - 30}{5} = \frac{18}{5} = 3.6 \).
12

Determine Percentage for Range 12 to 48

With \( k = 3.6 \): \( 1 - \frac{1}{(3.6)^2} = 1 - \frac{1}{12.96} \approx 0.9228 \). Hence, at least 92.28% of the data are within this range.

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Key Concepts

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

Understanding Standard Deviation
Standard deviation is a measure that indicates how much observations in a data set vary from the mean. It's a crucial concept in statistics as it provides insight into the data's spread.
A small standard deviation means the data points are close to the mean, while a large standard deviation indicates the data is spread out over a wider range. Consider this in terms of consistency. Data with a low standard deviation has little variability and can be seen as consistent. This consistency is especially important in quality control and risk assessment. For instance, if a factory produces parts with a low standard deviation of sizes, this implies each part is much like the others. Understanding how much data deviates from the mean helps in making informed decisions and understanding the reliability of data sets.
Decoding the Mean
The mean, often called the average, is simply the sum of all data points divided by the number of data points. It is a central value that provides a summary of the data set. In the context of Chebyshev's Theorem, the mean is essential as it helps to determine the center from which we measure the spread of data via standard deviations.
For example, a mean of 30 in a data set suggests that if you were to randomly pick a data point, on average it should be around 30. This makes the mean a straightforward approach to grasp the "typical" value within the data. But remember, while the mean gives a central point, it doesn't reflect the data's variability or distribution shape; this is where standard deviation and other statistical measures come into play.
Importance of Percentage of Data Range
The percentage of data range, as applied in Chebyshev's Theorem, helps to understand how much of the data falls within a specific interval around the mean. This is expressed as a percentage. Chebyshev's Theorem formula, given as \(1 - \frac{1}{k^2}\), where \(k\) is the number of standard deviations from the mean, gives us the minimum percentage for which data falls within \(k\) standard deviations.
This is useful when dealing with non-normally distributed data. Unlike the empirical rule, Chebyshev's Theorem holds for any distribution shape, whether it's skewed or otherwise.Evaluating the percentage of data range can greatly aid in risk assessment in fields like finance or quality control, where understanding consistency and variation is crucial.
Basics of Statistical Analysis
Statistical analysis involves collecting, reviewing, and drawing conclusions from data. It's a vital part of any field that relies on data-driven decisions. When applying Chebyshev's Theorem, statistical analysis is key. It aids in understanding the data's structure and what conclusions can be drawn based on the given parameters of mean and standard deviation. Through statistical analysis, one can determine whether Chebyshev's theorem is applicable and effective in providing insights into the data's behavior. Analysis tasks can include:
  • Calculating central tendencies like mean and median.
  • Evaluating consistency by assessing standard deviation and variance.
  • Identifying patterns or outliers in the data.
  • Deciphering data behavior using theorems like Chebyshev's.
Statistical analysis broadens the understanding of how data behaves, helping to guide strategies and make evidence-based decisions.

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Most popular questions from this chapter

The cost of consumer purchases such as housing, gasoline, Internet services, tax preparation, and hospitalization were provided in The Wall Street Journal, January 2, 2007. Sample data typical of the cost of tax-return preparation by services such as H\&R Block are shown here. \\[\begin{array}{lllll}120 & 230 & 110 & 115 & 160 \\ 130 & 150 & 105 & 195 & 155 \\ 105 & 360 & 120 & 120 & 140 \\\100 & 115 & 180 & 235 & 255\end{array}\\] a. Compute the mean, median, and mode. b. Compute the first and third quartiles. c. Compute and interpret the 90 th percentile.

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?

Annual sales, in millions of dollars, for 21 pharmaceutical companies follow. \\[\begin{array}{rrrrrr} 8408 & 1374 & 1872 & 8879 & 2459 & 11413 \\ 608 & 14138 & 6452 & 1850 & 2818 & 1356 \\ 10498 & 7478 & 4019 & 4341 & 739 & 2127 \\ 3653 & 5794 & 8305 & &\end{array}\\] a. Provide a five-number summary. b. Compute the lower and upper limits. c. Do the data contain any outliers? d. Johnson \& Johnson's sales are the largest on the list at \(\$ 14,138\) million. Suppose a data entry error (a transposition) had been made and the sales had been entered as \(\$ 41,138\) million. Would the method of detecting outliers in part (c) identify this problem and allow for correction of the data entry error? e. Show a box plot.

Consider a sample with a mean of 500 and a standard deviation of \(100 .\) What are the \(z\) -scores for the following data values: \(520,650,500,450,\) and \(280 ?\)

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?

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