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Suppose the data have a bell-shaped distribution with a mean of 30 and a standard deviation of 5. Use the empirical rule to determine the percentage of data within each of the following ranges: a. 20 to 40 b. 15 to 45 c. 25 to 35

Short Answer

Expert verified
a. 95%, b. 99.7%, c. 68%

Step by step solution

01

Understanding the Empirical Rule

The empirical rule, also known as the 68-95-99.7 rule, applies to bell-shaped or normal distributions. It states that approximately 68% of the data falls within one standard deviation of the mean, 95% within two standard deviations, and 99.7% within three standard deviations.
02

Calculating One Standard Deviation Range (20 to 40)

The mean (\(\mu\)) is 30 and the standard deviation (\(\sigma\)) is 5. One standard deviation from the mean spans from (\(30 - 5 = 25\)) to (\(30 + 5 = 35\)). But 20 to 40 covers (\(30 - 2\sigma\) to \(30 + 2\sigma\)), reaching from 20 to 40, which includes approximately 95% of the data according to the empirical rule.
03

Calculating Two Standard Deviation Range (15 to 45)

Two standard deviations from the mean extend from (\(30 - 2 \times 5 = 20\)) to (\(30 + 2 \times 5 = 40\)). However, 15 to 45 spans three standard deviations from 15 to 45, which encompass 99.7% of the data.
04

Calculating One Standard Deviation Inner Range (25 to 35)

This interval 25 to 35 is within one standard deviation of the mean since it goes from (\(30 - 5 = 25\)) to (\(30 + 5 = 35\)). According to the empirical rule, approximately 68% of the data falls within this range.

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Key Concepts

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

Normal Distribution
A normal distribution is a probability distribution that is symmetric around its mean, showing that data near the mean is more frequent in occurrence than data far from the mean. Imagine the shape of a bell curve, gently sloping upwards and downwards from the mean. This bell-shaped curve is a hallmark of normal distribution.

In the context of the problem, our data set is normally distributed with a mean of 30. This means that if you collected a lot of data from this distribution, most of the values would cluster around 30. Additionally, the symmetry means values further away from the mean occur less frequently. This is why normal distribution is so helpful in statistics: it provides a simple model for a wide range of phenomena.
  • Mean (\(ar{x}\)): This is the average of all the data points. Here, it's given as 30.
  • Symmetry: The distribution is perfectly symmetrical around the mean.
  • Bell Shape: Most of the data points lie close to the mean, creating a bell-shaped curve.
Standard Deviation
Standard deviation is a measure of the amount of variation or dispersion in a set of values. A low standard deviation means that the values tend to be close to the mean, while a high standard deviation means that the values are spread out over a wider range.

In our exercise, the standard deviation of the data is 5. This number helps us understand how spread out the numbers are around the mean of 30.

Let's break it down:
  • \(\sigma = 5\): Standard deviation means typical variability from the mean. A smaller standard deviation would mean data points are closer to 30.
  • Spans of Standard Deviation: Each increment of standard deviation (1σ, 2σ, etc.) represents a step of 5 units away from the mean.
  • Practical Utility: It helps in defining intervals such as 25 to 35, 20 to 40, etc., to contain predictable amounts of data as per normal distribution.
68-95-99.7 Rule
The 68-95-99.7 rule, also known as the empirical rule, is a handy guideline for understanding normal distributions. It tells us how much of our data lies within one, two, and three standard deviations from the mean.

Consider these key points:
  • 68% of data falls within one standard deviation (between 25 and 35 in this example).
  • 95% of data is contained within two standard deviations (from 20 to 40 here).
  • 99.7% of data will be within three standard deviations (which covers from 15 to 45 in our given problem).

This rule is significant because it allows us to understand and predict the dispersion of data in a bell-shaped normal distribution simply by knowing the mean and standard deviation. It makes it easy to see how concentrated or spread out the data can be.

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Most popular questions from this chapter

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

The National Association of Realtors provided data showing that home sales were the slowest in 10 years (Associated Press, December 24,2008 ). Sample data with representative sales prices for existing homes and new homes follow. Data are in thousands of dollars: $$\begin{array}{lllllllll} \text {Existing Homes} & 315.5 & 202.5 & 140.2 & 181.3 & 470.2 & 169.9 & 112.8 & 230.0 & 177.5 \\ \text {New Homes} & 275.9 & 350.2 & 195.8 & 525.0 & 225.3 & 215.5 & 175.0 & 149.5 & \end{array}$$ a. What is the median sales price for existing homes? b. What is the median sales price for new homes? c. Do existing homes or new homes have the higher median sales price? What is the difference between the median sales prices? d. A year earlier the median sales price for existing homes was \(\$ 208.4\) thousand and the median sales price for new homes was \(\$ 249\) thousand. Compute the percentage change in the median sales price of existing and new homes over the one-year period. Did existing homes or new homes have the larger percentage change in median sales price?

Consumer Reports provided overall customer satisfaction scores for AT\&T, Sprint, T-Mobile, and Verizon cell-phone services in major metropolitan areas throughout the United States. The rating for each service reflects the overall customer satisfaction considering a variety of factors such as cost, connectivity problems, dropped calls, static interference, and customer support. A satisfaction scale from 0 to 100 was used with 0 indicating completely dissatisfied and 100 indicating completely satisfied. The ratings for the four cellphone services in 20 metropolitan areas are as shown (Consumer Reports, January 2009 ). $$\begin{array}{lcccc} \text { Metropolitan Area } & \text { AT\&T } & \text { Sprint } & \text { T-Mobile } & \text { Verizon } \\ \text { Atlanta } & 70 & 66 & 71 & 79 \\ \text { Boston } & 69 & 64 & 74 & 76 \\ \text { Chicago } & 71 & 65 & 70 & 77 \\ \text { Dallas } & 75 & 65 & 74 & 78 \\ \text { Denver } & 71 & 67 & 73 & 77 \\ \text { Detroit } & 73 & 65 & 77 & 79 \\ \text { Jacksonville } & 73 & 64 & 75 & 81 \\ \text { Las Vegas } & 72 & 68 & 74 & 81 \\ \text { Los Angeles } & 66 & 65 & 68 & 78 \\ \text { Miami } & 68 & 69 & 73 & 80 \\ \text { Minneapolis } & 68 & 66 & 75 & 77 \\ \text { Philadelphia } & 72 & 66 & 71 & 78 \\ \text { Phoenix } & 68 & 66 & 76 & 81 \\ \text { San Antonio } & 75 & 65 & 75 & 80 \\ \text { San Diego } & 69 & 68 & 72 & 79 \\ \text { San Francisco } & 66 & 69 & 73 & 75 \\ \text { Seattle } & 68 & 67 & 74 & 77 \\ \text { St. Louis } & 74 & 66 & 74 & 79 \\ \text { Tampa } & 73 & 63 & 73 & 79 \\ \text { Washington } & 72 & 68 & 71 & 76 \end{array}$$ a. Consider T-Mobile first. What is the median rating? b. Develop a five-number summary for the T-Mobile service. c. Are there outliers for T-Mobile? Explain. d. Repeat parts (b) and (c) for the other three cell-phone services. e. Show the box plots for the four cell-phone services on one graph. Discuss what a comparison of the box plots tells about the four services. Which service did Consumer Reports recommend as being best in terms of overall customer satisfaction?

The following times were recorded by the quarter-mile and mile runners of a university track team (times are in minutes). $$\begin{array}{llllll} \text {Quarter-Mile Times:} & .92 & .98 & 1.04 & .90 & .99 \\ \text {Mile Times:} & 4.52 & 4.35 & 4.60 & 4.70 & 4.50 \end{array}$$ 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?

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