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8-7. Fitting to a Normal Distribution. Warm Up. Lesson Presentation. Lesson Quiz. Holt McDougal Algebra 2. Holt Algebra 2. Warm Up. Find the mean and standard deviation of each data set. 1. {2, 10, 5, 3}. mean: 5; std. dev. ≈ 3.08. 2. {2, 2, 2, 2,2}. mean: 2; std. dev. = 0.
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8-7 Fitting to a Normal Distribution Warm Up Lesson Presentation Lesson Quiz Holt McDougal Algebra 2 Holt Algebra 2
Warm Up Find the mean and standard deviation of each data set. 1. {2, 10, 5, 3} mean: 5; std. dev. ≈ 3.08 2. {2, 2, 2, 2,2} mean: 2; std. dev. = 0 3. Determine which data set has the greater standard deviation without calculating it. Explain.
Objectives Use tables to estimate areas under normal curves. Recognize data sets that are not normal.
Example 1: Finding Joint and Marginal Relative Frequencies Jamie can drive her car an average of 432 gallons per tank of gas, with a standard deviation of 36 miles. Use the graph to estimate the probability that Jamie will be able to drive more than 450 miles on her next tank of gas.
Example 1 : Continued The area under the normal curve is always equal to 1. Each square on the grid has an area of 10(0.001) = 0.01. Count the number of grid squares under the curve for values of x greater than 450. There are about 31 squares under the graph, so the probability is about 31(0.01) = 0.31 that she will be able to drive more than 450 miles on her next tank of gas.
You Try! Example 1 estimate the probability that Jamie will be able to drive less than 400 miles on her next tank of gas?
You Try! Example 1 continued There are about 19 squares under curve less than 400, so the probability is about 19(0.01) = 0.19 that she will be able to drive less than 400 miles on the next tank of gas.
First, find the standard normal value of 148, using μ = 160 and σ = 12. Example 2: Using Standard Normal Values Scores on a test are normally distributed with a mean of 160 and a standard deviation of 12. A. Estimate the probability that a randomly selected student scored less than 148. First, find the standard normal value of 148, using μ = 160 and σ = 12. µ X 148 160 Z = = = 1 12 σ
Example 2: Continued Use the table to find the area under the curve for all values less than 1, which is 0.16. The probability of scoring less than 148 is about 0.16. B. Estimate the probability that a randomly selected student scored between 154 and 184. Find the standard normal values of 154 and 184. Use the table to find the areas under the curve for all values less than z.
Example 2: Using Standard Normal Values continued Area=0.31 = = Area=0.98 Subtract the areas to eliminate where the regions overlap. The probability of scoring between 154 and 184 is about 0.98 – 0.31 = 0.67. µ µ X X 154 184 160 160 Z Z = = = = 2 0.5 12 12 σ σ
You Try! Example 2 Scores on a test are normally distributed with a mean of 142 and a standard deviation of 18. Estimate the probability of scoring above 106. First, find the standard normal value of 106, using μ = 142 and σ = 18. = µ X 106 142 Z = = 2 18 σ Use the table to find the area under the curve for all values less than –2, which is 0.02. The probability of scoring above 106 is 0.98.
Answers • .02 • .84 • .06 • .38 • 14. .5 • 15. .99 • 16. .38 • 17. .16 • 18. .92 • 19. .04
Determining Whether Data May Be Normally Distributed The lengths of the 20 snakes at a zoo, in inches, are shown in the table. The mean is 34.1 inches and the standard deviation is 10.5 inches. Does the data appear to be normally distributed? 2.6 13.1 23.6 34.1 44.6 55.1 65.6
Example 3: Continued Data does not appear to fit a normal curve
Another way to think about it is to plot the information and look at the shape.
You Try! Example 3 A random sample of salaries at a company is shown. If the mean is $37,000 and the standard deviation is $16,000, does the data appear to be normally distributed? No, the data does not appear to be normally distributed. 14 out of 18 values fall below the mean.
Check Point: Part I Scores on a test are normally distributed with a mean of 200 and a standard deviation of 12. Find each probability. 1. A randomly selected student scored less than 218. 0.93 2. A randomly selected student scored between 182 and 200. 0.43 3. A randomly selected student scored between 182 and 188. 0.09
Check Point: Part II 4. A randomly selected student scored above 224. 0.02 5. The weights, in grams, of 30 randomly chosen apples from a large bin are shown below. The mean weight is 110 grams and the standard deviation is 5.5 grams. Does the data appear to be normally distributed?
Check Point: Part III Yes; the projected number of values for each value of z is close to the actual number of data values.