1. © 2010 Pearson Prentice Hall. All rights reserved The Normal Distribution

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3. © 2010 Pearson Prentice Hall. All rights reserved Relative frequency histograms that are symmetric and bell-shaped are said to have the shape of a normal curve. 7-3

4. © 2010 Pearson Prentice Hall. All rights reserved If a continuous random variable is normally distributed, or has a normal probability distribution, then a relative frequency histogram of the random variable has the shape of a normal curve (bell-shaped and symmetric). 7-4

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8. As we saw in the previous slide, if the mean of the distribution changes, the center of the distribution shifts or translates across the axis. If the standard deviation changes, the distribution will either become more spread out and the maximum density will decrease or the distribution will become much more concentrated around the mean and the maximum density will increase.

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10. © 2010 Pearson Prentice Hall. All rights reserved EXAMPLEInterpreting the Area Under a Normal Curve The weights of giraffes are approximately normally distributed with mean μ = 2200 pounds and standard deviation σ = 200 pounds. (a)Draw a normal curve with the parameters labeled. (b)Shade the area under the normal curve to the left of x = 2100 pounds. (c)Suppose that the area under the normal curve to the left of x = 2100 pounds is 0.3085. Provide two interpretations of this result. (a), (b) (c) The proportion of giraffes whose weight is less than 2100 pounds is 0.3085 The probability that a randomly selected giraffe weighs less than 2100 pounds is 0.3085. 7-10

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14. © 2010 Pearson Prentice Hall. All rights reserved The table gives the area under the standard normal curve for values to the left of a specified Z-score, z o, as shown in the figure. 7-14

15. © 2010 Pearson Prentice Hall. All rights reserved Find the area under the standard normal curve to the left of z = -0.38. EXAMPLE Finding the Area Under the Standard Normal Curve Area left of z = -0.38 is 0.3520. 7-15

16. © 2010 Pearson Prentice Hall. All rights reserved Area under the normal curve to the right of z o = 1 – Area to the left of z o 7-16

17. © 2010 Pearson Prentice Hall. All rights reserved EXAMPLE Finding the Area Under the Standard Normal Curve Find the area under the standard normal curve to the right of Z = 1.25. Area right of 1.25 = 1 – area left of 1.25 = 1 – 0.8944 = 0.1056 7-17

18. © 2010 Pearson Prentice Hall. All rights reserved Find the area under the standard normal curve between z = -1.02 and z = 2.94. EXAMPLE Finding the Area Under the Standard Normal Curve Area between -1.02 and 2.94 = (Area left of z = 2.94) – (area left of z = -1.02) = 0.9984 – 0.1539 = 0.8445 7-18 In this case finding the area to the left of the z score of 2.94 is bigger than the area we are concerned with, and the area to the left of the z score of -1.04 contains area that we are not concerned with. To find the area we are interested in we will subtract the smaller area from the larger, which will give us the area that is between them.

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20. © 2010 Pearson Prentice Hall. All rights reserved Find the z-scores that separate the middle 80% of the area under the normal curve from the 20% in the tails. EXAMPLE Finding a z-score from a Specified Area Area = 0.8 Area = 0.1 z 1 is the z-score such that the area left is 0.1, so z 1 = -1.28. z 2 is the z-score such that the area left is 0.9, so z 2 = 1.28. 7-20

21. © 2010 Pearson Prentice Hall. All rights reserved Notation for the Probability of a Standard Normal Random Variable P(a < Z < b) represents the probability a standard normal random variable is between a and b P(Z > a)represents the probability a standard normal random variable is greater than a. P(Z < a) represents the probability a standard normal random variable is less than a. 7-21

22. © 2010 Pearson Prentice Hall. All rights reserved For any continuous random variable, the probability of observing a specific value of the random variable is 0. For example, for a standard normal random variable, P(a) = 0 for any value of a. This is because there is no area under the standard normal curve associated with a single value, so the probability must be 0. Therefore, the following probabilities are equivalent: P(a < Z < b) = P(a < Z < b) = P(a < Z < b) = P(a < Z < b) 7-22

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24. © 2010 Pearson Prentice Hall. All rights reserved It is known that the length of a certain steel rod is normally distributed with a mean of 100 cm and a standard deviation of 0.45 cm. * What is the probability that a randomly selected steel rod has a length less than 99.2 cm? * Based upon information obtained from Stefan Wilk. EXAMPLEFinding the Probability of a Normal Random Variable Interpretation: If we randomly selected 100 steel rods, we would expect about 4 of them to be less than 99.2 cm. 7-24

25. © 2010 Pearson Prentice Hall. All rights reserved It is known that the length of a certain steel rod is normally distributed with a mean of 100 cm and a standard deviation of 0.45 cm. What is the probability that a randomly selected steel rod has a length between 99.8 and 100.3 cm? Interpretation: If we randomly selected 100 steel rods, we would expect about 42 of them to be between 99.8 cm and 100.3 cm. EXAMPLEFinding the Probability of a Normal Random Variable 7-25

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27. © 2010 Pearson Prentice Hall. All rights reserved The combined (verbal + quantitative reasoning) score on the GRE is normally distributed with mean 1049 and standard deviation 189. (Source: http://www.ets.org/Media/Tests/GRE/pdf/994994.pdf.) EXAMPLEFinding the Value of a Normal Random Variable What is the score of a student whose percentile rank is at the 85 th percentile? The z-score that corresponds to the 85 th percentile is the z-score such that the area under the standard normal curve to the left is 0.85. This z-score is 1.04. x = µ + zσ = 1049 + 1.04(189) = 1246 Interpretation: 85% of all combined scores on the GRE will be lower than 1246, or that the student has a score that is higher than 85% of everyone that took the GRE exam. 7-27

28. © 2010 Pearson Prentice Hall. All rights reserved It is known that the length of a certain steel rod is normally distributed with a mean of 100 cm and a standard deviation of 0.45 cm. Suppose the manufacturer wants to accept 90% of all rods manufactured. Determine the length of rods that make up the middle 90% of all steel rods manufactured. EXAMPLEFinding the Value of a Normal Random Variable Area = 0.05 z 1 = -1.645 and z 2 = 1.645 x 1 = µ + z 1 σ = 100 + (-1.645)(0.45) = 99.26 cm x 2 = µ + z 2 σ = 100 + (1.645)(0.45) = 100.74 cm Interpretation: The length of steel rods that make up the middle 90% of all steel rods manufactured would have lengths between 99.26 cm and 100.74 cm. 7-28

29. © 2010 Pearson Prentice Hall. All rights reserved A normal probability plot plots observed data versus normal scores. A normal score is the expected Z-score of the data value if the distribution of the random variable is normal. The expected Z-score of an observed value will depend upon the number of observations in the data set. 7-29

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31. © 2010 Pearson Prentice Hall. All rights reserved If sample data is taken from a population that is normally distributed, a normal probability plot of the actual values versus the expected Z-scores will be approximately linear. 7-31

32. © 2010 Pearson Prentice Hall. All rights reserved The following data represent the time between eruptions (in seconds) for a random sample of 15 eruptions at the Old Faithful Geyser in California. Is there reason to believe the time between eruptions is normally distributed? EXAMPLEInterpreting a Normal Probability Plot 7-32

33. © 2010 Pearson Prentice Hall. All rights reserved The random variable time between eruptions is likely not normal. 7-33

34. © 2010 Pearson Prentice Hall. All rights reserved Suppose that seventeen randomly selected workers at a detergent factory were tested for exposure to a Bacillus subtillis enzyme by measuring the ratio of forced expiratory volume (FEV) to vital capacity (VC). NOTE: FEV is the maximum volume of air a person can exhale in one second; VC is the maximum volume of air that a person can exhale after taking a deep breath. Is it reasonable to conclude that the FEV to VC (FEV/VC) ratio is normally distributed? Source: Shore, N.S.; Greene R.; and Kazemi, H. Lung Dysfunction in Workers Exposed to Bacillus subtillis Enzyme, Environmental Research, 4 (1971), pp. 512 - 519. EXAMPLEInterpreting a Normal Probability Plot 7-34

35. © 2010 Pearson Prentice Hall. All rights reserved It is reasonable to believe that FEV/VC is normally distributed. 7-35