1. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. Chapter The Normal Probability Distribution 7

2. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. Section Properties of the Normal Distribution 7.1

3. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. Objectives 1.Use the uniform probability distribution 2.Graph a normal curve 3.State the properties of the normal curve 4.Explain the role of area in the normal density function 7-3

4. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. Objective 1 Use the Uniform Probability Distribution 7-4

5. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 7-5 Suppose that United Parcel Service is supposed to deliver a package to your front door and the arrival time is somewhere between 10 am and 11 am. Let the random variable X represent the time from 10 am when the delivery is supposed to take place. The delivery could be at 10 am (x = 0) or at 11 am (x = 60) with all 1-minute interval of times between x = 0 and x = 60 equally likely. That is to say your package is just as likely to arrive between 10:15 and 10:16 as it is to arrive between 10:40 and 10:41. The random variable X can be any value in the interval from 0 to 60, that is, 0 < X < 60. Because any two intervals of equal length between 0 and 60, inclusive, are equally likely, the random variable X is said to follow a uniform probability distribution. EXAMPLE Illustrating the Uniform Distribution

6. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. A probability density function (pdf) is an equation used to compute probabilities of continuous random variables. It must satisfy the following two properties: 1.The total area under the graph of the equation over all possible values of the random variable must equal 1. 2.The height of the graph of the equation must be greater than or equal to 0 for all possible values of the random variable. A probability density function (pdf) is an equation used to compute probabilities of continuous random variables. It must satisfy the following two properties: 1.The total area under the graph of the equation over all possible values of the random variable must equal 1. 2.The height of the graph of the equation must be greater than or equal to 0 for all possible values of the random variable. 7-6

7. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 7-7 The graph below illustrates the properties for the “time” example. Notice the area of the rectangle is one and the graph is greater than or equal to zero for all x between 0 and 60, inclusive. Because the area of a rectangle is height times width, and the width of the rectangle is 60, the height must be 1/60.

8. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 7-8 Values of the random variable X less than 0 or greater than 60 are impossible, thus the function value must be zero for X less than 0 or greater than 60.

9. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 7-9 The area under the graph of the density function over an interval represents the probability of observing a value of the random variable in that interval.

10. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 7-10 The probability of choosing a time that is between 15 and 30 minutes after the hour is the area under the uniform density function. 15 30 Area = P(15 < x < 30) = 15/60 = 0.25 EXAMPLEArea as a Probability

11. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. Objective 2 Graph a Normal Curve 7-11

12. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 7-12 Relative frequency histograms that are symmetric and bell-shaped are said to have the shape of a normal curve.

13. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 7-13 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).

14. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 7-14

15. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. Objective 3 State the Properties of the Normal Curve 7-15

16. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. Properties of the Normal Density Curve 1.It is symmetric about its mean, μ. 2.Because mean = median = mode, the curve has a single peak and the highest point occurs at x = μ. 3.It has inflection points at μ – σ and μ – σ 4.The area under the curve is 1. 5.The area under the curve to the right of μ equals the area under the curve to the left of μ, which equals 1/2. Properties of the Normal Density Curve 1.It is symmetric about its mean, μ. 2.Because mean = median = mode, the curve has a single peak and the highest point occurs at x = μ. 3.It has inflection points at μ – σ and μ – σ 4.The area under the curve is 1. 5.The area under the curve to the right of μ equals the area under the curve to the left of μ, which equals 1/2. 7-16

17. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 6.As x increases without bound (gets larger and larger), the graph approaches, but never reaches, the horizontal axis. As x decreases without bound (gets more and more negative), the graph approaches, but never reaches, the horizontal axis. 7.The Empirical Rule: Approximately 68% of the area under the normal curve is between x = μ – σ and x = μ + σ; approximately 95% of the area is between x = μ – 2σ and x = μ + 2σ; approximately 99.7% of the area is between x = μ – 3σ and x = μ + 3σ. 6.As x increases without bound (gets larger and larger), the graph approaches, but never reaches, the horizontal axis. As x decreases without bound (gets more and more negative), the graph approaches, but never reaches, the horizontal axis. 7.The Empirical Rule: Approximately 68% of the area under the normal curve is between x = μ – σ and x = μ + σ; approximately 95% of the area is between x = μ – 2σ and x = μ + 2σ; approximately 99.7% of the area is between x = μ – 3σ and x = μ + 3σ. 7-17

18. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 7-18

19. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. Objective 4 Explain the Role of Area in the Normal Density Function 7-19

20. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 7-20 The data on the next slide represent the heights (in inches) of a random sample of 50 two-year old males. (a) Draw a histogram of the data using a lower class limit of the first class equal to 31.5 and a class width of 1. (b) Do you think that the variable “height of 2- year old males” is normally distributed? EXAMPLE A Normal Random Variable

21. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 7-21 36.036.234.836.034.638.435.436.8 34.733.437.438.231.537.736.934.0 34.435.737.939.334.036.935.137.0 33.236.135.235.633.036.833.535.0 35.135.234.436.736.036.035.735.7 38.333.639.837.037.234.835.738.9 37.239.3

22. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 7-22

23. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 7-23 In the next slide, we have a normal density curve drawn over the histogram. How does the area of the rectangle corresponding to a height between 34.5 and 35.5 inches relate to the area under the curve between these two heights?

24. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 7-24

25. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 7-25

26. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. Area under a Normal Curve Suppose that a random variable X is normally distributed with mean μ and standard deviation σ. The area under the normal curve for any interval of values of the random variable X represents either the proportion of the population with the characteristic described by the interval of values or the probability that a randomly selected individual from the population will have the characteristic described by the interval of values. Area under a Normal Curve Suppose that a random variable X is normally distributed with mean μ and standard deviation σ. The area under the normal curve for any interval of values of the random variable X represents either the proportion of the population with the characteristic described by the interval of values or the probability that a randomly selected individual from the population will have the characteristic described by the interval of values. 7-26

27. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 7-27 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.

28. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 7-28 EXAMPLEInterpreting the Area Under a Normal Curve μ = 2200 pounds and σ = 200 pounds (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.