1 / 42

Chapter 7 The Normal Probability Distribution

Chapter 7 The Normal Probability Distribution. 7.1 Properties of the Normal Distribution. EXAMPLE Illustrating the Uniform Distribution

vail
Download Presentation

Chapter 7 The Normal Probability Distribution

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  1. Chapter 7The Normal Probability Distribution 7.1 Properties of the Normal Distribution

  2. EXAMPLE Illustrating the Uniform Distribution 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 from10 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.

  3. Probability Density Function A probability density function is an equation that is used to compute probabilities of continuous random variables that must satisfy the following two properties. Let f(x) be a probability density function. 1. The area under the graph of the equation over all possible values of the random variable must equal one,that is 2. The graph of the equation must be greater than or equal to zero for all possible values of the random variable. That is, the graph of the equation must lie on or above the horizontal axis for all possible values of the random variable, i.e. f(x)>=0

  4. The area under the graph of a density function over some interval represents the probability of observing a value of the random variable in that interval.

  5. EXAMPLE Area as a Probability Referring to the earlier example, what is the probability that your package arrives between 10:10 am and 10:20 am? (10:20-10:10)/60=10/60=1/6

  6. Relative frequency histograms that are symmetric and bell-shaped are said to have the shape of a normal curve.

  7. 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).

  8. Properties of the Normal Density Curve 7. The Empirical Rule: About 68% of the area under the graph is within one standard deviation of the mean; about 95% of the area under the graph is within two standard deviations of the mean; about 99.7% of the area under the graph is within three standard deviations of the mean.

  9. EXAMPLE A Normal Random Variable The following data represent the heights (in inches) of a random sample of 50 two-year old males. (a) Create a relative frequency distribution with the lower class limit of the first class equal to 31.5 and a class width of 1. (b) Draw a histogram of the data. (c ) Do you think that the variable “height of 2-year old males” is normally distributed?

  10. 36.0 36.2 34.8 36.0 34.6 38.4 35.4 36.8 34.7 33.4 37.4 38.2 31.5 37.7 36.9 34.0 34.4 35.7 37.9 39.3 34.0 36.9 35.1 37.0 33.2 36.1 35.2 35.6 33.0 36.8 33.5 35.0 35.1 35.2 34.4 36.7 36.0 36.0 35.7 35.7 38.3 33.6 39.8 37.0 37.2 34.8 35.7 38.9 37.2 39.3

  11. 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?

  12. EXAMPLE Interpreting the Area Under a Normal Curve The weights of pennies minted after 1982 are approximately normally distributed with mean 2.46 grams and standard deviation 0.02 grams. (a) Draw a normal curve with the parameters labeled. (b) Shade the region under the normal curve between 2.44 and 2.49 grams. (c) Suppose the area under the normal curve for the shaded region is 0.7745. Provide two interpretations for this area.

  13. EXAMPLE Relation Between a Normal Random Variable and a Standard Normal Random Variable The weights of pennies minted after 1982 are approximately normally distributed with mean 2.46 grams and standard deviation 0.02 grams. Draw a graph that demonstrates the area under the normal curve between 2.44 and 2.49 grams is equal to the area under the standard normal curve between the Z-scores of 2.44 and 2.49 grams.

  14. Chapter 7The Normal Probability Distribution 7.2 The Standard Normal Distribution

  15. Properties of the Normal Density Curve 7. The Empirical Rule: About 68% of the area under the graph is between -1 and 1; about 95% of the area under the graph is between -2 and 2; about 99.7% of the area under the graph is between -3 and 3.

  16. The table gives the area under the standard normal curve for values to the left of a specified Z-score, zo, as shown in the figure.

  17. EXAMPLE Finding the Area Under the Standard Normal Curve Find the area under the standard normal curve to the left of Z = -0.38.

  18. Area under the normal curve to the right of zo = 1 – Area to the left of zo

  19. EXAMPLE Finding the Area Under the Standard Normal Curve Find the area under the standard normal curve to the right of Z = 1.25. Look in the Normal distribution table: P(x>=1.25) =1-P(x<1.25)=1-0.8944=0.1056

  20. EXAMPLE Finding the Area Under the Standard Normal Curve Find the area under the standard normal curve between Z = -1.02 and Z = 2.94. P(-1.02<x<2.94)=P(x<2.94)-p(x<-1.02) =0.9984-0.1539 =0.8445

  21. EXAMPLE Finding a Z-score from a Specified Area to the Left Find the Z-score such that the area to the left of the Z-score is 0.68. i.e., find Z such that P(x<Z)=0.68 , Z =0.46

  22. EXAMPLE Finding a Z-score from a Specified Area to the Right Find the Z-score such that the area to the right of the Z-score is 0.3021. P(x>Z) =0.3021, so P(x<Z) = 0.6979 z=0.52

  23. EXAMPLE Finding a Z-score

  24. EXAMPLE Finding the Value of z Find the value of z0.25

  25. 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.

  26. EXAMPLE Finding Probabilities of Standard Normal Random Variables Find each of the following probabilities: (a) P(Z < -0.23) (b) P(Z > 1.93) (c) P(0.65 < Z < 2.10)

More Related