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# MTH120_Chapter7

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### MTH120_Chapter7

1. 1. 7- 1 Chapter Seven The Normal Probability DistributionGOALSWhen you have completed this chapter, you will be able to: ONE List the characteristics of the normal probability distribution. TWO Define and calculate z values. THREE Determine the probability an observation will lie between two points using the standard normal distribution. FOUR Determine the probability an observation will be above or below a given value using the standard normal distribution.
2. 2. 7- 2 Chapter Seven continued The Normal Probability Distribution GOALS When you have completed this chapter, you will be able to:FIVECompare two or more observations that are on differentprobability distributions.SIXUse the normal distribution to approximate the binomialprobability distribution.
3. 3. 7- 3 Characteristics of a Normal Probability Distribution• The normal curve is bell-shaped and has a single peak at the exact center of the distribution.The arithmetic mean, median, and mode of thedistribution are equal and located at the peak.Thus half the area under the curve is above themean and half is below it.
4. 4. 7- 4 Characteristics of a Normal Probability Distribution• The normal probability distribution is symmetrical about its mean. The normal probability distribution is asymptotic. That is the curve gets closer and closer to the X-axis but never actually touches it.
5. 5. 7- 5 r a l i t r b u i o n : µ = 0 , σ2 = 1 Characteristics of a Normal Distribution 0 . 4 Normal curve is 0 . 3 symmetrical 0 . 2 Theoretically, curve f ( x 0 . 1 extends to infinity . 0 - 5 a Mean, median, and x mode are equalMcGraw-Hill/Irwin Copyright © 2002 by The McGraw-Hill Companies, Inc. All rights reserved
6. 6. 7- 6 The Standard Normal Probability Distribution• The standard normal distribution is a normal distribution with a mean of 0 and a standard deviation of 1.• It is also called the z distribution.A z-value is the distance between a selected value,designated X, and the population mean , , divided bythe population standard deviation, t . The formula is: X −µ z = σ
7. 7. 7- 7 EXAMPLE 1• The bi-monthly starting salaries of recent MBA graduates follows the normal distribution with a mean of \$2,000 and a standard deviation of \$200. What is the z-value for a salary of \$2,200? X −µ z = σ \$2,200 − 2,000 \$ = =2.00 \$200
8. 8. 7- 8 EXAMPLE 1 continued What is the z-value of \$1,700. X −µ z = σ \$1,700 − 2,200 \$ = =− .50 1 \$200• A z-value of 1 indicates that the value of \$2,200 is one standard deviation above the mean of \$2,000. A z-value of –1.50 indicates that \$1,700 is 1.5 standard deviation below the mean of \$2000.
9. 9. 7- 9 Areas Under the Normal Curve• About 68 percent of the area under the normal curve is within one standard deviation of the mean. i• About 95 percent is within two standard deviations of the mean. o 22• Practically all is within three standard deviations of the mean. t 33
10. 10. 7- 10 Areas Under the Normal Curve r a l i t r b u i o n : µ = 0 , σ2 = 1 0 . 4 Between: B11 - 68.26% 0 . 3 22 - 95.44% 33 - 99.74% 0 . 2 f ( x 0 . 1 . 0 µ µ + 2σ - 5 µ − 2σ µ + 3σ x µ − 3σ µ −1σ µ +1σ Copyright © 2002 by The McGraw-Hill Companies, Inc.,.All rights reservedMcGraw-Hill/Irwin
11. 11. 7- 11 EXAMPLE 2 The daily water usage per person in New Providence, New Jersey is normally distributed with a mean of 20 gallons and a standard deviation of 5 gallons. About 68 percent of those living in New Providence will use how many gallons of water?• About 68% of the daily water usage will lie between 15 and 25 gallons.
12. 12. 7- 12 EXAMPLE 3• What is the probability that a person from New Providence selected at random will use between 20 and 24 gallons per day? X −µ 20 − 20 z = = = .00 0 σ 5 X −µ 24 − 20 z = = = .80 0 σ 5
13. 13. 7- 13 Example 3 continued• The area under a normal curve between a z-value of 0 and a z-value of 0.80 is 0.2881.• We conclude that 28.81 percent of the residents use between 20 and 24 gallons of water per day.• See the following diagram.
14. 14. 7- 14 r a l i EXAMPLE 3 t r b u i o n : µ = 0 , 0 . 4 0 . 3 P(0<z<.8) =.2881 0 . 2 0<x<.8 f ( x 0 . 1 . 0 - 5 -4 -3 -2 -1 x 0 1 2 3 4 Copyright ©2002 by The McGraw-Hill Companies, Inc.,.All rights reservedIrwin/McGraw-Hill
15. 15. 7- 15 EXAMPLE 3 continued• What percent of the population use between 18 and 26 gallons per day? X −µ 18 − 20 z = = = 0.40 − σ 5 X −µ 26 − 20 z = = = .20 1 σ 5
16. 16. 7- 16 Example 3 continued• The area associated with a z-value of –0.40 is . 1554.• The area associated with a z-value of 1.20 is . 3849.• Adding these areas, the result is .5403.• We conclude that 54.03 percent of the residents use between 18 and 26 gallons of water per day.
17. 17. 7- 17 EXAMPLE 4• Professor Mann has determined that the scores in his statistics course are approximately normally distributed with a mean of 72 and a standard deviation of 5. He announces to the class that the top 15 percent of the scores will earn an A. What is the lowest score a student can earn and still receive an A?
18. 18. 7- 18 Example 4 continued• To begin let X be the score that separates an A from a B.• If 15 percent of the students score more than X, then 35 percent must score between the mean of 72 and X.The z-value associated corresponding to35 percent is about 1.04.
19. 19. 7- 19 Example 4 continued• We let z equal 1.04 and solve the standard normal equation for X. The result is the score that separates students that earned an A from those that earned a B. X −72 1.04 = 5 X =72 + .04(5) =72 +5.2 =77.2 1Those with a score of 77.2 or more earn an A.
20. 20. 7- 20The Normal Approximation to the Binomial• The normal distribution (a continuous distribution) yields a good approximation of the binomial distribution (a discrete distribution) for large values of n.The normal probability distribution is generally agood approximation to the binomial probabilitydistribution when nn and n(1- n ) are both greaterthan 5.
21. 21. 7- 21The Normal Approximation continuedRecall for the binomial experiment:• There are only two mutually exclusive outcomes (success or failure) on each trial.• A binomial distribution results from counting the number of successes.• Each trial is independent.• The probability is fixed from trial to trial, and the number of trials n is also fixed.
22. 22. 7- 22 Continuity Correction Factor• The value .5 subtracted or added, depending on the problem, to a selected value when a binomial probability distribution (a discrete probability distribution) is being approximated by a continuous probability distribution (the normal distribution).
23. 23. 7- 23 EXAMPLE 5A recent study by a marketing research firmshowed that 15% of American households owneda video camera. For a sample of 200 homes, howmany of the homes would you expect to havevideo cameras? µ = nπ = (.15)(200) = 30This is the mean of a binomial distribution.
24. 24. 7- 24 EXAMPLE 5 continued• What is the variance? σ = nπ (1 − π ) = (30)(1−.15) = 25.5 2What is the standard deviation? σ = 25.5 = 5.0498
25. 25. 7- 25 Example 5 What is the probability that less than 40 homes in the sample have video cameras?• We use the correction factor, so X is 39.5.• The value of z is 1.88. X −µ 39.5 − 30.0 z= = = 1.88 σ 5.0498
26. 26. 7- 26 Example 5 continued• From Appendix D the area between 0 and 1.88 on the z scale is .4699.• So the area to the left of 1.88 is .5000 + .4699 = . 9699.• The likelihood that less than 40 of the 200 homes have a video camera is about 97%.
27. 27. 7- 27 µ σ2 EXAMPLE 5 r a l i t r b u i o n : = 0 , = 1 0 . 4 P(z<1.88) 0 . 3 =.5000+.4699 =.9699 0 . 2 f ( x 0 . 1 z=1.88 . 0 - 5 0 1 2 3 4 z Copyright ©2002 by The McGraw-Hill Companies, Inc., . All rights reservedMcGraw-Hill/Irwin