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# Normal distribution stat

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### Normal distribution stat

1. 1. Normal Distributions Reported by:onathan M. Pacuri
2. 2. 1-1 The Normal Distribution GOALS ONE List the characteristics of the normal distribution. TWO Define and calculate z values. THREE Determine the probability that an observation will lie between two points using the standard normal distribution. FOUR Determine the probability that an observation will be above or below a given value using the standard normal distribution. Jonathan M. Pacurib
3. 3. 1-1 GOALS FIVE Compare two or more observations that are on different probability distributions. Jonathan M. Pacurib
4. 4. 7-3 Characteristics of a Normal 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 the distribution are equal and located at the peak.  Half the area under the curve is above the peak, and the other half is below it.
5. 5. 7-4 Characteristics of a Normal Distribution  The normal distribution is symmetrical about its mean.  The normal distribution is asymptotic - the curve gets closer and closer to the x-axis but never actually touches it.
6. 6. 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 equal
7. 7. Properties of a Normal Distribution Inflection point Inflection point x • As the curve extends farther and farther away from the mean, it gets closer and closer to the x-axis but never touches it. • The points at which the curvature changes are called inflection points. The graph curves downward between the inflection points and curves upward past the inflection points to the left and to the right.
8. 8. Means and Standard Deviations Curves with different means, same standard deviation 10 11 12 13 14 15 16 17 18 19 20 Curves with different means, different standard deviations 9 10 11 12 13 14 15 16 17 18 19 20 21 22
9. 9. The Standard Normal Distribution
10. 10. 7-6 The Standard Normal Distribution A normal distribution with a mean of 0 and a standard deviation of 1 is called the standard normal distribution.  Z value: The distance between a selected value, designated X, and the population mean , divided by the population standard µ deviation, σ X −µ Z = σ
11. 11. The Standard Normal Distribution The standard normal distribution has a mean of 0 and a standard deviation of 1. Using z-scores any normal distribution can be transformed into the standard normal distribution. –4 –3 –2 –1 0 1 2 3 4 z
12. 12. The Standard Score The standard score, or z-score, represents the number of standard deviations a random variable x falls from the mean. The test scores for a civil service exam are normally distributed with a mean of 152 and a standard deviation of 7. Find the standard z-score for a person with a score of: (a) 161 (b) 148 (c) 152(a) (b) (c)
13. 13. From z-Scores to Raw ScoresTo find the data value, x when given a standard score, z:The test scores for a civil service exam are normallydistributed with a mean of 152 and a standard deviation of 7.Find the test score for a person with a standard score of:(a) 2.33 (b) –1.75 (c) 0 (a) x = 152 + (2.33)(7) = 168.31 (b) x = 152 + (–1.75)(7) = 139.75 (c) x = 152 + (0)(7) = 152
14. 14. Cumulative Areas The total area under the curve is one. –3 –2 –1 0 1 2 3 z• The cumulative area is close to 0 for z-scores closeto –3.49.• The cumulative area for z = 0 is 0.5000.• The cumulative area is close to 1 for z-scores close to3.49.
15. 15. References: Bluman. Allan G. Elementary Statistics. Step by Step. McGraw Hill international Edition, 2008. Pacho, Elsie et. Al. Fundamental Statistics. Trinitas Publishing House, Inc. Bulacan, 2003
16. 16. Thank you for Listening.
17. 17. 7-7 EXAMPLE 1  The monthly incomes of recent MBA graduates in a large corporation are normally distributed with a mean of \$2000 and a standard deviation of \$200. What is the Z value for an income of \$2200? An income of \$1700?  For X=\$2200, Z=(2200-2000)/200=1.
18. 18. 7-8 EXAMPLE 1 continued  For X=\$1700, Z =(1700-2000)/200= -1.5  A Z value of 1 indicates that the value of \$2200 is 1 standard deviation above the mean of \$2000, while a Z value of \$1700 is 1.5 standard deviation below the mean of \$2000.
19. 19. 7-9 Areas Under the Normal Curve  About 68 percent of the area under the normal curve is within one standard µ ± ofσ deviation 1 the mean.  About 95 percent is within two standard deviations of the mean. ± 2σ µ  99.74 percent is within three standard deviations of the mean. µ ± 3σ
20. 20. 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: 1.68.26% 0 . 3 2.95.44% 3.99.74% 0 . 2 f ( x 0 . 1 . 0 µ µ + 2σ - 5 µ − 2σ µ + 3σ x µ − 3σ µ −1σ µ +1σJonathan M. Pacurib
21. 21. 7-11  The daily water usage per person in New EXAMPLE 2 Providence, New Jersey is normally distributed with a mean of 20 gallons and a standard deviation of 5 gallons.  About 68% of the daily water usage per person in New Providence lies between what two values?  That is, about 68% of the daily water usage will lie between 15 and 25 gallons. µ ± 1σ = 20 ± 1(5).
22. 22. 7-12 EXAMPLE 3 atthat a personuse less than  What is the probability Providence selected random will from New 20 gallons per day?  The associated Z value is Z=(20-20)/5=0. Thus, P(X<20)=P(Z<0)=.5  What percent uses between 20 and 24 gallons?  The Z value associated with X=20 is Z=0 and with X=24, Z=(24-20)/5=.8. Thus, P(20<X<24)=P(0<Z<.8)=28.81%
23. 23. 7-13 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
24. 24. 7-14 EXAMPLE 3 continued  What percent of the population uses between 18 and 26 gallons?  The Z value associated with X=18 is Z=(18- 20)/5= -.4, and for X=26, Z=(26-20)/5=1.2. Thus P(18<X<26) =P(-.4<Z<1.2)=.1554+.3849=.5403
25. 25. 7-15  Professor Mann has determined that the EXAMPLE 4 final averages in his statistics course is normally distributed with a mean of 72 and a standard deviation of 5. He decides to assign his grades for his current course such that the top 15% of the students receive an A. What is the lowest average a student can receive to earn an A?  Let X be the lowest average. Find X such that P(X >X)=.15. The corresponding Z value is 1.04. Thus we have (X-72)/5=1.04, or X=77.2
26. 26. 7-16 r EXAMPLE 4 a l i t r b u i o n : µ = 0 , σ2 = 1 0 . 4 0 . 3 Z=1.04 0 . 2 f ( x 0 . 1 . 0 15% - 0 1 2 3 4
27. 27. 7-17  The amount of tip the servers in an exclusive EXAMPLE 5 restaurant receive per shift is normally distributed with a mean of \$80 and a standard deviation of \$10. Shelli feels she has provided poor service if her total tip for the shift is less than \$65. What is the probability she has provided poor service?  Let X be the amount of tip. The Z value associated with X=65 is Z= (65-80)/10= -1.5. Thus P(X<65)=P(Z<-1.5)=.5-.4332=.0668.
28. 28. 7-18 The Normal Approximation to  Using the normal distribution (a continuous the Binomial distribution) as a substitute for a binomial distribution (a discrete distribution) for large values of n seems reasonable because as n increases, a binomial distribution gets closer and closer to a normal distribution.  The normal probability distribution is generally deemed a good approximation to the binomial probability distribution when n and n(1- ) are both greater than 5. π π
29. 29. 7-19 The Normal Approximation continued  Recall 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.
30. 30. Binomial Distribution for an n of 37-20and 20,where π =.50 n=3 n=20 0.4 0.3 0.2 0.15 P(x) 0.2 P(x) 0.1 0.1 0.05 0 0 0 1 2 3 2 4 6 8 10 12 14 16 18 20 number of occurences number of occurences
31. 31. 7-21 Continuity Correction Factor  Thevalue .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).
32. 32. 7-22 EXAMPLE 6 A recent study by a marketing research firm showed that 15% of American households owned a video camera. A sample of 200 homes is obtained.  Of the 200 homes sampled how many would you expect to have video cameras? µ = nπ = (.15)(200) = 30
33. 33. 7-23  What is the variance? EXAMPLE 6 σ =nπ 1 − ) =( 30)(1− ) =25.5 2 ( π .15  What is the standard deviation? σ =  What is the.5 =5.0498 less than 40 homes 25 probability that in the sample have video cameras? We need P(X<40)=P(X<39). So, using the normal approximation, P(X<39.5) P[Z ≈ (39.5-30)/5.0498] = P(Z 1.8812) P(Z<1.88)=.5+.4699+.9699 ≤ ≤ ≈
34. 34. 7-24 µ σ2 EXAMPLE 6 r a l i t r b u i o n : = 0 , = 1 0 . 4 P(Z=1.88) 0 . 3 .5+.4699 =.9699 0 . 2 f ( x 0 . 1 Z=1.88 . 0 - 5 0 1 2 3 4 Jonathan M. Pacurib