1. Week 4 Dr. Jenne Meyer

2.  In order of last name  Miracle  Brandi  Sandy  LeAndrew  Bren  Christine  Brandon  Maria  Rose  Monique  Michelle  Jodi

3.  Discrete Variable – each value of X has its own probability P(X).  Continuous Variable – events are intervals and probabilities are areas underneath smooth curves. A single point has no probability.  Total area under curve = 1

4. Defined by two parameters,  and  Almost all area under the normal curve is included in the range  – 3  < X <  + 3 

5. McGraw-Hill/Irwin© 2007 The McGraw-Hill Companies, Inc. All rights reserved. All normal distributions have the same shape but differ in the axis scales. Diameters of golf balls  = 42.70mm  = 0.01mm CPA Exam Scores  = 70  = 10

6.  Since for every value of  and , there is a different normal distribution, we transform a normal random variable to a standard normal distribution with  = 0 and  = 1 using the formula: z =z =z =z = x –  

7. Appendix allows you to find the area under the curve from 0 to z.

8. .5000.5000 -.4750 =.0250 Now find P(Z < 1.96):

9. Now find P(-1.96 < Z < 1.96). Due to symmetry, P(-1.96 < Z) is the same as P(Z < 1.96). So, P(-1.96 < Z < 1.96) =.4750 +.4750 =.9500 or 95% of the area under the curve..9500

10. Some important Normal areas:

13. Suppose John took an economics exam and scored 86 points. The class mean was 75 with a standard deviation of 7. What percentile is John in (i.e., find P(X < 86)? z John = x –   = 86 – 75 7 = 11/7 = 1.57 So John’s score is 1.57 standard deviations about the mean.

14. Suppose John took an economics exam and scored 86 points. The class mean was 75 with a standard deviation of 7. What percentile is John in (i.e., find P(X < 86)?

16. John is approximately in the 94th percentile

17.  For example, let  = 2.040 cm and  =.001 cm, what is the probability that a given steel bearing will have a diameter between 2.039 and 2.042cm?  In other words, P(2.039 < X < 2.042)  Excel only gives left tail areas, so break the formula into two, find P(X < 2.039) and P(X < 2.042), then subtract them to find the desired probability:

18. P(X < 2.042) =.9773 P(X < 2.039) =.1587 P(2.039 < X < 2.042) =.9773 -.1587 =.8186 or 81.9%

20.  suppose we wanted the probability of selecting a foreman who earned less than $1,100. In probability notation we write this statement as P(weekly income < $1,100).

21.  =.8413

22.  suppose we wanted the probability of selecting a foreman who earned less than $1,100. In probability notation we write this statement as P(weekly income < $1,100).  =.8413

23.  The mean of a normal probability distribution is 500; the standard deviation is 10.  a. About 68 percent of the observations lie between what two values?  b. About 95 percent of the observations lie between what two values?  c. Practically all of the observations lie between what two values?

24.  The mean of a normal probability distribution is 500; the standard deviation is 10.  a. About 68 percent of the observations lie between what two values?  b. About 95 percent of the observations lie between what two values?  c. Practically all of the observations lie between what two values?  a. 490 and 510, found by 500 +/- 1(10).  b. 480 and 520, found by 500 +/- 2(10).  c. 470 and 530, found by 500 +/- 3(10).

25.  A normal distribution has a mean of 50 and a standard deviation of 4.  a. Compute the probability of a value between 44.0 and 55.0.  b. Compute the probability of a value greater than 55.0.  c. Compute the probability of a value between 52.0 and 55.0.

26.  a. 0.8276: First find z -1.5, found by (44 - 50)/4 and  z = 1.25 = (55 - 50)/4. The area between -1.5 and 0 is  0.4332 and the area between 0 and 1.25 is 0.3944, both  from Appendix D. Then adding the two areas we find  that 0.4332 + 0.3944 = 0.8276.  b. 0.1056, found by 0.5000 - 0.3994, where z = 1.25.  c. 0.2029: Recall that the area for z = 1.25 is 0.3944, and  the area for z = 0.5, found by (52 - 50)/4, is 0.1915.  Then subtract 0.3944 - 0.1915 and find 0.2029.

27.  Next weeks assignments.  Textbook Assignment: Complete Chapter 6 problems 17, 18, 19 (a, b, c), 20  Textbook Assignment: Complete Chapter 7 problems 11, 13  Moved Final Checkpoint to week 6