Chapter 5 Part 2

1. Chapter 5 Part 2 Continuous Probability Distributions (Continued)

2. Discussion of Normal Distribution Exercise The Normal Distribution exercise should have revealed the following: • The area under the normal curve within any interval (with respect to z scores-from Chapter 2) is always the same, regardless of the mean and standard deviation of the random variable. * For exercises B and C we had two different sets of mean and standard deviation values. * For exercise B in part a) we found the grade with a z score of 0.5 and in b) we found the area under the normal curve between the mean and a z score of 0.5. We did the same thing in exercise C and we had different values in part a), however, the area under the normal curve between the mean and a z score of 0.5 (i.e. part is identical.

3. Discussion of Normal Distribution Exercise (continued) * In exercise B part c) we found the grade with a z score of 1 and in d) we found the area under the normal curve between the mean and a z score of 1. We did the same thing in exercise C and we had different values in part c), however, the area under the normal curve between the mean and a z score of 1 (i.e. part d) is identical. * In exercise B and C part e) we found the area under the normal curve between a z score of 0.5 and a z score of 1 to be identical. • The normal curve is symmetric about the mean. * When we looked at identical intervals below the mean the areas were the same as the identical interval above the mean.

4. Normal Distribution • Since there is a common thread between all normal distributions (with respect to z scores), we can use a single source to determine area under the normal curves. • The Standard Normal Distribution is used to accomplish this task.

5. Standard Normal Distribution • If x is normally distributed with mean  and standard deviation , then x is the value of the corresponding z score. is normally distributed with mean 0 and standard deviation 1, and is called the standardnormal distribution. Clearly, the z score of the mean  is 0, since plugging  in the above z score formula for x yields a z value of zero. • We can think of z as a measure of the number of standard deviations x is from .

6. The Standard Normal Table The Standard Normal Table is in the appendix of your book. This table gives the area under a normal curve between the mean and any z score value that you look up. * If you want to know the area under the normal curve between z=0 (the mean) and a z score of 0.50, go to the row for 0.5 and the column for 0.00 and the row and column will meet at the corresponding area. On the following slide we see that the probability that a random variable takes on a value between the mean and 0.50 standard deviations above the mean is 0.1915. * If you want to know the area under the normal curve between z=0 (the mean) and a z score of 0.95.0, go to the row for 0.9 and the column for 0.05 and the row and column will meet at the corresponding area. From the table we see that the probability that a random variable takes on a value between the mean and 0.95 standard deviations above the mean is 0.3289.

7. Standard Normal Table

8. Areas under Normal Curves andNegative z Scores • Since the Normal Curve is symmetric about the mean, the area under the curve between the mean and + z is the same as the area under the curve between the mean and – z. • Thus area under the normal curve between z=0 (the mean) and a z score of -0.50 or P(0  z  -0.50) equals 0.1915. • Also P(0  z  -0.95) equals 0.3289.

9. Practice Problems 1 a. Determine P(-1  z  1) b. Determine P(-2  z  2) c. Determine P(-3  z  3) d. Determine P(-2.53  z  2.53) • Determine P(z  -1) • Determine P(z  1)

10. Practice Problem 1 Bowerman, et. al

11. Practice Problem 2 Calculating P(z  -1) Bowerman, et. al

12. Practice Problem 2 Calculating P(z  1) Bowerman, et. al

13. Steps for Determining Normal Probabilities • Formulate your problem in terms of x. • Restate your problem in terms of the corresponding z values. • Draw you Normal curve and shade in the region under the curve in the interval of interest. • Use the standard normal tables to find the indicated area under the normal curve.

14. Example 5.3 Pep Zone sells auto parts and supplies including a popular multi-grade motor oil. When the stock of this oil drops to 20 gallons, a replenishment order is placed. The store manager is concerned that sales are being lost due to stockouts while waiting for an order. It has been determined that demand is normally distributed with a mean of 15 gallons and a standard deviation of 6 gallons. The manager would like to know the probability of a stockout, P(x > 20). Anderson, Sweeney, and Williams

15. Following the steps for determining normal probabilities: • We are trying to find P(x > 20). • Restate the problem in terms of z. The z score for the value 20 is z= Thus P(x>20) can be restated as P(z>.83)

16. 3. Draw the curve and shade the appropriate region. x 15 20 z 0 .83

17. 4. Determine the shaded area. The Standard Normal table shows an area of .2967 for the region between z = 0 and z = .83 (See next slide). Since the area under half of the curve equals 0.5, P(x>20)=.5-.2967=.2033 .2967 .2033 x 15 20 z 0 .83

18. Standard Normal Table

19. Example 5.4 The grades on a statistics midterm exam were normally distributed with a mean of 72 and a standard deviation of 8. • What is the proportion of students received a B grade. • What is the probability that a randomly selected student received between a 65 and 85? • What is the proportion of students that failed the exam?

20. Example 5.4a Following the steps for determining normal probabilities: • We are trying to find P(80 < x < 90). • This can be restated as P(1< z < 2.25) Note: Recall that for a continuous random variable the probability that a random variable equals a specific value equals 0. Thus inserting  symbol above would not change the answer to the above stated problem. The z score for 90 is Since the z score for 80 is &

21. 3. Draw the curve and shade the appropriate region. 90 x 72 80 2.25 z 0 1

22. 4. Determine the shaded area The Standard Normal table shows an area of .3413 for the region between z=0 and z = 1.00. It also shows an area of .4878 for the region between z=0 and z = 2.25. .3413 P(80>x>90)=.4878-.3413= .1465 Thus 14.65% of the students received a B .4878 90 x 72 80 z 0 1 2.25

23. Example 5.4b Find P(65<x<85). This can be restated as P(-.88 < z < 1.63). The Standard Normal table shows an area of .3106 for the region between z=0 and z = +.88 and an area of .4484 for the region between z=0 and z = 1.63. P(65<x<85)=.3106+ .4484= .759 .3106 .4484 65 72 85 x 0 1.63 z -.88

24. Example 5.4c Find P(x < 60). This can be restated as P(z < -1.5). The Standard Normal table shows an area of .4332 for the region between z=0 and z = +1.5. P(x<60)=.5 - .4332= .0668 .3106 So Approximately 6.68% of the students will fail .4332 60 72 x 0 z -1.5

25. .05 Since the prob of falling in interval is .05 x? x z 0 Example 5.3 Revisited (Inverse Normal Problem) Refer to slide 14. If the manager of Pep Zone wants the probability of a stockout to be no more than .05, what should the reorder point be? Let’s call the reorder point x?. From above we know that P(x > x?)= 0.05. We drew the graph below because we knew that a stockout means that the demand exceeds the reorder point. Thus the interval of interest is all values that exceed the reorder point. We shaded the area in this interval. 15

26. Thus the area between the mean and x? is 0.45. Using the Standard Normal Probability Table we now look-up the .4500 area section to find the corresponding z value z = 1.645 is a reasonable estimate.

27. Example 5.3 Revisited (Inverse Normal Problem) If z=1.645, then . Thus x= 24.87. .05 Since the prob of falling in interval is .05 .4500 x? x 15 z 0