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Chapter 6 Part 2

Chapter 6 Part 2

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Chapter 6 Part 2

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  1. Chapter 6 Part 2 Continuous Probability Distributions (Continued) Dr. C. Lightner

  2. 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. Dr. C. Lightner

  3. 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 . Dr. C. Lightner

  4. The Standard Normal Table The Standard Normal Table is in the front of your book. This table gives the area under a normal curve to the left of any z score value that you look up. * If you want to know the area under the normal curve that is to the left of 1.57 , go to the row for 1.0 and the column for .07 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 less than 1.57 is 0.9418. * P(z< 1.57)= z 1.57

  5. Standard Normal Table P(Z<1.57) Dr. C. Lightner

  6. The Standard Normal Table If you want to know the area under the normal curve to the left of 1.90, go to the row for 1.9 and the column for 0.00 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 less than 1.90 is 0.9713. z 1.90

  7. Standard Normal Table P(Z<1.90) Dr. C. Lightner

  8. Areas under Normal Curves and Negative 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(-0.50  z  0) = P(0  z  0.5) = 0.1915. • Also P(0  z  0.95) = P(-0.95  z  0)= 0.3289. Dr. C. Lightner

  9. Practice Problems • Determine P(z < -1) • Determine P(z < 1) • Determine P(z  -1) • Determine P(z  1) • Determine P(0  z  2) • Determine P(-1.5  z  0) • 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) Dr. C. Lightner

  10. Practice Problem 1 Calculating P(z < -1) Steps 1-3 z Step 4 From chart, P(z<-1)=0.1587 z Dr. C. Lightner

  11. Practice Problem 2 Calculating P(z < 1) Steps 1-3 z Step 4 From chart, P(z<1)=0.8413 z z Dr. C. Lightner

  12. Practice Problem 3 Calculating P(z  -1) Step 4 Steps 1-3 a.) From chart, P(z<-1)=0.1587 z z b.) Since area under curve is 1 • P(z>-1)= 1-P(z<-1) =.8413 Dr. C. Lightner

  13. Practice Problem 4 Calculating P(z  1) Step 4 Steps 1-3 a.) From chart, P(z<1) = 0.8413 z z b.) Since area under curve is 1 • P(z>1)= 1-P(z<1) =0.1587 **Note that P(z>1) equals P(z<-1) Dr. C. Lightner

  14. Practice Problem 5 Calculating P(0<z < 2) Step 4 Steps 1-3 a.) From chart, P(z<2) = 0.9772 z z b.) Since area under curve is 1 and half the area under the curve is 0.5 • P(0<z < 2)= 0.9772-0.5 =0.4773 ***FINISH THE OTHER PRACTICE PROBLEMS Dr. C. Lightner

  15. 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. NOTE: The area under the normal curve calculated in step 4 equals P(a<x<b) Dr. C. Lightner

  16. Example 6.3 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- defined as P(x > 20). Anderson, Sweeney, and Williams Dr. C. Lightner

  17. 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) Dr. C. Lightner

  18. 3. Draw the curve and shade the appropriate region. x 15 20 0 .83 z Dr. C. Lightner

  19. 4. Determine the shaded area. The Standard Normal table shows an area of 0.7967 for the region less than z = 0.83 (See next slide). Since the area under the entire curve equals 1.0, P(x>20)=1.0-0.7967=.2033 0.7967 .2033 x 15 20 z Dr. C. Lightner 0 .83

  20. Standard Normal Probability Distribution Cumulative Probability Table for the Standard Normal Distribution P(z< .83) Anderson, Sweeney, and Williams P(z< .83)=0.7967 0.7967

  21. Example 6.4 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. b. What is the probability that a randomly selected student received between a 65 and 85? c. What is the proportion of students that failed the exam? Dr. C. Lightner

  22. Example 6.4a 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 & Dr. C. Lightner

  23. 3. Draw the curve and shade the appropriate region. 90 x 72 80 2.25 z Dr. C. Lightner 0 1

  24. 4. Determine the shaded area • The Standard Normal table shows an area of .8413 for the region less than z = 1.00. • It also shows an area of .9878 for the region between z = 2.25. P(80>x>90)=0.9878-0.8413= 0.1465 .8413 Thus 14.65% of the students received a B .9878 90 x 72 80 z Dr. C. Lightner 0 1 2.25

  25. Example 6.4b Find P(65<x<85). This can be restated as P(-.88 < z < 1.63). The Standard Normal table shows an area of 0.1894 for the region less than z = -0.88 and an area of .9484 for the region less than z = 1.63. P(65<x<85)=0.9484-0.1894= .759 .9484 0.1894 65 72 85 x 0 1.63 z -.88 Dr. C. Lightner

  26. Example 6.4c Find P(x < 60). This can be restated as P(z < -1.5). The Standard Normal table shows an area of 0.0668 for the region less than z = -1.5. P(x<60)= .0668 .3106 So Approximately 6.68% of the students will fail 0.668 60 72 x 0 z -1.5 Dr. C. Lightner

  27. .05 Since the prob of falling in interval is .05 x? x z 0 Example 6.3 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 Dr. C. Lightner

  28. Standard Normal Probability Distribution Solving for the Reorder Point Find the z-value that cuts off an area of .05 in the right tail of the standard normal distribution. We look up the complement of the tail area (1 - .05 = .95) Anderson, Sweeney, and Williams Dr. C. Lightner

  29. Example 6.3 Revisited (Inverse Normal Problem) If z=1.645, then . Thus, x= 24.87. .05 Since the prob of falling in interval is .05 .9500 x? x 15 z 0 Dr. C. Lightner

  30. *** The End *** Dr. C. Lightner