6. On the graph obtained in part (5), draw two horizontal lines—one at each of the BOUNCE scores determined in part (3) — During 8-hour shift golf-ball production golf ball

During an 8-hour shift of golf-ball production, one golf ball is randomly selected from each 2 minutes’ worth of output. The ball is then tested for “liveliness” by rolling it down a grooved, stainless- steel surface. At the bottom, it strikes a smooth iron block and bounces backward, up a gradually sloped, grooved surface inscribed with distance markings. The higher the ball gets on the rebound surface, the more lively it is.

1. A driving range has just purchased 100 golf balls. Use the computer and the binomial distribution in determining the individual and cumulative probabilities for x = the number of balls that scored at least 31.00 inches on the “bounce” test.

2. Repeat part (1), but use the normal approximation to the binomial distribution. Do the respective probabilities appear to be very similar?

3. Given the distribution of bounce test scores described above, what score value should be exceeded only 5% of the time? For what score value should only 5% of the balls do more poorly?

4. What is the probability that, for three consecutive balls, all three will happen to score below the lower of the two values determined in part (3)? Two hundred forty golf balls have been subjected to the bounce test during the most recent 8-hour shift. From the 1st through the 240th, their scores are provided in computer data file CDB07 as representing the con tinuous random variable, BOUNCE. The variable BALL is a sequence from 1 to 240.

5. Using the computer, generate a line graph in which BOUNCE is on the vertical axis and BALL is on the horizontal axis.

6. On the graph obtained in part (5), draw two horizontal lines—one at each of the BOUNCE scores determined in part (3).

7. Examining the graph and the horizontal lines drawn in part (6), does it appear that about 90% of the balls have BOUNCE scores between the two horizontal lines you’ve drawn, as the normal distribution would suggest when μ = 30.00 and σ = 2.00?

8. Comparing the BOUNCE scores when BALL = 1 through 200 to those when BALL = 201 through 240, does it appear that the process may have changed in some way toward the end of the work shift? In what way? Does it appear that the machine might be in need of repair or adjustment? If so, in what way should the adjustment alter the process as it appeared to exist at the end of the work shift?

SOLUTION

1. We must first determine the probability of "success" (x 31.00) on any given trial. This is based on the normal distribution with = 30.00 and = 2.00.

The corresponding z value is (31.00 -30.00)/2.00, or z = 0.50. Referring to the normal table, we find P(z 0.50) is 1.0000 - 0.6915, or 0.3085. Using Minitab, we obtain the following individual and cumulative binomial probabilities, which have been placed next to each other for purposes of clarity.

BINOMIAL WITH N = 100 P = 0.308500 BINOMIAL WITH N = 100 P = 0.308500

K P (X = K) K P (X LESS OR = K)

13 0.0000 13 0.0000

14 0.0001 14 0.0001

15 0.0001 15 0.0002

16 0.0003 16 0.0005

17 0.0007 17 0.0012

18 0.0014 18 0.0027

19 0.0028 19 0.0054

20 0.0050 20 0.0104

21 0.0085 21 0.0189

22 0.0136 22 0.0324

23 0.0205 23 0.0530

24 0.0294 24 0.0824

25 0.0399 25 0.1222

26 0.0513 26 0.1735

27 0.0627 27 0.2362

28 0.0730 28 0.3092

29 0.0808 29 0.3900

30 0.0853 30 0.4753

31 0.0859 31 0.5613

32 0.0827 32 0.6440

33 0.0760 33 0.7200

34 0.0668 34 0.7868

35 0.0562 35 0.8430

36 0.0453 36 0.8883

37 0.0349 37 0.9232

38 0.0258 38 0.9491

39 0.0183 39 0.9674

40 0.0125 40 0.9799

41 0.0081 41 0.9880

42 0.0051 42 0.9931

43 0.0031 43 0.9962

44 0.0018 44 0.9980

45 0.0010 45 0.9990

46 0.0005 46 0.9995

47 0.0003 47 0.9998

48 0.0001 48 0.9999

49 0.0001 49 1.0000

50 0.0000

2. Using the normal approximation to the binomial distribution, the mean and standard deviation for x = the number of balls out of 100 that will bounce at least 31.00 inches:

100(0.3085) = 30.85 balls

4.6187 balls

Using the normal approximation, cumulative probabilities for x when = 30.85 and = 4.6187: cumulative