business statistics ..probability

Midterm 2 Review (updated to current text & with answers) DSCI 245

A manufacturer took a simple random sample of two batteries from every hour’s production throughout an 8-hour shift and used them continuously until they were drained. The manufacturer suspects that there is a difference in battery life between the first half of the shift and the second half of the shift. One battery during the first half of the shift lasted less than 400 hours, while 4 batteries during the second half of the shift lasted less than 400 hours.

1. Make a contingency table using relative frequencies.

	First half	Second half	Total
Less than 400	1/16 = 0.06250	4/16 = 0.25000	5/16 = 0.31250
At least 400	7/16 = 0.43750	4/16 = 0.25000	11/16 = 0.68750
Total	8/16 = 0.50000	8/16 = 0.50000	16/16 = 1.00000

2. What is the probability that a randomly selected battery was produced during the first half of the shift?

P(first half) = 0.50000

3. What is the probability that a randomly selected battery lasts less than 400 hours?

P(less than 400) = 0.31250

4. What is the probability that a randomly selected battery was produced during the first half of the shift and that it lasts less than 400 hours?

P(first half AND less than 400) = 0.06250

5. What is the probability that a randomly selected battery produced during the first half of the shift will last less than 400 hours?

P(less than 400 GIVEN first half) = 0.0625/0.5000 = 0.12500

6. What is the probability that a randomly selected battery that last less than 400 hours was produced during the first half of the shift?

P(first half GIVEN less than 400) = 0.0625/0.3125 = 0.20000

7. Is a battery lasting less than 400 hours independent of which half of the shift it was produced in? Explain.

No, because P(less than 400 AND first half) does not equal P(less than 400)*P(first half)

Next, suppose that battery life is normally distributed with a mean of 500 hours and a standard deviation of 150 hours.

8. What proportion of all batteries last exactly 500 hours?

0, because the normal distribution is continuous

9. What proportion of all batteries will last less than 350 hours?

P(X < 350) = Normalcdf(-99999, 350, 500, 150) = 0.15866

10. What percentage of all batteries will last at least 300 hours?

P(X ( 300) = Normalcdf(300, 99999, 500, 150) = 0.90879

11. What is the probability that a randomly selected battery will last between 200 and 900 hours?

P(200 < X < 900) = Normalcdf(200, 900, 500, 150) = 0.97342

12. What is the battery life that will be exceeded by only 2% of all batteries?

P(X > x) = invNorm(0.98, 500, 150) = 808.06234

Next, suppose that the proportion of batteries that lasts less than 300 hours is 0.10.

13. What is the probability during an 8-hour shift of sampling exactly 3 batteries that last less than 300 hours?

P(X=3) = Binomialpdf(16, 0.1, 3) = 0.14234

14. What is the probability during an 8-hour shift of sampling at least 1 battery that lasts less than 300?

P(X ( 1) = 1 – Binomialpdf(16, 0.1, 0) = 0.81470

Next, suppose the table below contains the probability distribution of the life-length of the batteries rounded to the nearest hundred hours.

X = L1	300	400	500	600	700	1100	Total
P = L2	0.31250	0.06250	0.31250	0.12500	0.12500	0.06250	1.00000

15. What is the probability that a randomly selected battery will have a life (round to the nearest hundred) of at least 700 hours?

P(X ( 700) = 0.12500 + 0.06250 = 0.18750

16. What is the probability that a randomly selected battery will have a life (rounded to the nearest hundred) of less than 1100 hours?

P(X<1100) = 1 – P(X = 1100) = 1 – 0.06250 = 0.93750

17. What is the expected value of this distribution?

( = L1*L2(L3, get (L3 = 506.25000

18. What is the risk of this distribution?

( = (L1-()2*L2(L4, get (L4, then take the square root) = 204.53835

Next, suppose that battery lifetimes have a minimum of 200 and a maximum of 600 hours.

19. What is the probability that a randomly selected battery will last less than 300 hours?

P(200<X<300) = (300-200)/(600-200) = 0.25000

20. What is the mean and standard deviation of this distribution?

( = (600+200)/2 = 400.00000 hours

( = (600-200)/(12 = 115.47005 hours

Next, suppose that battery lifetimes have a minimum of 200 and a maximum of 600 hours with 500 hours being most likely.

21. What is the probability that a randomly selected battery will last less than 300 hours?

P(200<X<300) = (300-200)2/[(600-200)(500-200)] = 0.08333

Next, suppose there is an average of 3 battery failures every 300 hours and that one battery just failed.

22. How many batteries are expected to fail in the next 50 hours

E(X) = ( = (0.01 batteries/hr)(50 hrs) = 0.5 batteries

23. What is the probability that less than 2 batteries will fail in the next 200 hours?

( = 3/300*200 = 2, P(X<2) = Poissoncdf(2, 1) = 0.40601

24. What is the probability that the next failure will occur in the next 100 hours?

( = 3/300 = 0.01, P(0 ( X ( 100) = 1 – e-.01*100 = 0.63212

Finally, suppose a quality assurance test requires that, for a batch of 1000 batteries to be accepted, no more than 20% in a randomly selected sample of 5 fail in less than 700 hours. Now, suppose exactly 20% of the batteries in the batch will fail in less than 700 hours.

25. What is the probability that the batch will be rejected?

N = 1000, R = 0.2*1000 = 200, n = 5, X > 1

P(rejection) = 1 – P(acceptance) = 1 – {P(x = 0) + P(x = 1)}

= 1 – {(200 C 0)(800 C 5)/(1000 C 5) + (200 C 1)(800 C 4)/(1000 C 5)

= 0.26251