1. Review of distributions Question 1 The daily amount of coffee in litters, dispensed by a machine located in airport lobby is a random variable X having a continuous uniform distribution with A = 7 and B = 10. Find the brobability that on a given day the amount of coffee dispensed by this machine will be: a) At most 8.8 litters. b) More than 7.4 litters but less than 9.5 litters. c) At least 8.5 litters.

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5. Review of distributions Question 2: Given a standard normal distribution, find the value of k such that a) P( Z > k ) = 0.2946 b) P( Z < k ) = 0.0427 c) P(-0.93 < Z < k ) = 0.7235

6. Review of distributions Solution: a) P ( Z < k ) = 1 – 0.2946 = 0.7054, from table this area is equal to Z = 0.54 Then k = 0.54. b) P (Z < k) = 0.0427. Directly from table this area is equal to Z = -1.72 = k. c) P (- 0.93 < Z < k) = 0.7235. The total area equal to 0.7235 + area left of Z= - 0.93 = 0.7235 + 0.1762=0.8997 Now from table this area is equal to Z = 1.28=k

7. Review of distributions Q3: The finished inside diameter of a piston ring is normally distribution with a mean of 10 cm and standard deviation of 0.03 cm. a) What proportion of rings will have inside diameters exceeding 10.075 cm? b) What is the probability that a piston rings will have an inside diameter between 9.97 and 10.03 cm? c) Below what value of inside diameter will 15% of the piston rings fall?

8. Review of distributions Solution: P (Z > 2.5) = 1-P (Z < 2.5) =1 -.0.9938 = 0.0062 So that 0.62% of the rings has inside diameters exceeding 10.075 cm

9. Review of distributions Solution: Then P (9.97 < X< 10.03) = P (-1.0 < Z <1.0) = P(Z < 1.0) – P (Z < -1) = 0.8413 - 0.1587 = 0.6826

10. Review of distributions Solution: c) Area = 0.15 and from table the value of Z = -1.04 then Value of inside diameter will 15% of the piston rings fall below 9.9688 cm

11. Review of distributions Question 4: A lawyer commutes daily from his suburban home to his midtown office. The average time for a one-way trip is 24 minutes, with a standard deviation of 3.8 minutes. Assume the distribution of trip times to be normal distribution. a) What is the probability that a trip will take at least 0.5 hour?

12. Review of distributions Question 4: b) If the office opens at 9:00 A.M and the lawyer leaves his house at 8:45 A.M daily, what percentage of the time is he late for work? c) If he leaves the house at 8:35 A.M and coffee is served at the office from 8:50 A.M until 9:00 A.M, what is probability that he misses coffee? d) Find the length of time above which we find the slowest 15% of the trip?

13. Review of distributions Solution: P(X > 30) = P(Z > 1.58) = 1 – P (Z < 1.58) = 1 - 0.9428 = 0.0571 P( X > 15) = P( Z > -2.37 ) = 1-P( Z < -2.37 ) = (1-0.0089) = 0.9911 Therefore he is late at 99.11% of time.

14. Review of distributions Solution: P( X > 25) = P( Z > 0.26 ) = 1-P( Z < 0.26 ) = (1- 0.6026) = 0.3974 The probability that he misses coffee is 39.74% d) The length of time above which we find the slowest 15% of the trip is 1-.15=0.85,Then from the table Z is equal 1.04

15. Review of distributions Question 5: Twelve people are given two identical speakers, which they are asked to listen to for differences, if any. Suppose that these people answer simply by guessing, find the probability that 3 people claim to have heard a difference between the speakers.

16. Review of distributions Solution: We can solve this problem by binomial distribution with n=12 and p = 0.5 as: The probability that 3 people claim to have heard a difference between the speakers is 0.0537.

17. Review of distributions Question 6: Public opinion reported that 5% of Americans are afraid of being alone in a house at night. If a random sample of 20 Americans is selected, find these probabilities. a) There are exactly 5 people in the sample who are afraid of being alone at night. b) There are most 3 people in sample who are afraid of being alone at night. c) There are at least 3 people in the sample who are afraid of being alone at night.

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19. Review of distributions Solution b) At most 3 peoples are afraid means that P(0) +P(1)+P(2)+P(3)

20. Review of distributions Solution There are at least 3 people in the sample who are afraid of being alone at night. P (at least 3 are afraid) = 1 – [P (0) +P (1) +P (2)] = 1- [0.358+.0377+0.189) = 0.076