Statistics / Midterm Exam (Duration: 90 minutes) Name:

Q1. A long-distance taxi service owns four vehicles. These are of different ages and have different repair records. The probabilities that on any given day, each vehicle will be available for use are .95, .90, .90, and .80. Whether one vehicle is available is independent of whether any other vehicle is available.(a) Find the probability function for the number of vehicles available for use on a given day.(b) Find the expected number of vehicles available for use on a given day.

Find the standard deviation of the number of vehicles available for use on a given day.Answer: a) P(0)=0.0001, P(1)=0.0041, P(2)=0.0571, P(3)=0.3231, P(4)=0.6156 b) 3.55

c) 0.6225Solution: a) P(0) = (1-0.95)(1-0.9)(1-0.9)(1-0.8) = 0.0001P(1) = 0.95(1-0.9)(1-0.9)(1-0.8) + (1-0.95)0.9(1-0.9)(1-0.8) + (1-0.95)(1-0.9)0.9(1-0.8) +

+ (1-0.95)(1-0.9)(1-0.9)0.8 = 0.0041P(2) = (0.95)(0.9)(1-0.9)(1-0.8) + (0.95)(1-0.9)(0.9)(1-0.8) + (0.95)(1-0.9)(1-0.9)(0.8) + (1-0.95)

(0.9)(0.9)(1-0.8) + (1-0.95)(0.9)(1-0.9)(0.8) + (1-0.95)(1-0.9)(0.9)(0.8) = 0.0571P(3) = (0.95)(0.9)(0.9)(1-0.8) + (0.95)(0.9)(1-0.9)(0.8) + (0.95)(1-0.9)(0.9)(0.8) + (1-0.95)(0.9)

(0.9)(0.8) = 0.3231P(4) = (0.95)(0.9)(0.9)(0.8) = 0.6156b) E(X) = SUM(x.p(x)) = 3.55 STD(X) = SUM ((x-E(X))2p(x)) = 0.6225

Q2. A team of five analysts is about to examine the earnings prospects of twenty corporations. Each of the five analysts will study four of the corporations. These analysts are not equally competent. In fact, one of them is a star, having an excellent record of anticipating changing trends. Ideally, management would like to allocate to this analyst the four corporations whose earnings will deviate most from past trends. However, lacking this information, management allocates corporations to analysts randomly. What is the probability that at least two of the four corporations whose earnings will deviate most from past trends are allocated to the star analyst?Answer: 0.1808Solution: Let X be the number of the four corporations whose earnings will deviate most from past trends are allocated to the star analyst. Then X has a hypergeometric distribution: n=5 (number of trials), N=20 (population size) and s=4 (number of possible occurences). Need to find P(X≥2) = 1- P(0)- P(1) = 1 – 0.37564 – 0.46233 = 0.16202 Q3. It is estimated that amounts of money spent on gasoline by customers at a gas station follow

a normal distribution with standard deviation $2.50. It was also found that 10% of all customers spend more than $15. What percentage of customers spend less than $10?

Answer: 23.58%Solution: X=N(µ,2.52)We know P(X>15) = 0.1, and want to find P(X<10).P(Z>(15-µ)/2.5) = 0.1P(Z≤(15-µ)/2.5) = 0.9Hence (15-µ)/2.5 = 1.28, so µ = 11.8. Thus, X=N(11.8,2.52).P(X<10) = 0.2358. Thus, 23.58% of customers spend less than $10.

Q4. A consultant knows that it will cost her $10,000 to fulfill a particular contract. The contract is to be put out for bids, and she believes that the lowest bid, excluding her own, can be represented by a distribution that is uniform between $8,000 and $20,000. Therefore, if the ran-

dom variable X denotes the lowest of all other bids (in thousands of dollars), its probability density function is

(a) What is the probability that the lowest of the other bids will be less than the consultant's cost estimate of $10,000?

(b) If the consultant submits a bid of $12,000, what is the probability that she will secure the contract?

(c) The consultant decides to submit a bid of $12,000. What is her expected profit from this strategy?

(d) If the consultant wants to submit a bid so that her expected profit is as high as possible, discuss how she should go about making this choice.

Answer: a) 1/6 b) 2/3 c) $1,333.33 d) $15,000 Solution: X = U(8,20)a) P(X<10) = (10-8)/(20-8) = 2/12 = 1/6b) P(X>12) = (20-12)/12=8/12=2/3c) Expected profit = (12-10)(20-12)/12 = 4/3 = 1,333.33 (thousand US dollars)d) Let B denote the bid (in thousand of dollars). Then the expected profit is (B-10)(20-B)/12

thousand dollars. Choose B to maximize this, giving B=$15,000.

Q5. In a large department store, a customer complaints office handles an average of six complaints per hour about quality of service. The distribution is Poisson.(a) What is the probability that in any hour exactly six complaints will be received? (b) What is the probability that more than 20 minutes will elapse between successive

complaints?(c) What is the probability that less than 5 minutes will elapse between successive

complaints?(d) The store manager observes the complaints office for a 30-minute period, during

which no complaints are received. She concludes that a talk she gave to her staff on the theme "the customer is always right" has obviously had a beneficial effect. Suppose that in fact the talk had no effect. What is the probability of the manager's observing a period of 30 minutes or longer with no complaints?

Answer: a) 0.16 b) 0.135 c) 0.393 d) 0.05Solution: X=Poisson(6)

a) Probability of exactly 6 complaints will be received = f(6) = 66 . e-6 / 6! = 0.16

The length of the interval between two occurences Y follows exponential distribution, with mean µ=10 (minutes).b) P(Y>20) = e -2 = 0.135c) P(Y<5) = 1- e-1/2 = 0.393