AP Statistics Test 5

Transcription

1 AP Statistics Test 5 Name: Date: Period: ffl If X is a discrete random variable, the the mean of X and the variance of X are given by μ = E(X) = X xp (X = x); Var(X) = X (x μ) 2 P (X = x): ffl If X is the binomial random variable associated with n independent trials and success probability p, then X has mean μ = np and variance ff 2 = np(1 p). ffl If X 1 and X 2 are independent random variables, and if a 1 and a 2 are constants, then E(a 1 X 1 + a 2 X 2 )=a 1 E(X 1 )+a 2 E(X 2 ) and Var(a 1 X 1 + a 2 X 2 )=a 2 1 Var(X 1)+a 2 2 Var(X 2) Part 1, Multiple-Choice Questions 50 points. 1. Suppose that a television game has three payoffs with the probabilities tabulated at the right: Payoff($) ,000 Probability What are the mean (μ) and the standard deviation (ff) of the payoff variable? (A) μ = $150; ff =$50 (B) μ = $150; ff = $187; 500 (C) μ = $150; ff =$433:01 (D) μ =$2; 000; ff = $433:01 (E) μ = $150; ff = $414:43 2. A computer chip manufacturer sends out shipments of computer chips in lots of 10,500 each. If we assume that 0.3% of the chips are defective, then the mean μ and standard deviation ff of the number of defective chips in each lot is (A) μ =32; ff =5 (B) μ =10:5; ff =5:6 (C) μ =31:5; ff =5:6 (D) μ =31:5; ff =31:4 (E) μ = 315; ff =17:5

2 3. Suppose that we have a box with 10 marbles among which six are white and the remaining four are other colors. We randomly select four marbles without replacement from this box and let X be the random variable which counts the white marbles drawn. Consider the following statements: I. 0» X» 4 II. E(X) =4 III. X is not a binomial random variable. (A) I only is true. (B) II only is true. (C) III only is true. (D) I and II only are true. (E) I and III only are true. 4. Suppose that you are given a normally-distributed random variable X with mean μ and variance ff 2. If you are given that Then (A) μ ß 14:5 and ff ß 1 (B) μ ß 14 and ff ß 2 (C) μ ß 15 and ff ß 1 (D) μ ß 15:5 and ff ß :5 (E) μ ß 13 and ff ß 3 P (X» 15) = 69:1% and P (X 16) = 15:9% 5. Suppose that you are going to flip a fair coin 40,000 times, letting X be the total number of heads. Use the normal approximation to this binomial distribution to compute P (μ 2ff» X» μ +2ff), where E(X) =μ and Var(X) =ff 2. (A) P (μ 2ff» X» μ +2ff) ß 31:5% (B) P (μ 2ff» X» μ +2ff) ß 68% (C) P (μ 2ff» X» μ +2ff) ß 95% (D) P (μ 2ff» X» μ +2ff) ß 5% (E) This is impossible to determine without knowing μ and ff.

3 6. Jennifer often finds it difficult to arrive to class on time. In fact, it has been determined that if X measures the number of minutes that Jennifer is late for her statistics class, then X is a uniformly-distributed random variable with 2» X» 10: Based on this, on any given day we expect Jennifer to arrive (A) on time. (B) 2 minutes late. (C) 4 minutes late. (D) 6 minutes late. (E) not at all. 7. Continuing with the same assumptions as in Problem 6 above, the probability that Jennifer will arrive late is about (A) 17% (B) 50% (C) 83.3% (D) 97.5% (E) 91% 8. Suppose that we now consider Sunny s tardiness, which has a normal distribution with a mean tardiness of 4 minutes and a standard deviation of 3 minutes. The probability that on any given day Sunny will arrive early is approximately (A) 17% (B) 9% (C) 83% (D) 23% (E) 91% 9. We continue to assume that Jennifer s and Sunny s tardiness are given as described above. Assuming that they arrive independently of each other, what is the probability that neither one of them arrives late? (A) 14.1% (B) 9% (C) 15.3% (D) 1.5% (E) 98.5%

4 10. Suppose that an insurance company charges $1,000 annually for car insurance. The policy specifies that the company will pay $1,100 for a minor accident and $5,000 for a major accident. Assume that during a given year, your likelihood of having a minor accident is 20% and your likelihood of having a major accident is 5% How much profit per year does the insurance company expect to make from each client? (A) $500 (B) $510 (C) $520 (D) $530 (E) $540 Part 2, Free-Response Questions 50 points. 11. A company buys 44% of its stock of bolts from a manufacturer A and the rest from manufacturer B. The diameters of the bolts produced by each manufacturer follow a normal distribution with a standard deviation of 0.16 mm. The mean diameter of the bolts produced by manufacturer A is 1.56 mm. 24.2% of the bolts produced by manufacturer B have a diameter less than 1.52 mm. (a) (6 points) Find the mean diameter of the bolts produced by manufacturer B. (b) (8 points) Manufacturer B makes 8000 bolts in one day. It makes a profit of $1.50 on each bolt sold, on condition that its diameter measures between 1.52 and 1.83 mm. Bolts whose diameters measure less than 1.52 mm must be discarded at a loss of $0.85 per bolt. Bolts whose diameters measure over 1.83 mm are sold at a reduced profit of $0.50 per bolt. Find the expected profit for manufacturer B on a given day.

5 12. Let the random variable X represent the number of telephone lines in use by the technical support center of a software manufacturer at noon each day. The probability distribution of X is shown in the table below: x p(x) 0:35 0:20 0:15 0:15 0:10 0:05 (a) (5 points) Calculate the expected value (the mean) of X. (b) (5 points) Using past records, the staff at the technical support center randomly selected 20 days and found that an average of 1.25 telephone lines were in use at noon on those days. The staff proposes to select another random sample of 1,000 days, and compute the average number of telephone lines that were in use at noon on those days. How do you expect the average from this new sample to compare to that of the first sample? Justify your response. (c) (5 points) The median of a random variable is defined as any value x such that P (X» x) 0:5 and P (X x) 0:5. For the probability distribution shown in the table above, determine the median of X. (d) (3 points) In a sentence or two, comment on the relationship between the mean and the median relative to the shape of this distribution.

6 13. For an upcoming event, each customer may purchase up to 3 child tickets and 3 adult tickets. Let C be the number of child tickets purchased by a single customer. The probability distribution of the number of child tickets purchased by a single customer is given in the table to the right. Distribution of Child Tickets c P (C = c) 0:4 0:3 0:2 0:1 (a) (6 points) Compute the mean and the standard deviation of C. (b) (6 points) Suppose the mean and standard deviation for the number of adult tickets purchased by a single customer are 2 and 1.2, respectively. Assume that the numbers of child tickets and adult tickets purchased are independent random variables. Compute the mean and the standard deviation of the total number of adult and child tickets purchased by a single customer. (c) (6 points) Suppose each child ticket costs $15 and each adult ticket costs $25. Compute the mean and the stardard deviation of the total amount spent per purchase.