8 February 2001

Closed-book portion of test.

Question 1.
See lecture notes of 10 January elaborating on text material page 15.

Question 2.
See pages 28-31 of text, Paulos handout, and lectures notes of 10 January.

Question 3.
See section 2.5 of text and lecture notes of 24 January.

Question 4.
See pages 4-5 of text and lecture notes of 8 January.

Question 5.
See page 7 of text, lecture notes of 8 January, and assigned homework exercises 6 and 8 from section 1.1.

Question 6.
See sections 1.3, 1.4, and particularly formulas in boxes on pages 115 and 117 of the text. Assigned exercises 42 page 36 and 34 and 38, page 118 are also very pertinent.
Part (a): the variance and the standard deviation of the y_i's are unchanged from those of the x_i's.
Part (b): For the z_i's, the standard deviation of the x_i's, whatever it was, has been divided by itself. So the z_i's are standardized, with s² = s = 1.

Question 7.
Use Bayes's Theorem; this question is similar to exercise 66, pages 85-86.
P(line A) = 0.05*0.5/(0.05*0.5+0.1*0.35+0.15*0.15). P(line B) = 0.1*0.3 over the same denominator. P(line C) = 0.15*0.15 over the same denominator. These probabilities are approximately 0.30, 0.42, and 0.27; within rounding error, they sum to 1, as they must (the defective part came from some line).

Question 8.
This problem is ideal for use of the Poisson -- as mentioned in lecture 31 January, the Poisson fits cases where we have independence and we can't count non-occurrences -- how many flaws aren't there? As you did in assigned exercise 80, page 138, use the table pages 720-721. Since there's one flaw on average per 2 feet², and 10 feet² of paneling per automobile, lambda (average number of flaws per automobile interior) is 5. From the table, probability(3 or 4 flaws) = p(flaws ≤4) - p(flaws ≤2) = 0.440-0.125 = 0.315. By page 135, the mean is 5 and the standard deviation is the square root of 5.

Question 9.
This problem is appropriate for the binomial, and is similar to assigned exercises 54 and 60, pages 126-127. Use part (b) of the table on pages 718-719. Part (a): p(3) = p(≤3) - p(≤2) = 0.382-0.167 = 0.215. Part (b): p(at least one does not encounter a busy signal) = p(≤9) = 1 (within the accuracy of the table). By page 125, the mean is 10*0.4 = 4 and the standard deviation is the square root of the product 10*0.4*0.6 or the square root of 2.4, or 1.55.

Question 10.
This problem is appropriate for the negative binomial distribution, and is similar to assigned exercise 72 page 133. The probability that at most 3 attempts will be required is p(1 attempt) + p(2 attempts) + p(3 attempts) = 0.6 + 0.4*0.6 + 0.4*0.4*0.6, or 0.936. By page 132, the mean of the unsuccessful attempts is 0.4/0.6 = 2/3, and the standard deviation is the square root of 0.4/(0.6*0.6) or 1/3 the square root of 10, or 1.054.