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Math 235 Answers (Practice-Sept. 05) Answers to Additional Problems—click here 1. a) 40C5 = 40 * 39 * 38 * 37 * 36 Ans. 1/658,008 5! = 658,008 b) 40P5 = 40*39*38*37*36 = 78,960,960 Ans. 1/78,960,960 2. Mean 69.8458 Stan. Dev. 4.1628 (pop.) 4.1824 (sd) 3. P(r > 2) = 1 – P(r < 2) = 1 – [ P(0) + P(1) ] = 1 – [ (½)5 + 5 * (½)5 ] = 13/16 4. P (heart and spade) = P (heart on 1st, spade on 2nd) + P (heart on 2nd, spade on 1st) (Reason: the events are mutually exclusive) a) With replacement: ¼*¼+¼*¼=⅛ b) Without replacement: 1 12 1 12 3 3 6 2 * * 4 51 4 51 51 51 51 17 5. 6 * 10 * 4 = 240 6. The coefficient of variation (CV) for Group A is 2.3/35.4 * 100 = 6.5%. For Group B the CV is 1.4/8.1 * 100 = 17.3% So Group B has values which are more spread out. 7. – 8. x 0 1 2 3 4 Total P(x) 1/16 ¼ 3/8 ¼ 1/16 16/16 = 1 Answers to Additional Problems 1. a) Use the formula: Probability of r successes on n trials is: nCr (p)r (q)n–r Note n = 5, p = .45, q = .55, and r varies from 0 to 5. r P(r) 0.0503 0 0.2059 1 0.3369 2 0.27565 3 0.1128 4 0.0185 5 b) The fastest way is to use the formulas μ = np and σ = √(npq) Then μ = 5 ·0.45 = 2.25, σ = (5)(.45)(.55) = 1.1124 Alternately, you can work the values out using the formulas for mean and s.d.—see # 4 below. c) For a probability dist., the mean is called the expected value. The 2.25 represents the most likely number of heads obtained if the coin is tossed 5 times. I.e., if you repeat the experiment (tossing the coin five times) a large number of times, the number of heads obtained will average 2.25. 2. Use the same formula as in #1a: P(r = 10) = 20C10 (.7)10(.3)10 = 184,756 (.7)10(.3)10 = .0308 (about 3%) 3. The prob. of getting any one problem correct is 1/5 or 0.2, because the student is guessing. Thus p = 0.2, q = 0.8. There are 15 problems so n = 15. Note P(r > 3) = 1 – P(r < 3) = 1 – [ P(0) + P(1) + P(2) ] P(0) 15 C 0 (0.2) 0 (0.8)15 .0352 P(1) 15 C1 (0.2)1 (0.8)14 .1319 P(2) 15 C 2 (0.2) 2 (0.8)13 .2309 So the total probability is 1 – (.0352 + .1319 + .2309) =.602 (about 60%). 4. Age, years Percent of Patients Midpoint 20-29 30-39 40-49 50-59 60-69 70-79 Totals .02 .14 .18 .23 .24 .19 1 24.5 34.5 44.5 54.5 64.5 74.5 Midpoint · Frequency: x·P(x) .49 4.83 8.01 12.535 15.48 14.155 55.5 Deviation Dev.2 Dev.2 · Freq. -31 -21 -11 -1 9 19 961 441 121 1 81 361 19.22 61.74 21.78 0.23 19.44 68.59 191 The mean is μ = Σ [x · P(x)] = 55.5 The variance is σ2 = Σ[(x – μ)2 · P(x)] = 191 and the stan. dev. is √191 = 13.8203 5. We have r = number of calls out of n resulting in a sale, n = total number of calls, p = .23 (prob. that any one call will result in a sale). We want the smallest integer n such that P(r > 0) > .99 i.e., 1 – P(r = 0) > .99 1 – .99 > P(r = 0) .01 > P(r = 0) .01 > (.23)n log (.01) > log (.23)n -2 > n log (.23) -2 > n ·-.6383 (-2 / -.6383) < n (change the direction of the inequality since dividing by a negative) 3.133 < n The answer is 4 since that is the first integer satisfying 3.133 < n. Note that (.23)3 = .012 which is too large. Four calls are needed to ensure with a probability of at least 99% that one or more of the calls will result in a sale.