MGMT2262 Tutorial Sheet 3 Solutions

1) 0.84 = 1 – 1/k² from which we get 1/k² = 0.16; k² = 1/0.16 = 6.25; k = 2.5; The lower limit is 64.2 – 2.5(7.8) = 44.7; the upper limit = 64.2 + 2.5(7.8) = 83.7

2) 31.05 = 64.2 – 7.8k

97.35 = 64.2 + 7.8k

Using either equation, we get 33.15 = 7.8k and k = 4.25. Plugging this into either equation shows this to be the correct value of k. The minimum percentage is 1 – 1 / 4.25² = 0.9446 = 94.46%

3) m = 2(0.1) + 2.25(0.25) + 2.5(0.43) + 2.75(0.11) + 3(0.08) + 3.25(0.02) + 3.5(0.01) = 2.48

E(X²) = 2²(0.1) + 2.25²(0.25) + 2.5²(0.43) + 2.75²(0.11) + 3²(0.08) + 3.25²(0.02) + 3.5²(0.01)

= 6.23875; s² = 6.23875 – 2.48² = 0.08835; s = sqrt(0.08835) = 0.2972

m + 2s = 2.48 + 2(0.2972) = 3.0744 which rounds to 3 minutes. Since the survey times should be within 2 standard deviations of the mean, those with survey times over 3 minutes should receive additional training.

4) P(0) = (3C0)(21C6)/(24C6) = 54,264/134,596 = 40.32%. Then, P(at least 1) = 100% - P(0) = 59.68%

5) P(0) = (6C0)(42C12)/(48C12) = 15.87%. Then, P(at least 1) = 100% - 15.87% = 84.13%

6) P(2) = (20C2)(0.05)²(0.95)¹⁸ = 18.87%. Using Table A.1, we get 0.189 = 18.9%.

7) P(0) = (20C0)(0.05)⁰(0.95)²⁰ = 35.85%

P(1) = (20C1)(0.05)¹(0.95)¹⁹ = 37.74%

P(no more than 1) = 35.85% + 37.74% = 73.59%. Using Table A.2, we get 0.736 = 73.6%.

8) P(at least 2) = 100% - P(no more than 1) = 100% - 73.59% = 26.41%. Using Table A.2, we get 1 – 0.736 = 0.264 = 26.4%.

9) The number of trials = n = 5x20 = 100.

Then m = 100(0.05) = 5; s = sqrt(5(0.95)) = 2.18; m + 3s = 11.54 which rounds to 12. The maximum commission is 12 x $450 = $5400.

10) m = 500(1/5000) = 0.1. Then P(0) = e^-0.1(0.1)⁰/0! = 90.48%

11) If m = 0.1 for one shift, then m = 0.3 for three shifts.

P(0) = e^-0.3(0.3)⁰/0! = 74.08%. So P(at least 1) = 100% - 74.08% = 25.92%

12) If m = 0.3 for one day, then m = 2.1 for seven days.

P(0) = e^-2.1(2.1)⁰/0! = 12.25%

P(1) = e^-2.1(2.1)¹/1! = 25.72%

P(2) = e^-2.1(2.1)²/2! = 27.00%

P(3) = e^-2.1(2.1)³/3! = 18.90%

P(4) = e^-2.1(2.1)⁴/4! = 9.92%

P(5) = e^-2.1(2.1)⁵/5! = 4.17%

P(6) = e^-2.1(2.1)⁶/6! = 1.46%

Adding all these probabilities gives us a 99.42% of having no more than 6 scrap gears in a 7-day period. This appears to be a realistic expectation. It should be noted these probabilities can be found in Table A.3.

13) As mentioned in the previous question, m = 2.1. Then s = sqrt(2.1) = 1.45; m + 3s = 6.45 which we can round to 6. As we see in the previous question, the probability of having no more than 6 scrap gears in a 7-day period is 99.42%. This is clearly higher than the 88.9% probability under Chebyshev’s theorem.

14) m = 0.5. Then P(0) = 60.65%

15) m = 3; P(at least 2) = 100% - P(fewer than 2) = 100% - P(no more than 1) = 100% - 19.91% = 80.09%

16) m = 6. Then s = sqrt(6) = 2.45; m + 3s = 13.35 which we can round to 13. P(no more than 13) = 99.64%