STAT213 Tutorial sheet #4 Solutions

 

Question 1

Each probability is between 0 and 1 and they sum to 1.

 

Question 2

m = 1(0.691) + 2(0.178) + 3(0.049) = 1.194

E(X2) = 1(0.691) + 4(0.178) + 9(0.049) = 1.844; s2 = 1.844 – 1.1942 = 0.418364; s = sqrt(0.418364) = 0.6468

 

Question 3

We need m-2s = 1.194 – 2(0.6468) = -0.1 as well as m+2s = 2.5. The only values of the distribution that lie between these limits are 0, 1 and 2. P(0) + P(1) +P(2) = 8.2% + 69.1% + 17.8% = 95.1% which is well above the threshold of 75% under Chebyshev’s Theorem.

 

Question 4

Using Minitab, since we want P(2), we choose the probability radio button in the binomial dialogue box. The number of trials is 20, the probability of success is 0.05 and the input constant is 2. P(2) = 0.1887 = 18.87%

 

Question 5

Using Minitab, since we want P(no more than 1), we choose the cumulative probability radio button and the input constant is 1. P(no more than 1) = 0.7359 = 73.59%

 

Question 6

P(at least 2) = 100% - P(no more than 1) = 100% - 73.59% = 26.41%

 

Question 7

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.

 

Question 8

m = 500(1/5000) = 0.1. Using Minitab, since we want P(0), we choose the probability radio button in the Poisson dialogue box. The mean is 0.1 and the input constant is zero. P(0) = 0.9048 = 90.48%

 

Question 9

If m = 0.1 for one shift, then m = 0.3 for three shifts. Note that P(at least 1) = 1 – P(0). Using Minitab as with the previous question (except with the mean equal to 0.3), P(0) = 0.7408 = 74.08%. So, P(at least 1) = 100% - 74.08% = 25.92%

 

Question 10

If m = 0.3 for one day, then m = 2.1 for seven days. Using Minitab, since we want P(no more than 6), we choose the cumulative probability radio button, the mean is 2.1 and the input constant is 6. P(no more than 6) = 0.9942 = 99.42% which is the probability of having no more than 6 scrap gears in a 7-day period. This appears to be a realistic expectation.

 

Question 11

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.

 

Question 12

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

 

Question 13

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

 

Question 14

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%

 

Question 15

The first thing is to construct a crosstab letting cell phone = x and blackberry = y:

x | y

0

1

Total

0

0.02

0.03

0.05

1

0.09

0.86

0.95

Total

0.11

0.89

1.0

mx = 0.95; E(X2) = 0.95; sx2 = 0.95 – 0.952 = 0.0475

my = 0.89; E(Y2) = 0.89; sy2 = 0.89 – 0.892 = 0.0979

E(XY) = 0.86; Cov(X,Y) = 0.86 – (0.95)(0.89) = 0.0145

So mx + my = 0.95 + 0.89 = 1.84;

Var(X+Y) = 0.0475 + 0.0979 + 2(0.0145) = 0.1744; Stdev(X+Y) = sqrt(0.1744) = 0.4176;

r2 = 0.01452/[(0.0475)(0.0979)] = 4.52%