MGMT2262 Worksheet #3 Solutions

 

1)      First we need to solve for k: 1 – 1/k2 = 8/9 from which we get 1/k2 = 1/9 and so k=3. Then m - 3s = 14.4 and m + 3s = 33.6

2)      We have 2 equations:

Equation 1: 19.008 = 24 – 3.2k

Equation 2: 28.992 = 24 + 3.2k

Solving for k in either equation will give k = 1.56. Thus, the two limits are 1.56 standard deviations away from the mean of 24.

3)      First we need the mean and standard deviation of the data which are 5.0167 and 1.1832 respectively. The lower limit is 5.0167 – 2(1.1832) = 2.6503 and the upper limit is 5.0167 + 2(1.1832) = 7.3831. From the data, we see that all of the data except for 7.8 fall between these limits. Thus 29/30 = 96.67% of the data is within 2 standard deviations of the mean which is well above the 75% threshold under Chebyshev’s theorem.

4)      Each probability is between 0 and 1 and the probabilities sum to one.

5)      I’ll use Excel for this.

x

p(x)

x*p(x)

x2

x2*p(x)

0

0.2

0

0

0

1

0.28

0.28

1

0.28

2

0.37

0.74

4

1.48

3

0.15

0.45

9

1.35

1.47

3.11

0.9491

0.9742

I’m working on the assumption that the first zero is in cell A2. In cell c2 I type =a2*b2, then drag and drop to the remaining cells in column C. I then autosum in cell c6 to get the mean of 1.47.

Next, I square the x values in column D. In cell d2, I type = a2*a2 (or a2^2), then drag and drop through cell d5. Then in cell e2, I type =d2*b2, then drag and drop through cell e5. I then autosum in cell e6 to get Sx2p(x) = 3.11. Then, in cell e7, I type =e6-c6*c6 (or =e6-c6^2) to get  s2 = Sx2p(x) - m2 = 0.9491. Finally, in cell e8, I type = sqrt(e7) to get the value of s = 0.9742. To confirm Chebyshev’s theorem for k=2, m - 2s = 1.47 – 2(0.9742) = -0.4784 and m + 2s = 1.47 + 2(0.9742) = 3.4184. In this case, all the values of x = {0, 1, 2, 3} from the distribution lie between the limits. Thus 100% of the distribution lies within 2 standard deviations of the mean which is more than the 75% threshold under Chebyshev’s theorem.

6)      First, we use Excel to compute the mean:

x

p(x)

x*p(x)

0

0.02

0

10

0.22

2.2

20

0.34

6.8

30

0.24

7.2

40

0.12

4.8

50

0.06

3

24

Using the same procedure as in question 5, we find the average number of times a video is rented is 24. Since the store needs to bring in $95 ($80 + $15), the average price needs to be $95/24 = $3.9583 which we round up to $4.00

7)      P(no more than 2) = P(0) + P(1) + P(2)

P(0) = (9C0)(15C6)/(24C6) = 5005/134,596

P(1) = (9C1)(15C5)/(24C6) = (9 x 3003)/134,596 = 27,027/134,596

P(2) = (9C2)(15C4)/(24C6) = (36 x 1365)/134,596 = 49,140/134,596

P(no more than 2) = 81,172/134,596 = 60.31%

8)      P(at least 7) = P(7) + P(8) + P(9)

P(7) = (9C2)(15C7)/(24C9) = (36 x 6435)/1,307,504 = 231,660/1,307,504

P(8) = (9C1)(15C8)/(24C9) = (9 x 6435)/1,307,504 = 57,915/1,307,504

P(9) = (9C0)(15C9)/(24C9) = 5005/1,307,504

P(at least 7) = 294,580/1,307,504 = 22.53%

9)      P(1) = (4C1)(0.2)1(0.8)3 = 40.96%. Using Table A.1, we get 0.410 = 41%.

10)  P(fewer than 2) = P(0) + P(1)

P(0) = (10C0)(0.2)0(0.8)10 = 10.74%

P(1) = (10C1)(0.2)1(0.8)9 = 26.84%

P(fewer than 2) = 10.74% + 26.84% = 37.58%. Using Table A.2, we get 0.376 = 37.6%.

11)  P(at least 2) = 100% - P(fewer than 2)

P(0) = (25C0)(0.2)0(0.8)25 = 0.38%

P(1) = (25C1)(0.2)1(0.8)24 = 2.36%

P(fewer than 2) = 0.38% + 2.36% = 2.74%

P(at least 2) = 100% - 2.74% = 97.26%. In this case, since the sample size is more than 20, we cannot use Table A.2

12)  m = 300(0.2) = 60. So, his average commission is 60 x $2500 = $150,000

s = sqrt(60(0.8)) = 6.9282. So m + 3s = 80.7846 which rounds to 81. The maximum commission is then 81 x 2500 = $202,500.

13)  m = 10000(1/10000) = 1.

P(0) = e-1(1)0/0! = 36.79%. This is also the answer from Table A.3.

14)  m = 25000(1/10000) = 2.5.

P(at least 2) = 100% - P(fewer than 2)

P(0) = e-2.5(2.5)0/0! = 8.21%

P(1) = e-2.5(2.5)1/1! = 20.52%

P(fewer than 2) = 8.21% + 20.52% = 28.73%

P(at least 2) = 100% - 28.73% = 71.27%. We can also use Table A.3 to find P(0) and P(1).

15)  m = 5000(1/10000) = 0.5.

P(between 1 and 3 inclusive) = P(1) + P(2) + P(3)

P(1) = e-0.5(0.5)1/1! = 30.33%

P(2) = e-0.5(0.5)2/2! = 7.58%

P(3) = e-0.5(0.5)3/3! = 1.26%

P(between 1 and 3 inclusive) = 30.33% + 7.58% + 1.26% = 39.17%. We can also use Table A.3 to find P(1), P(2) and P(3).

16)  m = 3,500,000(1/10000) = 350; s = sqrt(350) = 18.7; m + 3s = 350 + 3(18.7) = 406 after rounding.

17)  m = 4; P(no more than 3) = P(0) + P(1) + P(2) + P(3) = 1.83% + 7.33% + 14.65% + 19.54% = 43.35%. We can also use Table A.3 to find P(0), P(1), P(2) and P(3).

18)  P(at least 5) = 100% - P(fewer than 5) = 1 – P(no more than 4). In the previous question, we found P(no more than 3). If we add P(4) = 19.54%, we get P(no more than 4) = 62.88%. Then P(at least 5) = 100% - 62.88% = 37.12%

19)  m = 2x4 = 8; P(at least 10) = 100% - P(fewer than 10) = 100% - P(no more than 9) = 100% - 71.66% = 28.34%. We can use Table A.3 to find P(0) through P(9) in order to add them.

20)  m = 2x4x8 = 64; s = sqrt(64) = 8; m + 3s = 88.

 

From the textbook

 

4.14 (page 125)

Part a

µ = 300(0.2) + 600(0.6) + 900(0.2) = 600. The average outcome of this investment is $600.

E(X2) = 3002(0.2) + 6002(0.6) + 9002(0.2) = 396,000

s2 = 396,000 – 6002 = 36,000 from which we get s = 189.74.

Part b

For k=3, Chebyshev’s theorem says that at least 8/9 of the distribution is between m - 3s and m + 3s. As s increases, the greater the possibility of either great gains or great losses.

Part c

Investment 3 since it has the largest standard deviation.

Part d

Investment A: cv = 10%; Investment B: cv = 31.67%. Investment B carries the greater risk.

Part e

Investment 1: cv = 3.33%; Investment 2: cv = 31.67%; Investment 3: cv = 50%. We have the same answer as part c since the standard deviations are all relative to the same mean.

 

4.23 (page 137)

Part a

P(x) = (6Cx)(0.3)x(0.7)6-x

Part b

P(at least 3) = 1 – P(no more than 2) = 1 – 0.744 = 0.256 using table A.2

P(no more than 2) = 0.744

P(at least 1) = 1 – P(0) = 1 – 0.118 = 0.882

 

4.33 (page 142)

Part a

µ = 4. P(no more than 5) = 0.0183 + 0.0733 + 0.1465 + 0.1954 + 0.1954 + 0.1563 = 0.7852 using table A.3

Part b

P(at least 6) = 1 – P(no more than 5) = 1 – 0.7852 = 0.2148

Part c

µ = 8. P(no more than 5) = 0.0003 + 0.0027 + 0.0107 + 0.0286 + 0.0573 + 0.0916 = 0.1912

Part d

µ = 6. P(at least 13) = 1 – P(no more than 12) = 1 – [P(0) + … + P(12)] = 1 – 0.9912 = 0.0088