MGMT2262 Worksheet #4 Solutions

 

Question 1

a)      P(x < 3.9) = (3.9 – 3.85)/(4.15 – 3.85) = 0.05/0.3 = 0.1667 = 16.67%

b)      P(x > 4.05) = (4.15 – 4.05)/(4.15 – 3.85) = 0.1/0.3 = 0.3333 = 33.33%

c)      m = (3.85 + 4.15)/2 = 4; s = (4.15 – 3.85)/sqrt(12) = 0.3/sqrt(12) = 0.0866

m - 0.5s = 4 – 0.5(0.0866) = 4 – 0.0433 = 3.9567

m + 0.5s = 4 + 0.0433 = 4.0433

P(3.9567 < x < 4.0433) = (4.0433 – 3.9567)/(4.15 – 3.85) = 0.0866/0.3 = 0.2887 = 28.87%

This does not seem like a realistic expectation.

 

Question 2

a)      2 customers per minute = 1 customer per 0.5 minute; m = 0.5; P(X > 2) = e-2/0.5 = e-4 = 0.0183 = 1.83%

b)      P(X < 1) = 1 – e-1/0.5 = 1 – e-2 = 1- 0.1353 = 0.8647 = 86.47%

 

Question 3

a)      m = 10: P(X > 10) = e-10/10 = e-1 = 0.3679 = 36.79%

b)      m = 10 x 6/3 = 20; P(X > 10) = e-10/20 = e-0.5 = 0.6065 = 60.65%

c)      m = 8 x 6/4 = 12; P(X > 15 | X > 10) = P(X > 5) = e-5/12 = 0.6592 = 65.92%

 

Question 4

a)      0.4429 from the table

b)      0.4993 from the table

c)      0.3997 + 0.2486 = 0.6483

d)     0.4945 – 0.4177 = 0.0768

e)      0.4761 – 0.008 = 0.4681

f)       We need the 85.08th percentile. From this we get P(0 < Z < x) = 0.3508. Looking for the probability 0.3508 in the Z table, we find the corresponding Z value to be 1.04

g)      Since P(-x < Z < x) = 0.9756, P(0 < Z < x) = 0.4878. As with part f, we find the corresponding Z value to be 2.25

 

Question 5

a)      Z = (7-5)/2 = 1. Then P(X < 7) = P(Z < 1) = 0.5 + 0.3413 = 0.8413.

b)      Z = (9.2-5)/2 = 2.1. Then P(X > 9.2) = P(Z > 2.1) = 0.5 – 0.4821 = 0.0179

c)      Z1 = (4-5)/2 = -0.5; Z2 = (6.4-5)/2 = 0.7. Then P(4 < X < 6.4) = P(-0.5 < Z < 0.7) = 0.1915 + 0.258 = 0.4495

d)     Since, we need the 93.7th percentile, we get P(0 < Z < x) = 0.437. The Z value is 1.53. Then 1.53 = (x-5)/2. Solving for x, x = 5 + 2(1.53) = 8.06.

e)      Using the empirical rule for symmetric data, m - 2s = 5 – 2(2) = 1; m + 2s = 5 + 2(2) = 9

 

Question 6

a)      Z = (750-800)/75 = -0.67. Then P(X < 750) = P(Z < -0.67) = 0.5 – 0.2486 = 0.2514

b)      Z = (900-800)/75 = 1.33. Then P(X > 900) = P(Z > 1.33) = 0.5 – 0.4082 = 0.0918

c)      We need the 95th percentile which is 1.645. Then 1.645 = (X – 800)/75 from which we get X = 800 + 1.645(75) = 923.38.

d)     Let x be the upper Z value. Then P(-x < Z < x) = 0.796 from which we get P(0 < Z < x) = 0.398. The upper Z value is 1.27 and the lower Z value is -1.27. Solving for the upper value, we have 1.27 = (X – 800)/75 from which we get X = 800 + 1.27(75) = 895.25. The lower value is 800 – 1.27(75) = 704.75.

 

Question 7

a)      Z = (10.01-10.2)/0.2 = -0.95. Then P(X < 10.01) = P(Z < -0.95) = 0.5 – 0.3289 = 0.1711

b)      Z = (10.3-10.2)/0.2 = 0.5. Then P(X > 10.3) = P(Z > 0.5) = 0.5 – 0.1915 = 0.3085

c)      In this case we need the 5th percentile of Z which is –1.645. Then –1.645 = (x-10.2)/0.2. Solving for x, x = 10.2 – 1.645(0.2) = 9.871 seconds.

d)     Using the empirical rule for symmetric data, m - 3s = 10.2 – 3(0.2) = 9.6; m + 3s = 10.2 + 3(0.2) = 10.8

 

From the textbook

 

5.9 (page 150)

Part a

µ = 6/2 = 3; s = 6/sqrt(12) = 1.7321; s2 = 36/12 = 3

Part b

µ - s = 1.2679, µ + s = 4.7321. P(1.2679 < X < 4.7321) = 3.4642/6 = 0.5774

 

5.29 (page 164)

Part a

Subpart 1: P(X < 959) = P(Z < 2.12) = 0.5 + 0.483 = 0.983

Subpart 2: P(X > 1004) = P(Z > 2.72) = 0.5 – 0.4967 = 0.0033

Subpart 3: P(X < 650) + P(X > 950) = P(Z < -2) + P(Z > 2) = 2P(Z > 2) = 2(0.5 – 0.4772) = 2(0.0228) = 0.0456

Part b

We need P(Z > k) = 0.025. Then k = 1.96 and 1.96 = (X – 800)/75 from which we get X = 800 + 1.96(75) = 947

 

5.57 (page 172)

Part a

Subpart 1: P(X > 2) = e-2 = 0.1353

Subpart 2: P(1 < X < 2) = P(X > 1) – P(X > 2) = e-1 – e-2 = 0.3679 – 0.1353 = 0.2326

Subpart 3: P(X < 0.25) = 1 – e-0.25 = 1 – 0.7788 = 0.2213

Part b

From a statistical viewpoint, there is better than 1 in 4 probability of successive accidents within a 1-week period and does not warrant investigation. From a humanitarian viewpoint, considering the possible pain and suffering caused to the employee and immediate family, an investigation would certainly be warranted.

 

6.13 (page 192)

Part a

Subpart 1: P(Xbar > 1.5) = P(Z > (1.5 – 1)/(1.3/sqrt(100)) = P(Z > 0.77) = 0.5 – 0.2794 = 0.2206

Subpart 2: P(Xbar > 1.5) = P(Z > (1.5 – 1)/(1.8/sqrt(100)) = P(Z > 2.78) = 0.5 – 0.4973 = 0.0027

Part b

Based on the probability from subpart 2, since it is highly unlikely that the sample mean is more than 1.5 if the actual population mean is 1.0, it would be reasonable to assume that the mean amount of lost unpaid time has increased.