Probability and Statistics for Engineering and the Sciences
Probability and Statistics for Engineering and the Sciences
9th Edition
ISBN: 9781305251809
Author: Jay L. Devore
Publisher: Cengage Learning
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Textbook Question
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Chapter 5.5, Problem 59E

Let X1, X2, and X3 represent the times necessary to perform three successive repair tasks at a certain service facility. Suppose they are independent, normal rv’s with expected values µ1, µ2, and µ3 and variances σ 1 2 σ 2 2 , and σ 3 2 , respectively.

  1. a. If µ = µ2 = µ3 = 60 and σ 1 2 = σ 2 2 = σ 3 2 = 15, calculate P(To ≤ 200) and P(150 ≤ To ≤ 200)?
  2. b. Using the µi’s and σi’s given in part (a), calculate both P(55 ≤ X) and P(58 ≤ X ≤ 62).
  3. c. Using the µi’s and σi’s given in part (a), calculate and interpret P(−10 ≤ X1 − .5X2 − .5X3 ≤ 5). d. If µ1 = 40, µ2 = 50, µ3 = 60, σ 1 2 = 10, σ 2 2 = 12, and σ 3 2 = 14, calculate P(X1 + X2 + X3 ≤ 160) and also P(X1 + X2 ≥ 2X3).

a.

Expert Solution
Check Mark
To determine

Calculate P(To200) and P(150To200) when μ1=μ2=μ3=60 and σ12=σ22=σ32=15.

Answer to Problem 59E

The value of P(To200) under the given condition is 0.9986.

The value of P(150To200) under the given condition is 0.9986.

Explanation of Solution

Given info:

The random variables X1, X2 and X3 denote the successive times required for repair tasks at a certain facility, having independent normal distributions with respective means μ1, μ2, μ3 and respective variances σ12, σ22, σ32.

Calculation:

The variable To denotes the total time for repairing, that is, To=X1+X2+X3. Being the sum of independent normal variables, To has a normal distribution with parameters:

μTo=μ1+μ2+μ3=3×60=180.

σ2To=V(X1+X2+X3)=V(X1)+V(X2)+V(X3)  (due to independence of the variables)=σ21+σ22+σ23=3×15

=45.

The value of P(To200) under the given condition is:

P(To200)=P(ToμToσTo200μToσTo)=P(Z20018045)=Φ(2.98).

From Table A.3, “Standard Normal Curve Areas”, Φ(2.98)=0.9986.

Thus, the value of P(To200) under the given condition is 0.9986.

The value of P(150To200) under the given condition is:

P(150To200)=P(150μToσToToμToσTo200μToσTo)=P(15018045Z20018045)=Φ(2.98)Φ(4.47).

It is known that Φ(3.5)0. Now, 4.47<3.35. Thus, Φ(4.47)=0.

Thus, the value of P(150To200) under the given condition is:

P(150To200)=0.99860=0.9986_.

b.

Expert Solution
Check Mark
To determine

Calculate P(55X¯) and P(58X¯62) when μ1=μ2=μ3=60 and σ12=σ22=σ32=15.

Answer to Problem 59E

The value of P(55X¯) under the given condition is 0.9875.

The value of P(58X¯62) under the given condition is 0.6266.

Explanation of Solution

Calculation:

The variable X¯ denotes the average time for repairing, that is, X¯=X1+X2+X33. Being a linear combination of independent normal variables, X¯ has a normal distribution with parameters:

μX¯=μ1+μ2+μ33=3×603=60.

σ2X¯=V(X1+X2+X33)=19[V(X1)+V(X2)+V(X3)]  (due to independence of the variables)=σ21+σ22+σ239=3×159

=5.

The value of P(55X¯) under the given condition is:

P(55X¯)=P(X¯μX¯σX¯55μX¯σX¯)=P(Z55605)=1P(Z2.24)=1Φ(2.24)

       =1[1Φ(2.24)]=Φ(2.24).

From Table A.3, “Standard Normal Curve Areas”, Φ(2.24)=0.9875.

Thus, the value of P(55X¯) under the given condition is 0.9875.

The value of P(58X¯62) under the given condition is:

P(58X¯62)=P(58μX¯σX¯X¯μX¯σX¯62μX¯σX¯)=P(58605Z62605)=Φ(0.89)Φ(0.89)=Φ(0.89)[1Φ(0.89)]

  =2Φ(0.89)1.

From Table A.3, “Standard Normal Curve Areas”, Φ(0.89)=0.8133.

Thus,

P(58X¯62)=2Φ(0.89)1=(2×0.8133)1=1.62661=0.6266_.

c.

Expert Solution
Check Mark
To determine

Calculate P(10X10.5X20.5X35) and interpret the same.

Answer to Problem 59E

The value of P(10X10.5X20.5X35) is 0.8364.

Explanation of Solution

Calculation:

The variable X10.5X20.5X3 is a linear combination of independent normal random variables. Thus, X10.5X20.5X3 has a normal distribution with parameters:

μX10.5X20.5X3=μ10.5μ20.5μ3=60(0.5×60)(0.5×60)=603030=0.

σ2X10.5X20.5X3=V(X10.5X20.5X3)=V(X1)+(0.5)2V(X2)+(0.5)2V(X3)=σ21+0.25σ22+0.25σ23=15+(0.25×15)+(0.25×15)

=15+3.75+3.75=22.5.

The value of P(10X10.5X20.5X35) under the given condition is:

P(10X10.5X20.5X35)=P(10μX10.5X20.5X3σX10.5X20.5X3X10.5X20.5X3μX10.5X20.5X3σX10.5X20.5X35μX10.5X20.5X3σX10.5X20.5X3)=P(10022.5Z5022.5)=Φ(1.05)Φ(2.13).

From Table A.3, “Standard Normal Curve Areas”, Φ(1.05)=0.8531 and Φ(2.13)=0.0166.

Thus, the value of P(10X10.5X20.5X35) under the given condition is:

P(10X10.5X20.5X35)=Φ(1.05)Φ(2.13)=0.85310.0166=0.8365_.

The obtained probability can be interpreted as follows:

In about 83.65% cases, the difference between the time taken to perform the first repair task, and half the time taken to perform the next two repair tasks will lie between –10 and 5.

d.

Expert Solution
Check Mark
To determine

Calculate P(X1+X2+X3160) and P(X1+X22X3) when μ1=40, μ2=50, μ3=60, σ12=10, σ22=12, σ32=14.

Answer to Problem 59E

The value of P(X1+X2+X3160) under the given condition is 0.9525.

The value of P(X1+X22X3) under the given condition is 0.0003.

Explanation of Solution

Calculation:

The variable X1+X2+X3 being a linear combination of independent normal variables, has a normal distribution with parameters:

μX1+X2+X3=μ1+μ2+μ3=40+50+60=150.

σ2X1+X2+X3=V(X1+X2+X3)=V(X1)+V(X2)+V(X3)  (due to independence of the variables)=σ21+σ22+σ23=10+12+14

=36.

The value of P(X1+X2+X3160)P(To200) under the given condition is:

P(X1+X2+X3160)=P(X1+X2+X3μX1+X2+X3σX1+X2+X3160μX1+X2+X3σX1+X2+X3)=P(Z16015036)=Φ(1.67).

From Table A.3, “Standard Normal Curve Areas”, Φ(1.67)=0.9525.

Thus, the value of P(X1+X2+X3160) under the given condition is 0.9525.

The value of P(X1+X22X3) under the given condition is:

P(X1+X22X3)=P(X1+X22X30).

The variable X1+X22X3 being a linear combination of independent normal variables, has a normal distribution with parameters:

μX1+X22X3=μ1+μ22μ3=40+50(2×60)=40+50120=30.

σ2X1+X22X3=V(X1+X22X3)=V(X1)+V(X2)+(2)2V(X3)  (due to independence of the variables)=σ21+σ22+4σ23=10+12+(4×14)

=10+12+56=78.

Thus,

P(X1+X22X30)=P(X1+X22X3μX1+X22X3σX1+X22X30μX1+X22X3σX1+X22X3)=1P(Z0(30)78)=1Φ(3.40).

From Table A.3, “Standard Normal Curve Areas”, Φ(3.40)=0.9997.

Thus,

P(X1+X22X3)=1Φ(3.40)=10.9997=0.0003_.

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Chapter 5 Solutions

Probability and Statistics for Engineering and the Sciences

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