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 ${\sigma }_1^2$ ${\sigma }_2^2$, and ${\sigma }_3^2$ , respectively.

1. a. If µ = µ₂ = µ₃ = 60 and ${\sigma }_1^2$ = ${\sigma }_2^2$ = ${\sigma }_3^2$ = 15, calculate P(T_o ≤ 200) and P(150 ≤ T_o ≤ 200)?
2. b. Using the µ_i’s and σ_i’s given in part (a), calculate both P(55 ≤ X) and P(58 ≤ X ≤ 62).
3. c. Using the µ_i’s and σ_i’s given in part (a), calculate and interpret P(−10 ≤ X₁ − .5X₂ − .5X₃ ≤ 5). d. If µ₁ = 40, µ₂ = 50, µ₃ = 60, ${\sigma }_1^2$ = 10, ${\sigma }_2^2$ = 12, and ${\sigma }_3^2$ = 14, calculate P(X₁ + X₂ + X₃ ≤ 160) and also P(X₁ + X₂ ≥ 2X₃).

To determine

Calculate $P\left(T_o\le 200\right)$ and $P\left(150\le T_o\le 200\right)$ when ${\mu }_1={\mu }_2={\mu }_3=60$ and ${\sigma }_1{}^2={\sigma }_2{}^2={\sigma }_3{}^2=15$.

Answer to Problem 59E

The value of $P\left(T_o\le 200\right)$ under the given condition is 0.9986.

The value of $P\left(150\le T_o\le 200\right)$ under the given condition is 0.9986.

Explanation of Solution

Given info:

The random variables $X_1$, $X_2$ and $X_3$ denote the successive times required for repair tasks at a certain facility, having independent normal distributions with respective means ${\mu }_1$, ${\mu }_2$, ${\mu }_3$ and respective variances ${\sigma }_1{}^2$, ${\sigma }_2{}^2$, ${\sigma }_3{}^2$.

Calculation:

The variable $T_o$ denotes the total time for repairing, that is, $T_o=X_1+X_2+X_3$. Being the sum of independent normal variables, $T_o$ has a normal distribution with parameters:

$\begin{array}{c}{\mu }_{T_o}={\mu }_1+{\mu }_2+{\mu }_3\\ =3\times 60\\ =180.\end{array}$

$\begin{array}{c}{\sigma }^2{}_{T_o}=V\left(X_1+X_2+X_3\right)\\ =V\left(X_1\right)+V\left(X_2\right)+V\left(X_3\right)\left(\text{due to independence of the variables}\right)\\ ={\sigma }^2{}_1+{\sigma }^2{}_2+{\sigma }^2{}_3\\ =3\times 15\end{array}$

$=45.$

The value of $P\left(T_o\le 200\right)$ under the given condition is:

$\begin{array}{c}P\left(T_o\le 200\right)=P\left(\frac{T_o-{\mu }_{T_o}}{{\sigma }_{T_o}}\le \frac{200-{\mu }_{T_o}}{{\sigma }_{T_o}}\right)\\ =P\left(Z\le \frac{200-180}{\sqrt{45}}\right)\\ =\Phi \left(2.98\right).\end{array}$

From Table A.3, “Standard Normal Curve Areas”, $\Phi \left(2.98\right)=0.9986$.

Thus, the value of $P\left(T_o\le 200\right)$ under the given condition is 0.9986.

The value of $P\left(150\le T_o\le 200\right)$ under the given condition is:

$\begin{array}{c}P\left(150\le T_o\le 200\right)=P\left(\frac{150-{\mu }_{T_o}}{{\sigma }_{T_o}}\le \frac{T_o-{\mu }_{T_o}}{{\sigma }_{T_o}}\le \frac{200-{\mu }_{T_o}}{{\sigma }_{T_o}}\right)\\ =P\left(\frac{150-180}{\sqrt{45}}\le Z\le \frac{200-180}{\sqrt{45}}\right)\\ =\Phi \left(2.98\right)-\Phi \left(-4.47\right).\end{array}$

It is known that $\Phi \left(-3.5\right)\approx 0$. Now, $-4.47<-3.35$. Thus, $\Phi \left(-4.47\right)=0$.

Thus, the value of $P\left(150\le T_o\le 200\right)$ under the given condition is:

$\begin{array}{c}P\left(150\le T_o\le 200\right)=0.9986-0\\ =\underset{\_}{0.9986}.\end{array}$

To determine

Calculate $P\left(55\le \overline{X}\right)$ and $P\left(58\le \overline{X}\le 62\right)$ when ${\mu }_1={\mu }_2={\mu }_3=60$ and ${\sigma }_1{}^2={\sigma }_2{}^2={\sigma }_3{}^2=15$.

Answer to Problem 59E

The value of $P\left(55\le \overline{X}\right)$ under the given condition is 0.9875.

The value of $P\left(58\le \overline{X}\le 62\right)$ under the given condition is 0.6266.

Explanation of Solution

Calculation:

The variable $\overline{X}$ denotes the average time for repairing, that is, $\overline{X}=\frac{X_1+X_2+X_3}{3}$. Being a linear combination of independent normal variables, $\overline{X}$ has a normal distribution with parameters:

$\begin{array}{c}{\mu }_{\overline{X}}=\frac{{\mu }_1+{\mu }_2+{\mu }_3}{3}\\ =\frac{3\times 60}{3}\\ =60.\end{array}$

$\begin{array}{c}{\sigma }^2{}_{\overline{X}}=V\left(\frac{X_1+X_2+X_3}{3}\right)\\ =\frac{1}{9}\left[V\left(X_1\right)+V\left(X_2\right)+V\left(X_3\right)\right]\left(\text{due to independence of the variables}\right)\\ =\frac{{\sigma }^2{}_1+{\sigma }^2{}_2+{\sigma }^2{}_3}{9}\\ =\frac{3\times 15}{9}\end{array}$

$=5.$

The value of $P\left(55\le \overline{X}\right)$ under the given condition is:

$\begin{array}{c}P\left(55\le \overline{X}\right)=P\left(\frac{\overline{X}-{\mu }_{\overline{X}}}{{\sigma }_{\overline{X}}}\ge \frac{55-{\mu }_{\overline{X}}}{{\sigma }_{\overline{X}}}\right)\\ =P\left(Z\ge \frac{55-60}{\sqrt{5}}\right)\\ =1-P\left(Z\le -2.24\right)\\ =1-\Phi \left(-2.24\right)\end{array}$

       $\begin{array}{l}=1-\left[1-\Phi \left(2.24\right)\right]\\ =\Phi \left(2.24\right).\end{array}$

From Table A.3, “Standard Normal Curve Areas”, $\Phi \left(2.24\right)=0.9875$.

Thus, the value of $P\left(55\le \overline{X}\right)$ under the given condition is 0.9875.

The value of $P\left(58\le \overline{X}\le 62\right)$ under the given condition is:

$\begin{array}{c}P\left(58\le \overline{X}\le 62\right)=P\left(\frac{58-{\mu }_{\overline{X}}}{{\sigma }_{\overline{X}}}\le \frac{\overline{X}-{\mu }_{\overline{X}}}{{\sigma }_{\overline{X}}}\le \frac{62-{\mu }_{\overline{X}}}{{\sigma }_{\overline{X}}}\right)\\ =P\left(\frac{58-60}{\sqrt{5}}\le Z\le \frac{62-60}{\sqrt{5}}\right)\\ =\Phi \left(0.89\right)-\Phi \left(-0.89\right)\\ =\Phi \left(0.89\right)-\left[1-\Phi \left(0.89\right)\right]\end{array}$

  $=2\Phi \left(0.89\right)-1.$

From Table A.3, “Standard Normal Curve Areas”, $\Phi \left(0.89\right)=0.8133$.

Thus,

$\begin{array}{c}P\left(58\le \overline{X}\le 62\right)=2\Phi \left(0.89\right)-1\\ =\left(2\times 0.8133\right)-1\\ =1.6266-1\\ =\underset{\_}{0.6266}.\end{array}$

To determine

Calculate $P\left(-10\le X_1-0.5X_2-0.5X_3\le 5\right)$ and interpret the same.

Answer to Problem 59E

The value of $P\left(-10\le X_1-0.5X_2-0.5X_3\le 5\right)$ is 0.8364.

Explanation of Solution

Calculation:

The variable $X_1-0.5X_2-0.5X_3$ is a linear combination of independent normal random variables. Thus, $X_1-0.5X_2-0.5X_3$ has a normal distribution with parameters:

$\begin{array}{c}{\mu }_{X_1-0.5X_2-0.5X_3}={\mu }_1-0.5{\mu }_2-0.5{\mu }_3\\ =60-\left(0.5\times 60\right)-\left(0.5\times 60\right)\\ =60-30-30\\ =0.\end{array}$

$\begin{array}{c}{\sigma }^2{}_{X_1-0.5X_2-0.5X_3}=V\left(X_1-0.5X_2-0.5X_3\right)\\ =V\left(X_1\right)+{\left(0.5\right)}^{2}V\left(X_2\right)+{\left(0.5\right)}^{2}V\left(X_3\right)\\ ={\sigma }^2{}_1+0.25{\sigma }^2{}_2+0.25{\sigma }^2{}_3\\ =15+\left(0.25\times 15\right)+\left(0.25\times 15\right)\end{array}$

$\begin{array}{l}=15+3.75+3.75\\ =\mathrm{22.5.}\end{array}$

The value of $P\left(-10\le X_1-0.5X_2-0.5X_3\le 5\right)$ under the given condition is:

$\begin{array}{c}P\left(-10\le X_1-0.5X_2-0.5X_3\le 5\right)\\ =P\left(-\frac{10-{\mu }_{X_1-0.5X_2-0.5X_3}}{{\sigma }_{X_1-0.5X_2-0.5X_3}}\le \frac{X_1-0.5X_2-0.5X_3-{\mu }_{X_1-0.5X_2-0.5X_3}}{{\sigma }_{X_1-0.5X_2-0.5X_3}}\le \frac{5-{\mu }_{X_1-0.5X_2-0.5X_3}}{{\sigma }_{X_1-0.5X_2-0.5X_3}}\right)\\ =P\left(\frac{-10-0}{\sqrt{22.5}}\le Z\le \frac{5-0}{\sqrt{22.5}}\right)\\ =\Phi \left(1.05\right)-\Phi \left(-2.13\right).\end{array}$

From Table A.3, “Standard Normal Curve Areas”, $\Phi \left(1.05\right)=0.8531$ and $\Phi \left(-2.13\right)=0.0166$.

Thus, the value of $P\left(-10\le X_1-0.5X_2-0.5X_3\le 5\right)$ under the given condition is:

$\begin{array}{c}P\left(-10\le X_1-0.5X_2-0.5X_3\le 5\right)=\Phi \left(1.05\right)-\Phi \left(-2.13\right)\\ =0.8531-0.0166\\ =\underset{\_}{0.8365}.\end{array}$

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.

To determine

Calculate $P\left(X_1+X_2+X_3\le 160\right)$ and $P\left(X_1+X_2\ge 2X_3\right)$ when ${\mu }_1=40$, ${\mu }_2=50$, ${\mu }_3=60$, ${\sigma }_1{}^2=10$, ${\sigma }_2{}^2=12$, ${\sigma }_3{}^2=14$.

Answer to Problem 59E

The value of $P\left(X_1+X_2+X_3\le 160\right)$ under the given condition is 0.9525.

The value of $P\left(X_1+X_2\ge 2X_3\right)$ under the given condition is 0.0003.

Explanation of Solution

Calculation:

The variable $X_1+X_2+X_3$ being a linear combination of independent normal variables, has a normal distribution with parameters:

$\begin{array}{c}{\mu }_{X_1+X_2+X_3}={\mu }_1+{\mu }_2+{\mu }_3\\ =40+50+60\\ =150.\end{array}$

$\begin{array}{c}{\sigma }^2{}_{X_1+X_2+X_3}=V\left(X_1+X_2+X_3\right)\\ =V\left(X_1\right)+V\left(X_2\right)+V\left(X_3\right)\left(\text{due to independence of the variables}\right)\\ ={\sigma }^2{}_1+{\sigma }^2{}_2+{\sigma }^2{}_3\\ =10+12+14\end{array}$

$=36.$

The value of $P\left(X_1+X_2+X_3\le 160\right)$$P\left(T_o\le 200\right)$ under the given condition is:

$\begin{array}{c}P\left(X_1+X_2+X_3\le 160\right)=P\left(\frac{X_1+X_2+X_3-{\mu }_{X_1+X_2+X_3}}{{\sigma }_{X_1+X_2+X_3}}\le \frac{160-{\mu }_{X_1+X_2+X_3}}{{\sigma }_{X_1+X_2+X_3}}\right)\\ =P\left(Z\le \frac{160-150}{\sqrt{36}}\right)\\ =\Phi \left(1.67\right).\end{array}$

From Table A.3, “Standard Normal Curve Areas”, $\Phi \left(1.67\right)=0.9525$.

Thus, the value of $P\left(X_1+X_2+X_3\le 160\right)$ under the given condition is 0.9525.

The value of $P\left(X_1+X_2\ge 2X_3\right)$ under the given condition is:

$P\left(X_1+X_2\ge 2X_3\right)=P\left(X_1+X_2-2X_3\ge 0\right)$.

The variable $X_1+X_2-2X_3$ being a linear combination of independent normal variables, has a normal distribution with parameters:

$\begin{array}{c}{\mu }_{X_1+X_2-2X_3}={\mu }_1+{\mu }_2-2{\mu }_3\\ =40+50-\left(2\times 60\right)\\ =40+50-120\\ =-30.\end{array}$

$\begin{array}{c}{\sigma }^2{}_{X_1+X_2-2X_3}=V\left(X_1+X_2-2X_3\right)\\ =V\left(X_1\right)+V\left(X_2\right)+{\left(2\right)}^{2}V\left(X_3\right)\left(\text{due to independence of the variables}\right)\\ ={\sigma }^2{}_1+{\sigma }^2{}_2+4{\sigma }^2{}_3\\ =10+12+\left(4\times 14\right)\end{array}$

$\begin{array}{l}=10+12+56\\ =78.\end{array}$

Thus,

$\begin{array}{c}P\left(X_1+X_2-2X_3\ge 0\right)=P\left(\frac{X_1+X_2-2X_3-{\mu }_{X_1+X_2-2X_3}}{{\sigma }_{X_1+X_2-2X_3}}\ge \frac{0-{\mu }_{X_1+X_2-2X_3}}{{\sigma }_{X_1+X_2-2X_3}}\right)\\ =1-P\left(Z\le \frac{0-\left(-30\right)}{\sqrt{78}}\right)\\ =1-\Phi \left(3.40\right).\end{array}$

From Table A.3, “Standard Normal Curve Areas”, $\Phi \left(3.40\right)=0.9997$.

Thus,

$\begin{array}{c}P\left(X_1+X_2\ge 2X_3\right)=1-\Phi \left(3.40\right)\\ =1-0.9997\\ =\underset{\_}{0.0003}.\end{array}$