Q.8.16 - A First Course in Probability

Question

A.J. has $20$jobs that she must do in sequence, with the times required to do each of these jobs being independent random variables with a mean of$50$ minutes and a standard deviation of$10$ minutes. M.J. has $20$jobs that he must do in sequence, with the times required to do each of these jobs being independent random variables with a mean of$52$ minutes and a standard deviation of $15$minutes.

$\left(a\right)$ Find the probability that A.J. finishes in less than $900$minutes.

$\left(b\right)$Find the probability that M.J. finishes in less than$900$ minutes.

$\left(c\right)$ Find the probability that A.J. finishes before M.J.

Step-by-Step Solution

Answer

Therefore,

$\left(a\right)$  $.0125$.

$\left(b\right) .0183$.

$\left(c\right) .6915$

 

Step 1 Given Information.

A.J. has$20$ jobs that she must do in sequence, with the times required to do each of these jobs being independent random variables with a mean of $50$minutes and a standard deviation of$10$ minutes. M.J. has $20$jobs that he must do in sequence, with the times required to do each of these jobs being independent random variables with a mean of $52$minutes and a standard deviation of$15$ minutes.

Step 2 Explanation.

Assume that$\mathbf{A}.\mathbf{J}.$ has$n_1=20$ jobs that she must do in sequence and let $X_i$represent the time (in minutes) required to do$i$ th job$i=1,\dots ,n_1$. It is given that these times are independent random variables with mean

${\mu }_{A.J.}=E\left[X_i\right]=50$

and variance

${\sigma }_{\text{A.J. }}^2$$=Var\left(X_i\right)={10}^2$

On the other side, assume that M.J. has$n_2=20$ jobs that he must do in sequence and let$Y_j$ represent the time (in minutes) required to do$j$ th job$j=1,\dots ,n_2$. It is given that these times are independent random variables with mean

${\mu }_{M.J.}=E\left[Y_j\right]=52$

and variance

${\sigma }_{M.J.}^2=Var\left(Y_j\right)={15}^2$

Step 3 Explanation.

Let $T_{A.J.}$. denote the total time for which A.J. finishes all $n_1$ jobs:

$T_{A.J.}=\sum _1^{20}{X}_i$

and let$T_{M.J.}$. denote the total time for which M.J. finishes all$n_2$ jobs:

$T_{M.J.}=\sum _1^{20}{Y}_j$

Then, since times are independent random variables, using the corresponding properties of expectation and variance we get:

$E\left[T_{\text{A.J. }.}\right]$$=20{\mu }_{\text{A.J. }}$$=1000,Var\left(T_{A.J.}\right)$$=20{\sigma }_{\text{A.J. }}^2=2000,$ 

$E\left[T_{M.J.]}\right]$$=20{\mu }_{M.J.}=1040,$ $Var\left(T_{M.J.}\right)=20{\sigma }_{M.J.}^2=4500.$

Step 4 part (a) Explanation.

$\left(a\right)$ The probability that A.J. finishes all$n_1$ jobs in less than $900$minutes is

$P\left\{T_{\text{A.J. }}\le 900\right\}$

To approximate this probability we use the central limit theorem:

$$P\{T_{\text{A.J. }.<900}<=P$\left\{\frac{T_{A . J .}-E\left[T_{\text {A.J. }}\right]}{\sqrt{\operatorname{Var}\left(T_{\text {A.J. }}\right)}}<\frac{900-E\left[T_{A . J .]}\right]}{\sqrt{\operatorname{Var}\left(T_{\text {A.J. } .}\right)}}\right\}\right.$ $=P\left\{\frac{T_{\text {A.J. }}-1000}{\sqrt{2000}}<\frac{900-1000}{\sqrt{2000}}\right\} \approx \Phi(-2.24) \stackrel{\Phi(-z)=1-\Phi(z)}{=} 1-\Phi(2.24)$ Table 5.1 (textbook, Chapter 5) $1-.9875=.0125 .$

(b) The probability that M.J. finishes all $n_{2}$ jobs in less than 900 minutes is

$$

P\left\{T_{M . J .} \leq 900\right\} .

$$

To approximate this probability we use the central limit theorem:

$$

P\left\{T_{M . J . J}<900\right\}=P\left\{\frac{T_{M . J .}-E\left[T_{M . J . J}\right]}{\sqrt{\operatorname{Var}\left(T_{M . J .}\right)}}<\frac{900-E\left[T_{M . J .}\right]}{\sqrt{\operatorname{Var}\left(T_{M . J . J}\right)}}\right\}

$$

$$

=P\left\{\frac{T_{M . J .}-1040}{\sqrt{4500}}<\frac{900-1040}{\sqrt{4500}}\right\} \approx \Phi(-2.09) \stackrel{\Phi(-z)=1-\Phi(z)}{=} 1-\Phi(2.09)

$$

Table 5.1 (textbook, Chapter 5) $1-.9817=. \mathbf{0 1 8 3}$.

Step 4 part (c) Explanation.

$\left(c\right)$ The probability that A.J. finishes all $n_1$ jobs before M.J. finishes all$n_2$ jobs is

$P\left\{T_{\text{A.J. }}<T_{M.J.}\right\}=P\left\{T_{A.J.}-T_{M.J.J.}<0\right\}. (*)$

Because of independence$T\_\{A . J .\}andT\_\{M . J . s\}$ we have:

$E\left[T_{A.J.}-T_{M.J.J}\right]=E\left[T_{A.J.}\right]-E\left[T_{M.J.}\right]=-40$

and

$Var\left(T_{A.J.}-T_{M.J.}\right)=Var\left(T_{A.J.}\right)+Var\left(T_{M.J.}\right)=6500.$

Finally, to approximate the desired probability$(*)$ we use the central limit theorem:

$P\left\{T_{\text{A.J. }}<T_{M.J.}\right\}=P\left\{T_{\text{A.J. }}-T_{M.J.}<0\right\}$ 

$=P\left\{\frac{\left(T_{\text{A.J. }}-T_{M.J.}\right)-E\left[T_{\text{A.J. }}-T_{M.J.}\right]}{\sqrt{Var\left(T_{A.J.}-T_{M.J.}\right)}}<\frac{0-E\left[T_{\text{A.J. }}-T_{M.J.J}\right]}{\sqrt{Var\left(T_{A.J.}-T_{M.J.}\right)}}\right\}$ 

$=P\left\{\frac{\left(T_{\text{A.J. }}-T_{M.J.}\right)-(-40)}{\sqrt{6500}}<\frac{40}{\sqrt{6500}}\right\}\approx \Phi (.5)$ 

$=.6915$ .