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 mean 50 minutes and standard deviation 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 mean 52 minutes and standard deviation 15 minutes.

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

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

(c) Find the probability that A.J. finishes before M.J.

Step-by-Step Solution

Answer

$$\begin{gathered} P\left(A.J.<900\right) \approx 0.01267 \\ P\left(M.J.<900\right) \approx 0.01845 \\ P\left(A.J.>M.J.\right) \approx 0.4565 \end{gathered}$$

Given information

The summary for A.J. is

$n=20, {𝜇}_1=50 and {𝜎}_1=10$

The summary for M.J. is

$m=20, {𝜇}_2=52 and {𝜎}_2=15$

Part (a)

$$\begin{gathered} P\left(A.J.<900\right) = P\left(\frac{A.J.-20*50}{10\sqrt{20}}<\frac{900-20*50}{10\sqrt{20}}\right) \\ \approx P(Z<-\sqrt{5})\approx 0.01267 \end{gathered}$$

Part (b)

$$\begin{gathered} P\left(M.J.<900\right) = P\left(\frac{A.J.-20*52}{15\sqrt{20}}<\frac{900-20*52}{15\sqrt{20}}\right) \\ \approx P(Z<-2.087)\approx 0.01845 \end{gathered}$$

Step (c)

$$\begin{gathered} P(A.J.>M.J.)=P(A.J.-M.J.>0) \\ =P\left\{\frac{(A.J.-M.J.)-({𝜇}_1-{𝜇}_2)}{\sqrt{{𝜎}_1^2+{𝜎}_2^2}}>\frac{-({𝜇}_1-{𝜇}_2)}{\sqrt{{𝜎}_1^2+{𝜎}_2^2}}\right\} \\ =P\left(Z>\frac{-(50-52)}{\sqrt{{10}^2+{15}^2}}\right) \\ =P\left(Z>\frac{2}{\sqrt{335}}\right) \\ \Rightarrow P(A.J.>M.J.)\approx 0.4565 \end{gathered}$$