Lamar and Hareesh run a two-person lawn-care service. They have been caring for Mr. Johnson’s very large lawn for several years, and they have found that the time $L$ it takes Lamar to mow the lawn on a randomly selected day is approximately Normally distributed with a mean of $105$ minutes and a standard deviation of $10$ minutes. The time $H$ it takes Hareesh to use the edger and string trimmer on a randomly selected day is approximately Normally distributed with a mean of $98$ minutes and a standard deviation of $15$ minutes. Assume that $L\mathrm{and}H$ are independent random variables. Find the probability that Lamar and Hareesh finish their jobs within $5$minutes of each other on a randomly selected day.

Given:

Mean,

${\mu }_L=105$minutes

${\mu }_H=98$minutes

Standard deviation,

${\sigma }_L=10$minutes

${\sigma }_H=15$minutes

Time taken, $x=L-H=-5\mathrm{or}5$(Within $5$minutes)

Because $\mathrm{L}\mathrm{and}\mathrm{H}$has a Normal distribution, their difference has Normal distribution as well.

The mean of the difference is now equal to the difference in the means of any two variables.

$$\begin{gathered} {\mu }_{L-H}={\mu }_L-{\mu }_H \\ =105-98 \\ =7\mathrm{minutes} \end{gathered}$$

The variance of the difference is equal to the sum of the variances of the random variables when they are independent.

role="math" localid="1654254992835" $$\begin{gathered} {{\sigma }^2}_{L-H}={{\sigma }^2}_L+{{\sigma }^2}_H \\ ={\left(10\right)}^2+({15)}^2 \\ =\mathbf{325}\mathrm{minutes} \end{gathered}$$

The standard deviation is equal to the square root of the variance, as we know.

$$\begin{gathered} {\sigma }_{L-H}=\sqrt{{{\sigma }^2}_{L-H}} \\ =\sqrt{325} \\ \approx \mathbf{18}\mathbf{.}\mathbf{0278} \end{gathered}$$

Calculate the $z-$score using the formula :

$$\begin{gathered} z=\frac{x-\mu }{\sigma } \\ =\frac{-5-7}{18.0278} \\ \approx -0.67 \end{gathered}$$

or

$$\begin{gathered} z=\frac{x-\mu }{\sigma } \\ =\frac{5-7}{18.0278} \\ \approx -0.11 \end{gathered}$$

To find the equivalent probability, use the normal probability table in the appendix. $P(z<-0.67)$is represented in the standard normal probability table by the row beginning with $-0.6$and the column beginning with $.07$.

$P(z<-0.11)$is represented in the standard normal probability table by the row beginning with $-0.1$and the column beginning with $.01.$

$$\begin{gathered} P(-5<L-H<5)=P(-0.67<z<-0.11) \\ =P(z-0.11)-P(z-0.67) \\ =0.4562-0.2514 \\ =\mathbf{}\mathbf{0}\mathbf{.}\mathbf{2048} \end{gathered}$$