Yard work 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 L amar to mow the law 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 elected day is approximately Normally distributed with a mean of 98 minutes and a standard deviation of 15 minutes. Assume that I, and H are independent random variables. F’ind the probability that Lamar and Hareesh finish their jobs within 5 minutes of each other on a randomly selected day
The Correct Answer and Explanation is:
To find the probability that Lamar and Hareesh finish within 5 minutes of each other, we are essentially calculating the probability that the absolute difference between their completion times is less than or equal to 5 minutes.
Let:
- L∼N(105,102)L \sim N(105, 10^2): Lamar’s mowing time is Normally distributed with mean 105 minutes and standard deviation 10 minutes.
- H∼N(98,152)H \sim N(98, 15^2): Hareesh’s edging time is Normally distributed with mean 98 minutes and standard deviation 15 minutes.
- Let D=L−HD = L – H: the difference in their times.
Step 1: Distribution of D=L−HD = L – H
Since LL and HH are independent and Normally distributed, the difference DD is also Normally distributed. The mean and variance of DD are:
- μD=μL−μH=105−98=7\mu_D = \mu_L – \mu_H = 105 – 98 = 7
- σD2=σL2+σH2=102+152=100+225=325\sigma_D^2 = \sigma_L^2 + \sigma_H^2 = 10^2 + 15^2 = 100 + 225 = 325
- σD=325≈18.03\sigma_D = \sqrt{325} \approx 18.03
So, D∼N(7,18.032)D \sim N(7, 18.03^2)
Step 2: Probability that Lamar and Hareesh finish within 5 minutes
We want: P(∣L−H∣≤5)=P(−5≤D≤5)P(|L – H| \leq 5) = P(-5 \leq D \leq 5)
Since D∼N(7,18.032)D \sim N(7, 18.03^2), we standardize this: Z=D−μDσD=D−718.03Z = \frac{D – \mu_D}{\sigma_D} = \frac{D – 7}{18.03}
Now convert the bounds:
- Lower bound: −5−718.03=−1218.03≈−0.665\frac{-5 – 7}{18.03} = \frac{-12}{18.03} \approx -0.665
- Upper bound: 5−718.03=−218.03≈−0.111\frac{5 – 7}{18.03} = \frac{-2}{18.03} \approx -0.111
Now find the probability: P(−0.665≤Z≤−0.111)P(-0.665 \leq Z \leq -0.111)
Using standard normal tables or a calculator:
- P(Z≤−0.111)≈0.4557P(Z \leq -0.111) \approx 0.4557
- P(Z≤−0.665)≈0.2524P(Z \leq -0.665) \approx 0.2524
P(−0.665≤Z≤−0.111)=0.4557−0.2524=0.2033P(-0.665 \leq Z \leq -0.111) = 0.4557 – 0.2524 = 0.2033
Final Answer:
0.2033 or 20.33%\boxed{0.2033} \text{ or } \boxed{20.33\%}
Explanation (Approx. 300 words)
In this problem, we are given two independent normally distributed variables representing the work times of Lamar and Hareesh. Lamar takes an average of 105 minutes to mow, with a standard deviation of 10 minutes. Hareesh takes an average of 98 minutes for edging, with a standard deviation of 15 minutes. Since both times are independent and Normally distributed, their difference (Lamar’s time minus Hareesh’s time) is also Normally distributed.
We define a new variable D=L−HD = L – H, where LL and HH are the times taken by Lamar and Hareesh, respectively. The mean of this difference is 105−98=7105 – 98 = 7 minutes, and the variance is the sum of the individual variances because the variables are independent: 102+152=32510^2 + 15^2 = 325, giving a standard deviation of 325≈18.03\sqrt{325} \approx 18.03.
We are interested in the probability that the absolute difference between their times is less than or equal to 5 minutes, which is written mathematically as P(∣L−H∣≤5)P(|L – H| \leq 5) or P(−5≤D≤5)P(-5 \leq D \leq 5). We then convert these bounds to standard normal zz-scores and use the standard normal distribution to find the area between them, which gives us the desired probability.
The final answer is approximately 0.2033, meaning there is about a 20.33% chance that Lamar and Hareesh finish their tasks within 5 minutes of each other on a randomly selected day.