Question

Lamar and lawrence 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 it takes lamar to mow the lawn itself is approximately normally distributed with a mean of 105 minutes and a standard deviation of 10 minutes. meanwhile, the time it takes for lawrence to use the edger and string trimmer to attend to details is also normally distributed with a mean of 98 minutes and a standard deviation of 15 minutes. they prefer to finish their jobs within 5 minutes of each other. what is the probability that this happens, assuming their finish times are independent?

Answer 1

The difference of finish times has a mean of
  μ = 105 -98 = 7 . . . . minutes
and a variance of
  σ² = 10² +15² = 325 . . . . minutes²

Using a probability calculator, we find the probability to be
  p(-5 < x < 5) ≈ 0.2030

Answer 2

Answer:

0.2892

Step-by-step explanation:

Let X be the time for Lamar to finish and Y for Lawrence

Given that X is N(105, 10)

Y N(98,15)

consider the random variable x-y

Since x and y are independent we have

x-y is normal with mean = 105-98 =7

and variance of x-y = varx +vary

= 100+225 =325

STd dev = sq rt 325 =18.03

X-Y = U is N(7, 18.03)

Required probability

= P(|x-y|<5)=P(-5<U<5)

Convert to std normal variate Z

-5 becomes (-5-7)/18.03 =-0.67

5 becomes  (5-7)/18.03 =-0.11

Hence required prob

P(-0.67<Z<0.11) = 0.2454+0.0438

=0.2892