The probability that they finish within 10 minutes of each other here their times are independent
i.e P( -10 < X₂ - X₁ < 10) where X₂ - X₁ = D Normally distributed, is 0.78 ..
Let X₁ be a random variable as taylor cleans the exterior of each car.
mean , μ₁ = 20 ,
standard deviations, σ₁ = 4.8
let X₂ be a random variable as sam cleans the interior of each car.
mean, μ₂ = 18
standard deviations, σ₂ = 6.4
If the two random variables are normal, then their difference , X₂ - X₁ also be normal.
As two normal variables here are normally distributed, therefore the distribution of the difference between the two variables here is given as: X₂ - X₁ = X
with mean, μ = μ₂ - μ₁ = 2
standard deviations, σ = σ₁² + σ₂²
= 6.4² + 4.8² = 64
To any bivariate Normal distribution of (X₁,X₂) Thus the variable
Z = (X−μ)/σ = (X₂−X₁ −(μ₂ - μ₁))/(√σ₁² + σ₂²)
= X - 2 /√64 = (X - 2 )/8
The probability here is computed as:
P( -10 < X₂ - X₁ < 10) = approx P(∣D∣<10)
where D is difference between their finishing times. Converting it to a standard normal variable, we have here:
= P( (-10 - 2) / 8 < Z < (10 - 2) / 8)
= P( -1.5 < Z < 1)
= P(Z < 1 ) - P(Z < -1.5)
Getting it from standard normal tables,
= 0.8431 - 0.0668
= 0.7763
Therefore 0.7763 is the required probability here.
To learn more about Normal probability distribution, refer:
https://brainly.com/question/6476990
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Complete question:
Sam and Taylor own and operate a car wash service. Sam cleans the interior of each car and Taylor cleans the exterior. The time it takes Sam to finish the interior has a mean of 20 minutes with a standard deviation of 6.4 minutes. The time it takes Taylor to finish the exterior has a mean of 18 minutes with a standard deviation of 4.8 minutes. Both of their finishing times are approximately normally distributed.
Suppose we select a car at random, and define the random variable D as the difference between their finishing times. We can assume that their times are independent.
Find the probability that they finish within 10minutes of each other.
You may round your answer to two decimal places.
P\left(|D|<10\right)\approxP(∣D∣<10)≈P
