Question

Carly commutes to work, and her commute time is dependent on the weather. When the weather is good, the distribution of her commute time is approximately normal with mean 20 minutes and standard deviation 2 minutes. When the weather is not good, the distribution of her commute times is approximately normal with mean 30 minutes and standard deviation 4 minutes. Suppose the probability that the weather will be good tomorrow is 0.9. What is the probability that Carly's commute time tomorrow will be greater than 25 minutes?

Answer 1

`0.950`

Explanation:

a) Probability that the commute time of Carly is greater than 25 minute when the weather is good

This is writtesn as
`P( x > A)`, where A represents a numerical value or a condition.

Substituting the value of A, we get -

`P( x > 25)`

Here mean
`= 20` and standard deviation
`= 2`

Now we will find the z value for give x value -

As we know -

`z = (x - u) / sigma`

Substituting the available values , we get -

`A =  (25-20) / 2 =  2.5 \\\\`

Now, probability that the commute time of Carly is greater than 25 minute when the weather is good is equal to probability value as compared to the corresponding z values

Thus, probability that the commute time of Carly is greater than 25 minute when the weather is good
`= 0.0062`

b) Probability that the commute time of Carly is greater than 25 minute when the weather is not good

`z = (x - u) / sigma`

Substituting the available values , we get -

`A =  (25-30) / 4 =  -1.25 \\\\`

Now, probability that the commute time of Carly is greater than 30 minute when the weather is good is equal to probability value as compared to the corresponding z values

Thus, probability that the commute time of Carly is greater than 25 minute when the weather is good
`= 0.0062`

`P(x > 25) \\= P( z > (25-30) / 4) \\= P(z > -1.25) \\= 0.8944`

Final Probability -

`P( x > 25) \\= P( x > 25, weather good) + P( x > 25 , weather not good) \\= P( x > 25) \\= (0.0062)(0.9) + (0.8944)(0.10) \\= 0.0950`