A Civil engineer commutes daily from his suburban home to his midtown office. The average time for a one-way trip is 25 minutes, with a standard deviation of 4.2 minutes. Assume the distribution of trip times to be normally distributed. a) What is the probability that a trip will take at least 35 minutes? (0.5 points) b) If the office opens at 9:00 A.M. and the engineer leaves his house at 8:40 A.M. daily, what percentage of the time is he late for work? (0.5 points) c) If he leaves the house at 8:32 A.M. and coffee is served at the office from 8:50 A.M. until 9:00 A.M., what is the probability that

The time it takes a civil engineer to commute to work is a random variable with normal distribution and mean μ= 25 min and standard deviation δ= 4.2 min.

To calculate the probabilities you have to standardize the values of X using Z= (X-μ)/δ~N(0;1) and then look for the corresponding probabilities in the Z-tables.

a)

What is the probability that a trip will take at least 35 minutes?

Symbolically:

P(X≥35) = 1 - P(X<35)= 1 - P(Z<(35-25)/4.2)= 1 - P(Z<2.86)= 1 - 0.99788= 0.00212

The probability of the trip taking at least 35 min is 0.00212.

b)

If the office opens at 9:00 A.M. and the engineer leaves his house at 8:40 A.M. daily, what percentage of the time is he late for work?

If he needs to be at work at nine, this means that whenever the trip takes more than 20 minutes, symbolically:

P(X≥20)= 1- P(X<20)= 1 - P(Z<(20-25)/4.2)= 1 - P(Z<-1.19)= 1 - 0.11702= 0.88298

If he leaves at 8:40 AM daily, he will be late 88.298% of the time.

c.

If he leaves the house at 8:32 A.M. and coffee is served at the office from 8:50 A.M. until 9:00 A.M., what is the probability that he misses coffee?

If coffee is only served from 8:50 to 9:00 A.M., the engineer has to arrive after 9:00 A.M. If he leaves at 8:32, then the trip has to take more than 32 minutes for him to miss coffee:

P(X>32)= 1 - P(X≤32)= 1 - P(Z≤(32-25)/4.2)= 1 - P(Z≤1.67)= 1 - 0.95254= 0.04746

If you wanted, for example, to know what is the probability of him arriving in time to make coffee, with the same starting time, this means that his trip has to take between 18 min (8:50 - 8: 32) and 32 min, symbolically:

P(18≤X≤32)= P(X≤32) - P(X≤18)

P(Z≤((32-25)/4.2) - P(Z≤18-25)/4.2)

P(Z≤1.67) - P(Z≤-1.67)= 0.95254 - 0.04746= 0.90508

d.

Find the length of time above which we find the slowest 15% of the trips.

You need to find a value x₀ that marks 15% of the lowest trip time, symbolically:

P(X≤x₀)= 0.15

For this, you have to look for the probability on the Z-table and see which is the corresponding Z-value z₀= -1.04.

Then using the standard normal distribution you clear the value of x₀:

z₀= (x₀-μ)/δ

x₀= (z₀*δ)+μ

x₀= (-1.04*4.2)+25= 20.632 min

I hope you have a SUPER day!