Answer to Question #327124 in Statistics and Probability for Lily

Question

John's commute from home to work is normally distributed with mean 45 minutes, deviation

standard 3 minutes.

a) What is the probability that John travels from home to work in 30 to 40 minutes?

b) If the starting time at the company is 9:00 am, what time should John start going from home?

so that the probability of getting to work on time is 95%?

Answer 1

$a:\\P\left( 30<X<40 \right) =P\left( \frac{30-45}{3}<\frac{X-45}{3}<\frac{40-45}{3} \right) =\\=P\left( -5<Z<-1.666667 \right) =\varPhi \left( -1.666667 \right) -\varPhi \left( -5 \right) =\\=0.04779-2.87\cdot 10^{-7}=0.0477897\\b:\\P\left( X<C \right) =0.95\Rightarrow P\left( \frac{X-45}{3}<\frac{C-45}{3} \right) =0.95\Rightarrow \\\Rightarrow \varPhi \left( \frac{C-45}{3} \right) =0.95\Rightarrow \frac{C-45}{3}=z_{0.95}\Rightarrow \\\Rightarrow C=45+3z_{0.95}=45+3\cdot 1.6449=49.9347\\He\,\,should\,\,take\,\,50 \min utes, so\,\,he\,\,should\,\,start\,\,going\,\,from\,\,home\,\,at\,\,08:10$