We have been given for a normal distribution the mean time it takes to walk to the bus stop is 8 minutes with a standard deviation of 2 minutes. And the mean time it takes for the bus to get to school is 20 minutes with a standard deviation of 4 minutes.
(a) Average time that it would take reach school can be obtained by adding the average times.
8+20 = 28 minutes.
(b) Standard deviation of the trip to school can be found as:
[tex]\sigma =\sqrt{2^{2}+4^{2}}=\sqrt{4+16}=\sqrt{20}=4.47[/tex]
Therefore, standard deviation of the entire trip is 4.47 minutes.
(c) Let us first find z score corresponding to 30 minutes.[tex]z=\frac{x-\mu }{\sigma }=\frac{30-28}{4.47}=0.447[/tex]
We need to find the probability such that [tex]P(x>30)=P(z>0.447)=0.67[/tex]
Therefore, the required probability is 0.67.
(d) If average time to walk to school is 10 minutes, then overall average time for the trip will be 10+20 = 30 minutes.
(e) Standard deviation won't change it will remain 4.47
(f) The new probability will be:
[tex]z=\frac{x-\mu }{\sigma }=\frac{30-90}{4.47}=0[/tex]
[tex]P(x>30)=P(z>0)=0.5[/tex]
Therefore, probability will be 0.50.
