Question

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). the distributions are normal.
a. how long will it take (in minutes), on average, to get to school?
b. what is the standard deviation of the trip to school?
c. what is the probability that it will take longer than 30 minutes to get to school? due to a miscalculation, you realize it actually takes an average of 10 minutes to walk to the bus stop.
d. how long will it take (in minutes), on average, to get to school?
e. what is the standard deviation of the trip to school? f. what is the probability that it will take longer than 30 minutes to get to school

Answer 1

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:

$\sigma =\sqrt{2^{2}+4^{2}}=\sqrt{4+16}=\sqrt{20}=4.47$

Therefore, standard deviation of the entire trip is 4.47 minutes.

(c) Let us first find z score corresponding to 30 minutes.$z=\frac{x-\mu }{\sigma }=\frac{30-28}{4.47}=0.447$

We need to find the probability such that $P(x>30)=P(z>0.447)=0.67$

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:

$z=\frac{x-\mu }{\sigma }=\frac{30-90}{4.47}=0$

$P(x>30)=P(z>0)=0.5$

Therefore, probability will be 0.50.

Answer 2

A) 20 + 8 = 28 minutes
b) 4.47
c) 0.327
d) 30 minutes
e) the std deviation will not change