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

(a) Average time to get to school
Average time (minutes) = Summation of the two means = mean time to walk to bus stop + mean time for the bust to get to school = 8+20 = 28 minutes

(b) Standard deviation of the whole trip to school
Standard deviation for the whole trip = Sqrt (Summation of variances)

Variance = Standard deviation ^2
Therefore,
Standard deviation for the whole trip = Sqrt (2^2+4^2) = Sqrt (20) = 4.47 minutes

(c) Probability that it will take more than 30 minutes to get to school
P(x>30) = 1-P(x=30)
Z(x=30) = (mean-30)/SD = (28-30)/4.47 ≈ -0.45
Now, P(x=30) = P(Z=-0.45) = 0.3264
Therefore,
P(X>30) = 1-P(X=30) = 1-0.3264 = 0.6736 = 67.36%

With actual average time to walk to the bus stop being 10 minutes;

(d) Average time to get to school
Actual average time to get to school = 10+20 = 30 minutes

(e) Standard deviation to get to school
Actual standard deviation = Previous standard deviation = 4.47 minutes. This is due to the fact that there are no changes with individual standard deviations.

(f) Probability that it will take more than 30 minutes to get to school
Z(x=30) = (mean - 30)/Sd = (30-30)/4.47 = 0/4.47 = 0

From Z table, P(x=30) = 0.5
And therefore, P(x>30) = 1- P(X=30) = 1- P(Z=0.0) = 1-0.5 = 0.5 = 50%