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:

Therefore, standard deviation of the entire trip is 4.47 minutes.
(c) Let us first find z score corresponding to 30 minutes.
We need to find the probability such that 
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:


Therefore, probability will be 0.50.
The shadow of the tree is about 17.955 feet long.
Answer:
The estimate is 
Step-by-step explanation:
From the question we are told that
The sample size is n = 522
The sample proportion of students would like to talk about school is 
Given that the confidence level is 90 % then the level of significance can be mathematically evaluated as



Next we obtain the critical value of
from the normal distribution table, the value is

Generally the margin of error can be mathematically represented as

=> 
=> 
Generally the estimate the proportion of all teenagers who want more family discussions about school at 90% confidence level is

substituting values
