Daily water intake (including water used in drinks such as coffee, tea and juice) for Canadian adults follows a normal distribution with mean 1.5 litres and standard deviation 0.31 litres.

Question

(d) Using the 68-95-99.7 Rule, approximately 95% of samples of 22 Canadian adults will have mean daily water intakes between and .

(e) If we take a random sample of 11 Canadian adults, there is an 11% chance that their mean daily water intake will be greater than

I've tried these multiple times and seem to be missing a step, any help would be much appreciated.

for question D since 95% is 2 s.d away i multiplied 0.31 by 2 and added it to the mean then subtracted it from the mean to get two values on either side of the mean but i'm unsure how to use the n=22 information

for question E i attempted to approach it as a revers look up problem (turning 11% into 0.11 and looking up its Z score, then plugging it into the standardizing equation, but this isn't working.)

Answer 1

D) since this is average question with given sample size, you need to use (mean, sigma/root sample size).
This is normal distribution, so you don't need to worry about sample size being greater than 30.

Therefore, you should do = mean - 2 (0.31/squre root 22) and mean + 2(0.31/square root 22)
Then you will get the answer.