SOLUTION: The mean concentration of carbon dioxide is 365 parts per million with a standard deviation of 100 parts per million. Take a sample from 25 cities. Describe the sampling distribu
Algebra ->
Probability-and-statistics
-> SOLUTION: The mean concentration of carbon dioxide is 365 parts per million with a standard deviation of 100 parts per million. Take a sample from 25 cities. Describe the sampling distribu
Log On
Question 377628: The mean concentration of carbon dioxide is 365 parts per million with a standard deviation of 100 parts per million. Take a sample from 25 cities. Describe the sampling distribution of the sample mean. Find the probability that the sample mean concentration will be above 385 parts per million.
So far, I am able to figure
(385-365)/(100/(sqrt 25)) = 0.0527
At this point, checking the Z table I get the value of 0.06031. I'm not sure where I go from here in finishing this problem. Any help would be greatly appreciated. Answer by ewatrrr(24785) (Show Source):
Hi
(385-365)/(100/(sqrt 25)) = 20/20 = 1 Note:
Z table or NORMSDIST(1)= .8413
1 - .8413 = .1587 Or 15.87% is the probability that the sample mean concentration will be above 385 parts per million.
