Thread: Finding Probability, which z-score should I use?

I am looking for someone to help me on a problem. Here it is:
1. A medication that helps reduce seizures in children is claimed to have an average of 25mg with a standard deviation of .5mg. In a sample of 35 pills we find that the average amount is 25.3mg.

a) Find the probability of observing an average of at least this much.

b) what does this say about the claim?


So for a, I want to use either the z-score with the standard deviation divided by the square root of n in the denominator, or just the standard deviation in the denom. Basicily, is the the average in a) a regular average or the average of the 35 pills?????

Using just the standard deviation in the denom of the z-score results in a probability of .2743 that it is at least this much. The other method gives 0%.

I am torn, which one is it? My instinct says it is .2743 is the right answer, but the more i read the problem I wonder if 0 is the answer to 1)a. But maybe I am overthinking things.

Opinions????

Thanks!!!

We know sample average is 25.3mg and . You need to standardize it by using the Central Limit Theory:

and P(Z>z*).

Originally Posted by statmajor

We know sample average is 25.3mg and . You need to standardize it by using the Central Limit Theory:

and P(Z>z*).

Alright, thank you!