# the normal curve

• May 19th 2010, 12:25 PM
statisticsStudent
the normal curve
I got halfway through this problem but am stumped on D and E.
The amount of active ingredient in a certain type of cold remedy tablet has mean 400.0 milligrams and standard deviation 4.5 milligrams. If a sample of 72 tablets is randomly selected, what is the probability that the mean amount of active ingredients in this sample is more than 401 milligrams?

A) Let x denote the amount of active ingredient in the cold remedy tablets. What’s known about the shape of the distribution of X?
It is a bell curve
B) Let X denote the mean amount of active ingredient in a sample of 72 of these cold remedy tablets. What is known about the shape of X-bar and why?
i. It is a normal distribution because the sample size, n, is greater than 30.
C) What is the mean of X-bar?
Mean of x-bar equals 400.0 because M of x-bar equals the M of x
D) What is the standard deviation of X-bar?
E) Solve the problem
• May 19th 2010, 02:46 PM
awkward
D) If the population has standard deviation $\sigma$, then the standard deviation of the mean of a sample of size n is $\sigma / \sqrt{n}$.
• May 19th 2010, 04:26 PM
statisticsStudent
Thanks, awkward.

I calculated the standard deviation as standard deviation of x-bar = standard deviation of x divided by the square root of n, or 4.5 divided by the square root of 72, and got .530

I also calculated the Probability by converting the 401 milligrams to a z-score (1.89) and the calculating the normalcdf on my calculator (1.89, 1000) to come up with .0293, or probability of about 3 %

Does this sound accurate?
• May 20th 2010, 02:57 PM
awkward
Correct!