Question 1210452
<pre> 
The number of treats Addie the cat beats per minute is normally distributed with
a mean of 400 treats and a standard deviation of 120 treats. Determine the
following: If 21 minutes are selected at random, what is the probability that
the mean number of treats consumed exceeded 440 treats?

Calculate the standard error of the mean First, we need to find the standard
error of the mean, which is the standard deviation of the sampling distribution.
The formula is 

{{{sigma[xbar]}}}{{{""=""}}}{{{sigma/sqrt(n)}}} where {{{sigma}}} is the population standard deviation and {{{n}}} is the sample size.

{{{sigma[xbar]}}}{{{""=""}}}{{{120/sqrt(21)}}}{{{""=""}}}{{{26.186}}}  approximately.

Calculate the Z-score
{{{Z}}}{{{""=""}}}{{{(xbar-mu)/sigma[xbar])}}},

where xbar is the sample mean, {{{mu}}} is the population mean, and {{{sigma[xbar]}}} is the standard error.  

{{{Z}}}{{{""=""}}}{{{(440-400)/26.186=40/26.186=1.527}}}, approximately

We find the probability that the mean number of treats exceeded 440, which
corresponds to the area to the right of the calculated z-score. Using a 
standard normal distribution table or calculator, the probability of a 
z-score being less than 1.527 is approximately 0.9366. Therefore, the
probability of it being greater than 1.527 is:

{{{P(Z>1.527)}}}{{{""=""}}}{{{1-P(Z<1.527)}}}{{{""=""}}}{{{1-0.9366=0.0634}}}

Answer: 0.0634

Edwin</pre>