Can any body help with this problem?
Given:
Suppose a random sample of size 36 is drawn from a population with a mean of 278. If 86% of the time the sample mean is less than 280, what is the population Std Deviation?
Can any body help with this problem?
Given:
Suppose a random sample of size 36 is drawn from a population with a mean of 278. If 86% of the time the sample mean is less than 280, what is the population Std Deviation?
Here you are meant to assume that sample means of samples of size $\displaystyle 36$ can be treated as though it were normally distributed. $\displaystyle 86\%$ of the normal distribution falls below a z-score of $\displaystyle 1.08$.
So for the sample mean you have its mean is $\displaystyle 278$, and its sd is $\displaystyle 2/1.08$. Now the sample means sd is $\displaystyle \sigma/\sqrt{n}$ , where $\displaystyle \sigma$ is the population sd and $\displaystyle n$ is the sample size.
RonL
Well here is a possible solution
If 86% of the time the sample means fall below 280 the Z scores could be found by 86% -50% ( which the area under the curve, the right half) so we get 0.36 look up for the 0.36 on the Ztable the value associated is 1.11 now simply put this into the z formula to determine standard deviation ( the values we have are Z=1.11, sample mean = 278, possible population mean 280, n=36 the solution should be 11.1 approx) I hope this is correct but i am not sure.