C.I. X bar from mu, sigma and n Suppose that the size of Wendy's single hamburger is normally distributed with mean 70 grams and standard deviation of 5 grams. If 4 hamburgers are selected at random, find the probability that the same mean is less than 77 grams. 
This section is on confidence intervals, and I know the following formulas. 
$$P[\bar X<{\frac{Z_{\alpha/2}\sigma_X}{n}+\mu_X}]$$
if the right hand equals 77, then $Z_{\alpha/2}$ = 5.6
But the answer for the probability is .9974. 
How exactly do I solve this problem? 
The formula for the confidence interval didn't help me with a solution either. Thanks for the help. 
 A: *

*The standard error ($\text{SE}$) of the sampling distribution is $\frac{\sigma}{\sqrt{n}}$. It is less than the population standard deviation $\sigma =5$, because taking the mean of a group smooths out extreme values.

*The sample mean is a consistent estimator. If they had asked you for the probability of the sample mean being less than $70$, the answer would have been $1/2$ - the lower one-half of the bell's curve.

*But the question is "less than $77$". So you have to find the cumulative probability for the quantile $\frac{77-\mu}{\text{SE}}$, corresponding to the normalized $77$ value.

Does it make sense? Yes, you don't even need a calculator to see that this is the case: The $\text{SE}=2.5$ grams, and the value you are interested in is $7$ grams above the mean. So how many standard errors is it above the mean? Almost $3$, right? What is the probability that if you sample a group of $4$ people and measure their height, their average will be shorter than an NBA player, whose height is $3$ standard deviations above the mean? Would you consider betting on it?
