For finding the probability of an individual data point, z = (x-u)/s

For computing the probability of a particular **sample mean**, the z score formula is slightly different:

z = (x - u) / (**s/sqrt(n)**)

where **s/sqrt(n) = standard error of the means** --> the standard deviation of sample means

z = (x - u) / (s/sqrt(n))

z = (50.7 - 53) / (24/sqrt(59))

= -2.3 / 3.125

= -0.736

Using the normal distribution table, find the probability that z **<** -0.736 since the question asks for the probability of getting a mean of < 50.7

you should get 23.1%