Question

Help please

This 1 pt 31 of 40 (0 complede) The10 scoes of auts ae maly dstibuted with a mean 100 and astandd deviation of 15 I a grop of 64 akults is randomly selected, what is the probability ha ther mean 1Q) will be at last 95

Answer 1

Here' the answer to the question. Please don't hesitate to give a "thumbs up" in case you're satisfied with the answer

We have been given normal distribution parameters:

Mean = 100
stdev = 15
n = 64
X = 95

So,
P(X>95)
= P(Z> (x-Mu)/(stdev/sqrt(n))
= P(Z> (95-100)/(15/sqrt(64))
= P(Z> -5*8/15)
= P(Z> -40/15)
= P(Z> -2.67) [we use the Z tables here to convert Z value to probability]
= 1-0.0038
= .9962

A is correct