Question

A population is normally distributed with a mean of 100 and a standard deviation of 10, for samples of size 25, what is the probability of randomly sampling and getting a sample mean of 103 or more?

Answer 1

let us consider

r ~N(100, 102)

and we have given n = 25

we know that $x\sim N(\mu,\sigma^{2})$

then $\bar{x}\sim N(\mu,\frac{\sigma}{\sqrt{n}})$

P (
> 103
) =
(103 - 100) P(z > P10/V25

P (
> 103
) =
P(z 1.5)

P () = 0.0668

> 103