Question

What is the p(x>x')=0.24 when mean is 12 standard deviation is 4

Answer 1

Assuming
`X` is normally distributed, you have


`\mathbb P(X>x')=\mathbb P\left(\frac{X-12}4>\frac{x'-12}4\right)=\mathbb P\left(Z>\frac{x'-12}4\right)=0.24`

The right-tail probability of
`0.24` corresponds to a
`z` score of approximately
`0.7063`.

In terms of
`X`, this critical value is


`0.7063=\frac{x'-12}4\implies x'\approx14.83`