Answer
$P(\bar{x} > 24.3) = 0.0082$
It is unusual for the mean to be greater than 24.3
Work Step by Step
n = 100
$\sigma$ = 1.25
$\mu$ =24
Want to find P($\bar{x}$ > 24.3):
i) Find the z score corresponding to 24.3
z = $\frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt n}}$
z = $\frac{24.3 - 24}{\frac{1.25}{\sqrt 100}}$
z = 2.4
ii) $P( z > 2.4) = 1 - P( z < 2.4)$
= 1 - 0.9918
= 0.0082
iii) Therefore $P(\bar{x} > 24.3) = 0.0082$
It is unusual for the mean to be greater than 24.3. We can infer this because the corresponding z-score is greater than 2.
