## Elementary Statistics: Picturing the World (6th Edition)

$P(\bar{x} > 24.3) = 0.0082$ It is unusual for the mean to be greater than 24.3
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.