Question

mean is 95.3 standard deviation of 15.4 find the probability that a randomly selected adult has an iIQ greater than 119.8

Answer 1

First, we find the z-score

`\begin{gathered} Z=(x-\mu)/(\sigma) \\ Z=(119.8-95.3)/(15.4)_{} \\ Z=(24.5)/(15.4) \\ Z\approx1.6 \end{gathered}`

Then, we find P(Z > 1.6). According to a z-score table, we have

`P(Z>1.6)=1-0.9452=0.0548`

Hence, the probability of an IQ greater than 119.8