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

$\begin{gathered} Z=\frac{x-\mu}{\sigma} \\ Z=\frac{119.8-95.3}{15.4}_{} \\ Z=\frac{24.5}{15.4} \\ Z\approx1.6 \end{gathered}$

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