Answer
$P(120 \leq x \leq 180) = 0.6625$
Work Step by Step
$\mu =138.5$, $\sigma = 29$
Want to find $P(120 \leq x \leq 180)$
i) Convert 120 to a z-score:
z = $\frac{x - \mu}{\sigma}$
z = $\frac{120 - 138.5}{29}$
z= -0.64
ii) Convert 180 to a z score:
z = $\frac{180 - 138.5}{29}$
z = 1.43
iii) $P(120 \leq x \leq 180) = P(-0.64 < z < 1.43)$
$P(-0.64 < z < 1.43) = P(x < 1.43) - P(x < -0.64) = 0.9236 - 0.2611 = 0.6625$
Therefore $P(120 \leq x \leq 180) = 0.6625$
