Answer
$68\%$, or (H)
Work Step by Step
So, to solve this problem we need to figure out how many standard deviations away from the mean 120 and 160 are, i.e. find the z-scores.
$z = \frac{x-\mu}{\sigma}$, where $\mu$ = mean, $\sigma$ = standard deviation
$z(120) = \frac{120-140}{20} = -1$
$z(160) = \frac{160-140}{20} = 1$
So, we are between $z = -1$ and $z = 1$. I.e. this range is within one standard deviation of the mean.
Using the $68-95-97.5$ rule for 1, 2, and 3 standard deviations away from the mean, the probability is $68\%$.
