Answer
No, it would not be unusual for a car to spend more than 3 minutes in Wendy’s drive through. This is because 3 minutes (180 seconds) is within two standard deviations of the mean (as can be seen by the z-score is 1.43). Furthermore, the probability of spending greater than 3 minutes in the drive-through is greater than 0.05, which is also an indication that it is not unusual for a car to spend more than 3 minutes in Wendy's drive-through.
Work Step by Step
$\mu$ = 138.5, $\sigma$ = 29
Want to find $P( x > 180 )$
i) Convert 180 to a z-score:
z = $\frac{x - \mu}{\sigma}$ = $\frac{180 - 138.5}{29}$ = 1.43
ii) $P(x > 180) = P(z > 1.43)$
$P(z > 1.43)$
$= 1 - P( z < 1.43)$
$= 1- 0.9236 $
$= 0.0764$
