Answer
a) Yes; 9.75
b) 187.5
c) .1401; No
d) 14 passangers
Work Step by Step
a) It should be corrected, for the finite correction factor can only be ignored if the population is above 300, which is not the case in this example. The value is:
$=\frac{40}{\sqrt{16}}(\frac{300-16}{300-1})^{.5}=9.75$
b) We find that the weight is:
$=\frac{3000}{16}=187.5$
c) We use the z-score to find:
$z=\frac{187.5-177}{9.75}=1.08$
Thus, using the table of z-scores, we find that this corresponds to a probability of $1-.8599=.1401$
While this is a low probability, it should be much lower if you want to be sure that the elevator is safe, for you want extremely low probabilities when safety is at risk.
d. The z-score would need to be at least 3.1. Thus, it follows:
$=\frac{3000}{177+(3.1)(9.75)}=14$
