Answer
a) Yes, $9.75$
b) $187.5$
c) $0.1401$, No
d) $14$ passengers
Work Step by Step
a) It should be corrected, because the finite correction factor can only be ignored if the population is more $300$, which is not true here.
The value is:
$\frac{40}{\sqrt{16}}(\frac{300−16}{300−1})^{0.5}=9.75$
b) We can find that the weight is:
$\frac{3000}{16}=187.$5
c) We use the z-score to find:
$z=\frac{187.5−177}{9.75}=1.08$
Hence, using the table of z-scores, we can find that the corresponding probability is $1−0.8599=0.1401$.
Because this is a low probability, it should be much lower if you want to be sure that the elevator is safe, because you want extremely low probabilities when safety is at risk.
d. The z-score should be at least $3.1$. Hence:
$\frac{3000}{177+(3.1)(9.75)}=14$
