Answer:
9.34%
Step-by-step explanation:
p = 4%, or 0.04
n = Sample size = 667
u = Expected value = n * p = 667 * 0.04 = 26.68
SD = Standard deviation = [tex]\sqrt{np(1-p)} =\sqrt{667*0.04*(1-0.04)}[/tex] = 5.06
Now, the question is if the manager is correct, what is the probability that the proportion of flops in a sample of 667 released films would be greater than 5%?
This statement implies that the p-vlaue of Z when X = 5% * 667 = 33.35
Since,
Z = (X - u) / SD
We have;
Z = (33.35 - 26.68) / 5.06
Z = 1.32
From the Z-table, the p-value of 1.32 is 0.9066
1 - 0.9066 = 0.0934, or 9.34%
Therefore, the probability that the proportion of flops in a sample of 667 released films would be greater than 5% is 9.34%.
