Answer :
Answer:
the probability that the proportion of flops in a sample of 442 released films would differ from the population proportion by greater than 4% is 0.0042
Step-by-step explanation:
Given that :
A film distribution manager calculates that 9% of the films released are flops
Let p be the probability for the movies that were released are flops;
[tex]\mu_p = P = 0.9[/tex]
If the manager is right, what is the probability that the proportion of flops in a sample of 442 released films would differ from the population proportion by greater than 4%
now; we know that our sample size = 442
the standard deviation of the variance is [tex]\sigma_p= \sqrt{\dfrac{p(1-p)}{n}}[/tex]
[tex]\sigma_p= \sqrt{\dfrac{0.9(1-0.9)}{442}}[/tex]
[tex]\sigma_p= \sqrt{\dfrac{0.9(0.1)}{442}}[/tex]
[tex]\sigma_p= \sqrt{\dfrac{0.09}{442}}[/tex]
[tex]\sigma_p= \sqrt{2.0361991 \times 10^{-4}}[/tex]
[tex]\sigma _p = 0.014[/tex]
So; if the manager is right; the probability that the proportion of flops in a sample of 442 released films would differ from the population proportion by greater than 4% can be calculated as:
[tex]P(|p-P|>0.04)=1 -P(p-P|<0.04)[/tex]
[tex]P(|p-P|>0.04)=1 -P(-0.04 \leq p-P \leq 0.04)[/tex]
[tex]P(|p-P|>0.04)=1 -P( \dfrac{-0.04}{\sigma_p} \leq \dfrac{ p-P}{\sigma_p} \leq \dfrac{0.04}{\sigma_p})[/tex]
[tex]P(|p-P|>0.04)=1 -P( \dfrac{-0.04}{0.014} \leq Z\leq \dfrac{0.04}{0.014})[/tex]
[tex]P(|p-P|>0.04)=1 -P( -2.8571 \leq Z\leq 2.8571)[/tex]
[tex]P(|p-P|>0.04)=1 -[P(Z \leq 2.8571) -P (Z\leq -2.8571)[/tex]
[tex]P(|p-P|>0.04)=1 -(0.9979 -0.0021)[/tex]
[tex]P(|p-P|>0.04)=1 -0.9958[/tex]
[tex]\mathbf{P(|p-P|>0.04)=0.0042}[/tex]
∴
the probability that the proportion of flops in a sample of 442 released films would differ from the population proportion by greater than 4% is 0.0042