A film distribution manager calculates that 7% of the films released are flops. If the manager is correct, what is the probability that the proportion of flops in a sample of 404 released films would be greater than 4%? Round your answer to four decimal places.

Respuesta :

Answer:

0.9909 = 99.09% probability that the proportion of flops in a sample of 404 released films would be greater than 4%

Step-by-step explanation:

I am going to use the binomial approximation to the normal to solve this question.

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

[tex]E(X) = np[/tex]

The standard deviation of the binomial distribution is:

[tex]\sqrt{V(X)} = \sqrt{np(1-p)}[/tex]

Normal probability distribution

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

When we are approximating a binomial distribution to a normal one, we have that [tex]\mu = E(X)[/tex], [tex]\sigma = \sqrt{V(X)}[/tex].

In this problem, we have that:

[tex]p = 0.07, n = 404[/tex]. So

[tex]\mu = E(X) = np = 404*0.07 = 28.28[/tex]

[tex]\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{404*0.07*0.93} = 5.13[/tex]

If the manager is correct, what is the probability that the proportion of flops in a sample of 404 released films would be greater than 4%?

This is the pvlaue of Z when X = 0.04*404 = 16.16. So

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{16.16 - 28.28}{5.13}[/tex]

[tex]Z = -2.36[/tex]

[tex]Z = -2.36[/tex] has a pvalue of 0.0091

1 - 0.0091 = 0.9909

0.9909 = 99.09% probability that the proportion of flops in a sample of 404 released films would be greater than 4%