Mathematics, 30.06.2021 19:50 dsporski

A film distribution manager calculates that 9% of the films released are flops. If the manager is right, what is the probability that the proportion of flops in a sample of 469 released films would be greater than 6%

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Answer from: miklff

0.9884 = 98.84% probability that the proportion of flops in a sample of 469 released films would be greater than 6%.

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

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 p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean $\mu = p$ and standard deviation $s = \sqrt{\frac{p(1-p)}{n}}$

A film distribution manager calculates that 9% of the films released are flops.

This means that p = 0.09

Sample of 469

This means that n = 469

Mean and standard deviation:

$\mu = p = 0.09$

$s = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.09*0.91}{469}} = 0.0132$

What is the probability that the proportion of flops in a sample of 469 released films would be greater than 6%?

1 subtracted by the p-value of Z when X = 0.06. So

$Z = \frac{X - \mu}{\sigma}$

By the Central Limit Theorem

$Z = \frac{X - \mu}{s}$

$Z = \frac{0.06 - 0.09}{0.0132}$

Z = -2.27

Z = -2.27 has a p-value of 0.0116

1 - 0.0116 = 0.9884

0.9884 = 98.84% probability that the proportion of flops in a sample of 469 released films would be greater than 6%.

Answer from: Quest

the answers c

Answer from: Quest

i think is g(1)=-1.

hope that i can

Which statements about the function are true? check all that apply. g(0) = 0 (1) (-1) = 1 u 9(4) =

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