Question

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%

Answer 1

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

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 = (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 = (\sigma)/(√(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{(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{(p(1-p))/(n)} = \sqrt{(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 = (X - \mu)/(\sigma)`

By the Central Limit Theorem

`Z = (X - \mu)/(s)`

`Z = (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%.