A film distribution manager calculates that 9% of the films released are flops. If the manager is right, what is the probability that the

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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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Khoii Minh 4 years 2021-08-08T08:46:45+00:00 1 Answers 39 views 0

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  1. Nem
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    2021-08-08T08:47:49+00:00
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    Answer:

    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%.

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