Suppose a large shipment of televisions contained 9% defectives. If a sample of size 393 is selected, what is the probability that the sa

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Suppose a large shipment of televisions contained 9% defectives. If a sample of size 393 is selected, what is the probability that the sample proportion will differ from the population proportion by less than 3%

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Ngọc Diệp 6 months 2021-08-03T10:26:34+00:00 1 Answers 5 views 0

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    2021-08-03T10:27:44+00:00

    Answer:

    0.9624 = 96.24% probability that the sample proportion will differ from the population proportion by less than 3%

    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}}

    Suppose a large shipment of televisions contained 9% defectives

    This means that p = 0.09

    Sample of size 393

    This means that n = 393

    Mean and standard deviation:

    \mu = p = 0.09

    s = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.09*0.91}{393}} = 0.0144

    What is the probability that the sample proportion will differ from the population proportion by less than 3%?

    Proportion between 0.09 – 0.03 = 0.06 and 0.09 + 0.03 = 0.12, which is the p-value of Z when X = 0.12 subtracted by the p-value of Z when X = 0.06.

    X = 0.12

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

    By the Central Limit Theorem

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

    Z = \frac{0.12 - 0.09}{0.0144}

    Z = 2.08

    Z = 2.08 has a p-value of 0.9812

    X = 0.06

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

    Z = \frac{0.06 - 0.09}{0.0144}

    Z = -2.08

    Z = -2.08 has a p-value of 0.0188

    0.9812 – 0.0188 = 0.9624

    0.9624 = 96.24% probability that the sample proportion will differ from the population proportion by less than 3%

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