A telephone exchange operator assumes that 7% of the phone calls are wrong numbers. If the operator is accurate, what is the probability

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A telephone exchange operator assumes that 7% of the phone calls are wrong numbers. If the operator is accurate, what is the probability that the proportion of wrong numbers in a sample of 459 phone calls would differ from the population proportion by more than 3%

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Philomena 3 years 2021-08-03T02:14:47+00:00 1 Answers 13 views 0

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  1. bonexptip
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    2021-08-03T02:15:59+00:00
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    Answer:

    0.0118 = 1.18% probability that the proportion of wrong numbers in a sample of 459 phone calls would differ from the population proportion by more 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}}

    A telephone exchange operator assumes that 7% of the phone calls are wrong numbers.

    This means that p = 0.07

    Sample of 459 phone calls:

    This means that n = 459

    Mean and standard deviation:

    \mu = p = 0.07

    s = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\sqrt{\frac{0.07*0.93}{459}}} = 0.0119

    What is the probability that the proportion of wrong numbers in a sample of 459 phone calls would differ from the population proportion by more than 3%?

    Proportion below 0.07 – 0.03 = 0.04 or above 0.07 + 0.03 = 0.1. Since the normal distribution is symmetric, these probabilities are the same, which means that we find one of them and multiply by 2.

    Probability the proportion is below 0.04.

    p-value of Z when X = 0.04. So

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

    By the Central Limit Theorem

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

    Z = \frac{0.04 - 0.07}{0.0119}

    Z = -2.52

    Z = -2.52 has a p-value of 0.0059

    2*0.0059 = 0.0118

    0.0118 = 1.18% probability that the proportion of wrong numbers in a sample of 459 phone calls would differ from the population proportion by more than 3%

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