The manager of a donut store believes that 35% of the customers are first-time customers. A random sample of 150 customers will be used to e

Question

The manager of a donut store believes that 35% of the customers are first-time customers. A random sample of 150 customers will be used to estimate the proportion of first-time customers. Assuming this belief is correct, what is the probability that the sample proportion will be between 0.2 and 0.4

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Mộc Miên 3 years 2021-08-06T04:25:29+00:00 1 Answers 7 views 0

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  1. kimcuc
    0
    2021-08-06T04:27:04+00:00
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    Answer:

    0.8996 = 89.96% probability that the sample proportion will be between 0.2 and 0.4

    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}}

    The manager of a donut store believes that 35% of the customers are first-time customers.

    This means that p = 0.35

    Sample of 150 customers

    This means that n = 150

    Mean and standard deviation:

    \mu = p = 0.35

    s = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.35*0.65}{150}} = 0.0389

    What is the probability that the sample proportion will be between 0.2 and 0.4?

    p-value of Z when X = 0.4 subtracted by the p-value of Z when X = 0.2.

    X = 0.4

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

    By the Central Limit Theorem

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

    Z = \frac{0.4 - 0.35}{0.0389}

    Z = 1.28

    Z = 1.28 has a p-value of 0.8997

    X = 0.2

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

    Z = \frac{0.2 - 0.35}{0.0389}

    Z = -3.85

    Z = -3.85 has a p-value of 0.0001

    0.8997 – 0.0001 = 0.8996

    0.8996 = 89.96% probability that the sample proportion will be between 0.2 and 0.4

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