In 2019, 15.9% of Broadway actors were acting in their first role on Broadway. Suppose we took a survey of 38 Broadway actors and found that

Question

In 2019, 15.9% of Broadway actors were acting in their first role on Broadway. Suppose we took a survey of 38 Broadway actors and found that 18.4% of the actors we surveyed were first-timers. What are the mean and standard deviation for the sampling distribution of pÌ p^? 1. Mean: 0.159, Standard Deviation: 0.0592. Mean: 0.159, Standard Deviation: 0.36573. Mean: 0.184, Standard Deviation: 0.0634. Mean: 0.184, Standard Deviation: 0.0595. Mean: 0.159, Standard Deviation: 0.063In 2019 the CDC reported that 14.0% of US adults are smokers. Suppose you take a random sample of 30 smokers and find that the proportion of them who are current smokers is 16.7%.What is the mean and the standard deviation of the sampling distribution of pÌ p^ ?1. mean = 0.140, standard deviation = 0.0682. mean = 0.167, standard deviation = 0.0633. mean = 0.140, standard deviation = 0.0634. mean = 0.167, standard deviation = 0.068

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Thiên Di 3 years 2021-08-05T07:27:27+00:00 1 Answers 41 views 0

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  1. ngockhue
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    2021-08-05T07:29:20+00:00
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    Answer:

    For the Broadway actors acting in their first role on Broadway, mean: 0.184, Standard Deviation: 0.063.

    For the proportion of smokers, mean = 0.167, standard deviation = 0.068

    Step-by-step explanation:

    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 we took a survey of 38 Broadway actors and found that 18.4% of the actors we surveyed were first-timers.

    This means that p = 0.184, n = 38

    What are the mean and standard deviation for the sampling distribution of p^?

    Mean:

    \mu = p = 0.184

    Standard deviation:

    s = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.184*0.816}{38}} = 0.063

    Suppose you take a random sample of 30 smokers and find that the proportion of them who are current smokers is 16.7%.

    This means that n = 30, p = 0.167

    What is the mean and the standard deviation of the sampling distribution of p^ ?

    Mean:

    \mu = p = 0.167

    Standard deviation:

    s = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.167*0.833}{30}} = 0.068

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