The number of years of education of self‑employed individuals in the United States has a population mean of 13.6 years and a population stan

Question

The number of years of education of self‑employed individuals in the United States has a population mean of 13.6 years and a population standard deviation of 3 years. If we survey a random sample of 100 self‑employed people to determine the average number of years of education for the sample, what is the mean of the sampling distribution of ¯ x , the sample mean? 0.3 years 3 years 13.6 years

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Delilah 1 week 2021-10-08T05:33:04+00:00 1 Answer 0

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    2021-10-08T05:34:38+00:00

    Answer:

    Since the sample size is large enough n =100 >30. From the central limit theorem we know that the distribution for the sample mean \bar X is given by:

    \bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})

    And we know that:

     \mu_{\bar X} = 13.6

     \sigma_{\bar X} = \frac{3}{\sqrt{100}}=0.3

    Step-by-step explanation:

    Previous concepts

    Normal distribution, is a “probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean”.

    The Z-score is “a numerical measurement used in statistics of a value’s relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean”.  

    Solution to the problem

    Let X the random variable that represent the number of years of education of a population and we know the following parameters

    Where \mu=13.6 and \sigma=3

    The central limit theorem states that “if we have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large”.

    Since the sample size is large enough n =100 >30. From the central limit theorem we know that the distribution for the sample mean \bar X is given by:

    \bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})

    And we know that:

     \mu_{\bar X} = 13.6

     \sigma_{\bar X} = \frac{3}{\sqrt{100}}=0.3

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