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The daily revenue at a university snack bar has been recorded for the past five years. Records indicate that the mean daily revenue is $1500 and the standard deviation is $500. The distribution is skewed to the right due to several high volume days (football game days). Suppose that 100 days are randomly selected and the average daily revenue is computed. The sampling distribution has a mean of $1,500 and a standard deviation of $____.

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Answer:

sample standard deviation is $50

Step-by-step explanation:

Given a  population standard deviation of $500 and a random sample of size, n=100.

-The sample standard deviation is calculated using the formula:

[tex]s=\sigma/\sqrt{n}[/tex]

Where:

  • [tex]s[/tex] is the sample standard deviation.
  • [tex]n[/tex] is the sample size.
  • [tex]\sigma[/tex] is the population standard deviation

Therefore:

[tex]s=\sigma/\sqrt{n}\\\\=500\sqrt{100}\\\\=50[/tex]

Hence, the sampling distribution has a mean of $1,500 and a standard deviation of $50

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