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The mean annual income for adult women in one city is $28,520 and the standard deviation of the incomes is $5100. The distribution of incomes is skewed to the right. Determine the sampling distribution of the mean for samples of size 71.

Respuesta :

Answer:

sampling distribution is 605

Explanation:

given data

mean = 28520

SD = 5100

sample n = 71

solution

here [tex]\mu[/tex] = 28520

and sampling distribution will be

sampling distribution = [tex]\frac{\sigma }{\sqrt{n}}[/tex]   ..................1

put here value

sampling distribution = [tex]\frac{5100 }{\sqrt{71}}[/tex]  

sampling distribution = 605.25 = 605

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