Question

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.

Answer 1

sampling distribution is 605

Step-by-step explanation:

given data

mean = 28520

SD = 5100

sample n = 71

solution

here
`\mu` = 28520

and sampling distribution will be

sampling distribution =
`(\sigma )/(√(n))` ..................1

put here value

sampling distribution =
`(5100 )/(√(71))`

sampling distribution = 605.25 = 605