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Answer: Mean = $36,000 and standard deviation = $1000.
Step-by-step explanation:
The mean and the standard deviation of the sampling distribution of the sample mean:
[tex]Mean=\mu\\\\Standard\ deviation = \dfrac{\sigma}{\sqrt{n}}[/tex] , where n =sample size.
Here, we are given
[tex]\mu=\$36000,\ \ \sigma=\$3,000[/tex]
n= 9
Then, the mean and the standard deviation of the sampling distribution of the sample mean:
[tex]Mean=\$36000,\ \ Standard\ deviation=\dfrac{3000}{\sqrt{9}}=\dfrac{3000}{3}=\$1000[/tex]
Hence, Mean = $36,000 and standard deviation = $1000.