Answer:
Less than .0001
Step-by-step explanation:
The population and sample mean family income are the same: $34,000.
The sample standard deviation is given by:
[tex]SD=\frac{\sigma}{\sqrt n}\\SD=\frac{\$5,000}{\sqrt 200}\\SD=\$353.55[/tex]
For any give income, X, the correspondent z-score of the distribution is given by:
[tex]z=\frac{X-M}{SD}[/tex]
For X = $37,000:
[tex]z=\frac{37,000-34,000}{353.55}\\z= 8.485[/tex]
Traditional z-score tables present values up to 3.49 at the 99.98-th percentile. A z-score of 8.485, which is way above the upper represented limit, denotes an almost zero probability that the sample mean exceeds $37,000.
It is fair to say that the probability that the sample mean exceeds $37,000 is less than .0001
The mean income and standard deviation of $34,000 and $5,000, gives
the probability the sample mean exceeds $37,000 is approximately 0.
The given parameters are;
Mean family income = $34,000
The standard deviation = $5,000
The sample size = 200
Required:
The probability that the sample mean exceeds $37,000
Solution:
The z-score is found as follows;
[tex]z = \dfrac{37,000 - 34,000}{\dfrac{5,000}{\sqrt{200} } } \approx \mathbf{8.485}[/tex]
The value, 8.485 is larger than the largest value on the z-table which is
3.49, such that we have;
Learn more about the z-score table here:
https://brainly.com/question/6096474
