Solution: There is nothing wrong with your calculation's. But the question wants to round the z score to two decimal places.
Let me show you how to get 86.21%.
We are given:
[tex]\mu=125,000[/tex]
[tex]\sigma=23,000[/tex]
We have to find [tex]P(x>100,000)[/tex]
Using the z-score formula, we have:
[tex]z=\frac{x-\mu}{\sigma}[/tex]
[tex]=\frac{100,000-125,000}{23,000}[/tex]
[tex]=-1.09[/tex] rounded to two decimal places
Now we have to find [tex]P(z>-1.09)[/tex]
Using the standard normal table, we have:
[tex]P(x>100,000)=P(z>-1.09) = 0.8621[/tex] rounded to 4 decimal places
[tex]=86.21\%[/tex]
Therefore, the percentage of homes in the county that are valued over $100,000 is 86.21%
