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Assume the annual day care cost per child is normally distributed with a mean of $8000 and a standard deviation of $1000. What percent of day care costs are more than $7100 annually?Click here to view page 1 of the standard normal distribution tableClick here to view page 2 of the standard normal distribution table,% (Round to two decimal places as needed.)

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Answer:

81.59%

Explanation:

To find the percentage, we first need to standardize $7100. To standardize, we subtract the mean and then divide by the standard deviation. So, $7100 is equivalent to:

[tex]\begin{gathered} z=\frac{x-\operatorname{mean}}{std\text{ deviation}} \\ z=\frac{7100-8000}{1000}=-0.9 \end{gathered}[/tex]

Now, the percentage of day care costs that are more than $7100 is equivalent to the probability that z is greater than -0.9, so:

P(x > $7100) = P(z > -0.9)

Finally, we can use the standard normal distribution table to get:

P(z > -0.9) = 0.8159

So, the answer is 81.59%