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A random sample of vehicle mileage expectancies has a sample mean of x¯=169,200 miles and sample standard deviation of s=19,400 miles. Use the Empirical Rule to estimate the percentage of vehicle mileage expectancies that are more than 188,600 miles. Round your answer to the nearest whole number (percent).

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MrRoyal

The percentage of vehicle mileage expectancies that are more than 188,600 miles is 93%

How to determine the percentage?

The given parameters are:

Mean = 169200

Standard deviation = 19400

x = 188600

Calculate the z value using:

[tex]z = \frac{x - \mu}{\sigma}[/tex]

So, we have:

[tex]z = \frac{188600 - 169200}{19400}[/tex]

Evaluate

z = 1

Using the Empirical Rule, we have:

P(x > 188600) = P(z > 1.5)

From z table of probability, we have:

P(x > 188600) = 0.93319

Express as percentage

P(x > 188600) = 93.319%

Approximate

P(x > 188600) = 93%

Hence, the percentage of vehicle mileage expectancies that are more than 188,600 miles is 93%

Read more about Empirical Rule at:

https://brainly.com/question/10093236

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