The percentage of vehicle mileage expectancies that are more than 188,600 miles is 93%
<h3>How to determine the percentage?</h3>
The given parameters are:
Mean = 169200
Standard deviation = 19400
x = 188600
Calculate the z value using:
So, we have:
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:
brainly.com/question/10093236
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