This is a question on a study guide for a test tomorrow. How do you do it? Please help
According to the Federal Highway Administration, teenage males drive an average of 10,718 miles each year. Assume the mileage totals are normally distributed with a standard deviation of 3573 miles. For a randomly selected teenage male, find the probability that he drives less than 9,000 miles in a year.
Z=(mean-x)/σ
=(10718-9000)/3573
=0.4808
From tables, probability for x≤9000
=0.5-0.184
=0.316
To find the probability that a randomly selected teenage male drives less than 9,000 miles in a year, we need to use the concept of standard normal distribution.
The first step is to standardize the given value using the formula:
z = (x - μ) / σ
where:
z is the standard score
x is the given value (9,000 miles in this case)
μ is the mean (10,718 miles)
σ is the standard deviation (3,573 miles)
Plugging in these values, we get:
z = (9,000 - 10,718) / 3,573
Calculating this gives us:
z = -0.4822
Once we have the standard score (z-value), we can use a Z-table or a calculator to find the probability corresponding to that value. The Z-table provides the cumulative probability up to a specific z-value.
Looking up a Z-table or using a calculator, we find that the probability of having a z-value of -0.4822 is approximately 0.3144.
Therefore, the probability that a randomly selected teenage male drives less than 9,000 miles in a year is approximately 0.3144, or 31.44%.