Drivers pay an average of (mean $690) per year for automobile insurance the distribution of insurance payments is approximately normal with a standard deviation of 110 dollars. What proportion of drivers pay more than 100 dollars per year for insurance?
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Check for typo in your post. If there are no typos, here's how you would proceed.
Standardize:
P(X>100)
=1-P(X≤100)
=1-P(Z≤(100-690)/110)
=1-P(Z≤-5.36)
Look up a probability table for
P(Z≤-5.36) and calculate the answer accordingly.
To find the proportion of drivers who pay more than $100 per year for insurance, we need to calculate the area under the normal distribution curve to the right of $100.
First, let's standardize the value of $100 using the Z-score formula:
Z = (X - μ) / σ
Where:
X = $100 (the value we want to standardize)
μ = $690 (mean)
σ = $110 (standard deviation)
Z = (100 - 690) / 110
Z = -5.36
Now, we need to find the area to the right of Z = -5.36 on the standard normal distribution curve. This represents the proportion of drivers who pay more than $100 per year for insurance.
Using a standard normal distribution table or a calculator, we can find that the proportion of drivers who pay more than $100 per year for insurance is approximately 1.02 x 10^-7 (or 0.000000102).
Therefore, a very small proportion of drivers (about 0.0000102%) pay more than $100 per year for insurance.