Question

A study shows that annual premiums paid for car insurance are

normally distributed, and that people spend a mean of $1280
on car insurance per year with a standard deviation of $420.
What is the probability that someone will spend less than
$1700 or more than $2120 in a year?

Answer 1

To find the probability that someone will spend less than $1700 or more than $2120 on car insurance in a year, we need to calculate the z-scores for these values and then use the standard normal distribution table. The probability is approximately 0.9135, or 91.35%.

Step-by-step explanation:

To find the probability that someone will spend less than $1700 or more than $2120 on car insurance in a year, we need to calculate the z-scores for these values and then use the standard normal distribution table. The formula for calculating the z-score is z = (x - μ) / σ, where x is the value, μ is the mean, and σ is the standard deviation.

For $1700:

z = (1700 - 1280) / 420 = 1.2381

For $2120:

z = (2120 - 1280) / 420 = 2

Using the standard normal distribution table, we can find the probabilities associated with these z-scores:

For z = 1.2381, the probability is 0.8907

For z = 2, the probability is 0.9772

Since we want to find the probability of spending less than $1700 or more than $2120, we can add these probabilities together:

0.8907 + (1 - 0.9772) = 0.8907 + 0.0228 = 0.9135

Therefore, the probability that someone will spend less than $1700 or more than $2120 in a year is approximately 0.9135, or 91.35%.