Final answer:
To find the probability that an individual has auto insurance costs between $800 and $1200, we calculate the z-scores for these values and use the standard normal distribution to determine that there is approximately a 50.22% chance of an individual falling within this range.
Step-by-step explanation:
Probability of Auto Insurance Costs
The problem at hand concerns the mean annual cost of auto insurance, which is reported to be $1001, and follows a normal distribution with a standard deviation of $289. We want to find the probability that a randomly selected individual has an annual auto insurance cost between $800 and $1200.
Firstly, we convert the dollar amounts into z-scores using the formula: Z = (X - μ) / σ, where X is the value, μ is the mean, and σ is the standard deviation. So for $800, Z800 = (800 - 1001) / 289 = -0.696, and for $1200, Z1200 = (1200 - 1001) / 289 = 0.688.
Using the standard normal distribution table, we find the probability for the z-scores. The probability for Z being less than 0.688 is approximately 0.7454, and the probability for Z being less than -0.696 is approximately 0.2432. To find the probability of a cost between $800 and $1200, we take the difference: 0.7454 - 0.2432 = 0.5022, which is about 50.22%.
Thus, there is approximately a 50.22% chance that a randomly selected individual will have an annual auto insurance cost between $800 and $1200.