Question

the figure illustrates a normal distribution for the prices paid for a particular model of a new car. the mean is $15,000 and the standard deviation is $500. use the 68 - 95 - 99.7 rule to find the percentage of buyers who paid between $13,500 and $15,000. what percentage of buyers paid between $13,500 and $15,000?

Explanation:

Step1: Determine number of standard - deviations

First, find how many standard deviations $13500$ is from the mean. The formula for the z - score is $z=\frac{x - \mu}{\sigma}$, where $\mu$ is the mean, $\sigma$ is the standard deviation, and $x$ is the value. Given $\mu = 15000$, $\sigma=500$, and $x = 13500$. Then $z=\frac{13500 - 15000}{500}=\frac{- 1500}{500}=-3$.

Step2: Apply the 68 - 95 - 99.7 rule

The 68 - 95 - 99.7 rule states that about 99.7% of the data lies within 3 standard deviations of the mean, about 95% lies within 2 standard deviations, and about 68% lies within 1 standard deviation of the mean. The normal distribution is symmetric about the mean. The percentage of data between $z=-3$ and $z = 0$ is half of the percentage of data between $z=-3$ and $z = 3$. Since the percentage of data between $z=-3$ and $z = 3$ is 99.7%, the percentage of data between $z=-3$ and $z = 0$ is $\frac{99.7\%}{2}=49.85\%$.

Answer:

49.85