Question

the figure illustrates a normal distribution for the prices paid for a particular model of a new car. the mean is $14,000 and the standard deviation is $500. use the 68 - 95 - 99.7 rule to find the percentage of buyers who paid between $12,500 and $14,000. what percentage of buyers paid between $12,500 and $14,000? %

Explanation:

Step1: Determine number of standard - deviations from the mean

The mean $\mu = 14000$ and the standard deviation $\sigma=500$. The value $x = 12500$. Calculate the z - score $z=\frac{x - \mu}{\sigma}=\frac{12500 - 14000}{500}=\frac{- 1500}{500}=-3$. The value $14000$ is the mean, so its z - score is $z = 0$.

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. Since the normal distribution is symmetric, the percentage of data between $z=-3$ and $z = 0$ is $\frac{99.7\%}{2}=49.85\%$.

Answer:

49.85