Question

the figure illustrates a normal distribution for the prices paid for a particular model of a new car. the mean is $16,000 and the standard deviation is $500. use the 68 - 95 - 99.7 rule to find the percentage of buyers who paid between $14,500 and $16,000. what percentage of buyers paid between $14,500 and $16,000?

Explanation:

Step1: Calculate number of standard - deviations

The mean $\mu = 16000$ and the standard deviation $\sigma=500$. For $x = 14500$, the z - score is $z=\frac{14500 - 16000}{500}=\frac{- 1500}{500}=-3$. The value $16000$ is the mean (z - score = 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%