Question

the figure illustrates a normal distribution for the prices paid for a particular model of a new car. the mean is $21,000 and the standard deviation is $500. use the 68 - 95 - 99.7 rule to find the percentage of buyers who paid between $19,500 and $21,000. what percentage of buyers paid between $19,500 and $21,000? %

Explanation:

Step1: Calculate the number of standard deviations from the mean

The mean is $\mu = 21000$ and the standard deviation is $\sigma=500$.
For $x = 19500$, we calculate $z=\frac{\mu - x}{\sigma}=\frac{21000 - 19500}{500}=\frac{1500}{500}=3$.

Step2: Use the 68 - 95 - 99.7 rule

The 68 - 95 - 99.7 rule states that:

• Approximately 68% of the data lies within $1$ standard deviation of the mean ($\mu\pm\sigma$)
• Approximately 95% of the data lies within $2$ standard deviations of the mean ($\mu\pm2\sigma$)
• Approximately 99.7% of the data lies within $3$ standard deviations of the mean ($\mu\pm3\sigma$)

The interval from $\mu - 3\sigma$ to $\mu$ is half of the interval from $\mu - 3\sigma$ to $\mu+ 3\sigma$.
Since the data is symmetric about the mean in a normal distribution, the percentage of data from $\mu - 3\sigma$ to $\mu$ is $\frac{99.7\%}{2}=49.85\%$

Answer:

$49.85$