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: Identify the mean and standard deviation

The mean (\(\mu\)) is \(\$15,000\) and the standard deviation (\(\sigma\)) is \(\$500\).

Step2: Determine the number of standard deviations from the mean

For \(\$14,500\):
\(z = \frac{14500 - 15000}{500} = \frac{-500}{500} = -1\)
For \(\$16,000\):
\(z = \frac{16000 - 15000}{500} = \frac{1000}{500} = 2\)

Step3: Apply the 68-95-99.7 Rule

• The 68-95-99.7 Rule states that:
• Approximately 68% of data lies within \(\mu \pm \sigma\) (i.e., between \(-1\sigma\) and \(+1\sigma\)).
• Approximately 95% of data lies within \(\mu \pm 2\sigma\) (i.e., between \(-2\sigma\) and \(+2\sigma\)).
• Approximately 99.7% of data lies within \(\mu \pm 3\sigma\) (i.e., between \(-3\sigma\) and \(+3\sigma\)).

We need the percentage between \(z = -1\) and \(z = 2\).

First, find the percentage between \(z = -1\) and \(z = 0\) (mean):
Since the normal distribution is symmetric, the percentage between \(z = -1\) and \(z = 0\) is half of the percentage between \(z = -1\) and \(z = +1\). So, \(\frac{68\%}{2} = 34\%\).

Next, find the percentage between \(z = 0\) and \(z = 2\):
The percentage between \(z = -2\) and \(z = +2\) is 95%, so the percentage between \(z = 0\) and \(z = 2\) is half of that, which is \(\frac{95\%}{2} = 47.5\%\).

Step4: Sum the two percentages

To find the percentage between \(z = -1\) and \(z = 2\), we add the percentage between \(z = -1\) and \(z = 0\) (34%) and the percentage between \(z = 0\) and \(z = 2\) (47.5%):
\(34\% + 47.5\% = 81.5\%\)

Answer:

\(81.5\%\)