QuestionFind the percentage of buyers who paid between \$22,500 and \$24,500 using the normal distribution with mean \$22,500 and SD \$1000.

Studdy Solution

STEP 1

Assumptions1. The distribution of car prices is normal. . The mean price is \ $22,500.<br />3. The standard deviation is \$1,000.<br />4. We are using the$ 68-95-99.7 $Rule, which states that for a normal distribution, approximately$ 68\% $of data falls within one standard deviation of the mean,$ 95\% $falls within two standard deviations, and$ 99.7\%$ falls within three standard deviations.
5. We want to find the percentage of buyers who paid between \$22,500 and \$24,500.

STEP 2

First, we need to determine how many standard deviations away \$24,500 is from the mean (\$22,500). We can do this by subtracting the mean from \$24,500 and dividing by the standard deviation.
$Z = \frac{X - \mu}{\sigma}$

STEP 3

Now, plug in the given values for X (the price we're interested in), μ (the mean), and σ (the standard deviation) to calculate the Z score.
$Z = \frac{\$24,500 - \$22,500}{\$1,000}$

STEP 4

Calculate the Z score.
$Z = \frac{\$24,500 - \$22,500}{\$1,000} =2$

STEP 5

Now that we have the Z score, we can use the $68-95-99.7$ Rule to find the percentage of buyers who paid between \ $22,500 and \$24,500. According to the rule, approximately$ 95\%$ of data falls within two standard deviations of the mean. However, this includes both sides of the mean, so we need to divide this percentage by2 to get the percentage for one side of the mean.
$Percentage = \frac{95\%}{2}$

STEP 6

Calculate the percentage.
$Percentage = \frac{95\%}{2} =47.5\%$ So, approximately $47.5\%$ of buyers paid between \$22,500 and \$24,500.

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