Question

a dealer finds that the average price of a previously owned vehicle is $8,256. he decides to sell to the middle 60% of the market in terms of price. find the maximum and minimum prices the dealer will sell. the standard deviation is $1,150 and the variable is normally distributed.

Explanation:

Step1: Determine the z - scores

The middle 60% of a normal distribution leaves 20% in the lower - tail and 20% in the upper - tail. Looking up in the standard normal distribution table, the z - score corresponding to the lower 20% (or 0.20) is approximately $z_1=- 0.84$ and the z - score corresponding to the upper 80% (1 - 0.20) is approximately $z_2 = 0.84$.

Step2: Use the z - score formula

The z - score formula is $z=\frac{x-\mu}{\sigma}$, where $\mu$ is the mean, $\sigma$ is the standard deviation, and $x$ is the value from the data set. We know that $\mu = 8256$ and $\sigma=1150$.
For the minimum value $x_1$:
\[

$$\begin{align*} -0.84&=\frac{x_1 - 8256}{1150}\\ x_1-8256&=-0.84\times1150\\ x_1-8256&=-966\\ x_1&=8256 - 966\\ x_1&=7290 \end{align*}$$

\]
For the maximum value $x_2$:
\[

$$\begin{align*} 0.84&=\frac{x_2 - 8256}{1150}\\ x_2-8256&=0.84\times1150\\ x_2-8256&=966\\ x_2&=8256 + 966\\ x_2&=9222 \end{align*}$$

\]

Answer:

The minimum price is $\$7290$ and the maximum price is $\$9222$.