Question 1168212: If the average price of a new one-family home is $246,300 with a standard deviation of $15,000, find the minimum and maximum prices of the houses that a contractor will build to satisfy the middle 88% of the market. Assume that the variable is normally distributed. Round z-value calculations to 2 decimal places and final answers to the nearest dollar.
Minimum Price
Maximum Price
Answer by CPhill(1987)   (Show Source):
You can put this solution on YOUR website! Let $X$ be the random variable representing the price of a new one-family home. We are given that $X$ follows a normal distribution with a mean ($\mu$) of $246,300 and a standard deviation ($\sigma$) of $15,000.
We want to find the minimum and maximum prices that satisfy the middle 88% of the market. This means that the area in the tails is $100\% - 88\% = 12\%$, with $6\%$ in each tail. We need to find the z-scores that correspond to the 6th percentile and the 94th percentile of the standard normal distribution.
Looking up the z-value for a cumulative probability of $0.06$ in a standard normal distribution table, we get approximately $z_1 = -1.55$.
Looking up the z-value for a cumulative probability of $0.94$ in a standard normal distribution table, we get approximately $z_2 = 1.55$.
Now, we use the z-score formula to convert these z-values back to the original price scale:
$z = \frac{x - \mu}{\sigma}$
For the minimum price ($x_{min}$), using $z_1 = -1.55$:
$-1.55 = \frac{x_{min} - 246,300}{15,000}$
$x_{min} - 246,300 = -1.55 \times 15,000$
$x_{min} - 246,300 = -23,250$
$x_{min} = 246,300 - 23,250$
$x_{min} = 223,050$
Rounding to the nearest dollar, the minimum price is $223,050$.
For the maximum price ($x_{max}$), using $z_2 = 1.55$:
$1.55 = \frac{x_{max} - 246,300}{15,000}$
$x_{max} - 246,300 = 1.55 \times 15,000$
$x_{max} - 246,300 = 23,250$
$x_{max} = 246,300 + 23,250$
$x_{max} = 269,550$
Rounding to the nearest dollar, the maximum price is $269,550$.
Final Answer: The final answer is $\boxed{Minimum Price: 223050, Maximum Price: 269550}$
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