the selling prices for homes in a certain community are approximately normally distributed with a mean of…

Explanation:

Step1: Calculate z - scores

The formula for the z - score is $z=\frac{x-\mu}{\sigma}$, where $\mu = 321000$ is the mean and $\sigma=38000$ is the standard deviation.

Step2: For part (a)

For $x = 283000$, $z_1=\frac{283000 - 321000}{38000}=\frac{- 38000}{38000}=-1$. For $x = 359000$, $z_2=\frac{359000 - 321000}{38000}=\frac{38000}{38000}=1$. Using the empirical rule for a normal distribution, the percentage of data within $z=-1$ and $z = 1$ is approximately $68%$.

Step3: For part (b)

For $x = 397000$, $z=\frac{397000 - 321000}{38000}=\frac{76000}{38000}=2$. The percentage of data above $z = 2$ is $\frac{100 - 95}{2}=2.5%$ (since about $95%$ of the data is within $z=-2$ and $z = 2$ in a normal distribution).

Step4: For part (c)

Since for $x = 283000$, $z=-1$, and the percentage of data below $z=-1$ is $\frac{100 - 68}{2}=16%$ (because about $68%$ of the data is within $z=-1$ and $z = 1$).

Step5: For part (d)

For $x = 207000$, $z_1=\frac{207000 - 321000}{38000}=\frac{-114000}{38000}=-3$. For $x = 359000$, $z_2 = 1$. The percentage of data within $z=-3$ and $z = 1$: The percentage of data within $z=-3$ and $z = 3$ is about $99.7%$, and within $z=-1$ and $z = 1$ is $68%$. The percentage of data within $z=-3$ and $z=-1$ is $\frac{99.7 - 68}{2}=15.85%$, so the percentage within $z=-3$ and $z = 1$ is $68+15.85 = 83.85%$.

Answer:

(a) $68$ (b) $2.5$ (c) $16$ (d) $83.85$