The prices of a certain model of new homes are normally distributed with a mean of $150,000 and a standard deviation of $900. What percentage of homes exceed $152,700?
The correct answer and explanation is :
To determine what percentage of homes exceed \$152,700, we need to first standardize the value and then use the standard normal distribution to find the corresponding percentage.
Step 1: Standardize the Value
We can standardize the value of \$152,700 using the formula for the z-score:
$$
Z = \frac{X – \mu}{\sigma}
$$
Where:
- $X = 152,700$ (the value we are interested in),
- $\mu = 150,000$ (the mean price of homes),
- $\sigma = 900$ (the standard deviation of home prices).
Plugging the values into the formula:
$$
Z = \frac{152,700 – 150,000}{900} = \frac{2,700}{900} = 3
$$
So, the z-score for \$152,700 is 3.
Step 2: Find the Percentage from the Standard Normal Distribution
Now that we have the z-score, we can use a standard normal distribution table or a calculator to determine the cumulative probability for $Z = 3$. The cumulative probability for a z-score of 3 is approximately 0.99865, which means that about 99.865% of the home prices fall below \$152,700.
Step 3: Calculate the Percentage Above \$152,700
Since the cumulative probability tells us the percentage of homes that cost less than \$152,700, the percentage of homes that cost more than \$152,700 is the complement of this value:
$$
1 – 0.99865 = 0.00135
$$
Thus, the percentage of homes that exceed \$152,700 is approximately 0.135%.
Final Answer:
0.135% of homes exceed \$152,700.
Explanation:
This result shows that the distribution of home prices is tightly clustered around the mean, with most homes priced close to \$150,000. Only a very small percentage of homes have prices significantly higher than this value, such as those exceeding \$152,700. This illustrates the properties of a normal distribution, where extreme values (far from the mean) are rare.