The average selling price of homes in a certain city is $356,300. Assume the variable is normally distributed with a standard deviation of $64,600. If 396 homes are for sale, how many homes will sell for more than $325,000? (Round up to the next whole number.)
I don't understand how to do this problem. I have a test this weekend and want to understand. It won't be the same question, but one similar. Thanks.
To solve this problem, we need to use the concept of the standard normal distribution.
1. Calculate the z-score for the given value of $325,000 using the formula:
z = (x - μ) / σ
where:
x = $325,000 (the value we want to find the probability for)
μ = $356,300 (the average selling price)
σ = $64,600 (the standard deviation)
Plugging in the values, we get:
z = (325000 - 356300) / 64600
z = -0.4841 (rounded to four decimal places)
2. Look up the cumulative probability corresponding to the calculated z-score in the standard normal distribution table or use a calculator. The cumulative probability represents the proportion of the distribution that is less than or equal to the given value (in this case, $325,000).
3. Once you have found the cumulative probability, subtract it from 1 to get the proportion of homes that will sell for more than $325,000.
4. Multiply the proportion by the total number of homes for sale (396) to get the estimated number of homes that will sell for more than $325,000. Round up to the next whole number.
Let's calculate this step by step:
1. Calculate the z-score:
z = (325000 - 356300) / 64600
z = -0.4841 (rounded to four decimal places)
2. Look up the cumulative probability for -0.4841 in the standard normal distribution table or use a calculator. The cumulative probability is 0.3148.
3. Subtract the cumulative probability from 1:
1 - 0.3148 = 0.6852
4. Multiply the proportion by the total number of homes for sale and round up to the next whole number:
0.6852 * 396 ≈ 271
Therefore, we can estimate that approximately 271 homes will sell for more than $325,000 in the given city.