The selling prices of houses in a city during one year were normally ditributed with a mean of $259,000 and a standard deviation of $15,000. Which of the following is closest to the percentage of the homes that had a selling Price between $244,000 and $289,000?

a. 81.8%
b. 47.7%
c. 95.4%
d. 68.2%
Please help! So stuck..

To find the percentage of homes that had a selling price between $244,000 and $289,000, we can use the concept of the standard normal distribution.

First, we need to calculate the z-scores for both selling prices using the formula:
z = (x - μ) / σ

Where:
x = individual selling price
μ = mean selling price
σ = standard deviation

For $244,000:
z1 = (244,000 - 259,000) / 15,000
= -1,000 / 15,000
= -0.0667

For $289,000:
z2 = (289,000 - 259,000) / 15,000
= 30,000 / 15,000
= 2

Now, we can calculate the percentage using the z-table or a calculator that provides the cumulative probability.

Looking up the z-scores in the table:
For z1 = -0.0667, the cumulative probability is 0.4750 (approximately).
For z2 = 2, the cumulative probability is 0.9772 (approximately).

To find the percentage of values between these two z-scores, we subtract the cumulative probability of z2 from z1:
Percentage = 0.9772 - 0.4750
= 0.5022

To convert this to a percentage, we multiply by 100:
Percentage = 0.5022 * 100
= 50.22%

Therefore, the answer is closest to 50.22%, so the correct option is b. 47.7%.

To find the percentage of homes that had a selling price between $244,000 and $289,000, we need to calculate the z-scores for these two values and then find the area under the normal curve between those two z-scores.

The z-score formula is: z = (x - μ) / σ

where x is the value, μ is the mean, and σ is the standard deviation.

So, for $244,000:
z1 = (244000 - 259000) / 15000 = -1

And for $289,000:
z2 = (289000 - 259000) / 15000 = 2

Next, we need to find the area under the normal curve between these two z-scores. We can use a standard normal distribution table or a graphing calculator to find the corresponding probabilities.

Using a standard normal distribution table, we can find that the area to the left of z = -1 is approximately 0.1587, and the area to the left of z = 2 is approximately 0.9772.

To find the area between these two z-scores, we subtract the smaller area from the larger area:

0.9772 - 0.1587 = 0.8185

Since the question asks for the closest percentage, we can multiply this result by 100 to get the percentage:

0.8185 * 100 = 81.85%

Therefore, the closest answer is option a. 81.8%.

Note: In some cases, rounding errors may occur, so it's always a good practice to select the closest option in these types of questions.

Z = (score-mean)/SD

Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/probability related to the Z scores. Multiply by 100 for percentage.

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