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.