All the houses in Stanton County have a mean size of 1,650 square feet with a standard deviation of 320. Assume the distribution of their size follows a normal distribution.
If you inspected the houses in Stanton County, about what percentage would be between 1,330 and 1,970 square feet?
a) 99.7%
b) 95%
c) 34%
d) 68%
To find the percentage of houses that would be between 1,330 and 1,970 square feet, we can use the concept of the standard normal distribution.
Step 1: Convert the values to z-scores.
To do this, we use the formula:
z = (x - μ) / σ
Where:
x = the value we want to convert to a z-score (in this case, 1,330 and 1,970)
μ = the mean of the distribution (1,650)
σ = the standard deviation of the distribution (320)
For 1,330:
z1 = (1,330 - 1,650) / 320
For 1,970:
z2 = (1,970 - 1,650) / 320
Step 2: Look up the corresponding area under the standard normal distribution.
We can use a standard normal distribution table or a calculator to find the area to the left of each z-score.
Using a standard normal distribution table, we find that the area to the left of z1 is approximately 0.1977 and the area to the left of z2 is approximately 0.7475.
Step 3: Calculate the percentage between the two z-scores.
To find the percentage between the two z-scores, we subtract the smaller area from the larger area and multiply by 100.
Percentage = (larger area - smaller area) * 100
Percentage = (0.7475 - 0.1977) * 100 = 54.98%
Therefore, the percentage of houses in Stanton County that would be between 1,330 and 1,970 square feet is approximately 54.98%.
None of the answer choices provided match the calculated value.
To determine the percentage of houses that fall between 1,330 and 1,970 square feet, we can utilize the properties of a normal distribution.
First, let's standardize the values using the formula:
Z = (X - μ) / σ
Where:
Z is the standard score (also known as z-score)
X is the value or observation we want to standardize
μ is the mean of the distribution
σ is the standard deviation of the distribution
In this case, the mean (μ) is 1,650 square feet, and the standard deviation (σ) is 320 square feet.
For 1,330 square feet:
Z1 = (1,330 - 1,650) / 320
For 1,970 square feet:
Z2 = (1,970 - 1,650) / 320
Next, we can use a standard normal distribution table (also known as the Z-table) to find the percentage of houses within these standard scores.
Looking up the Z-values in the Z-table:
For Z1, the Z-value is approximately -1
For Z2, the Z-value is approximately 1.03
To determine the percentage between these two values, we subtract the percentage corresponding to Z2 from the percentage corresponding to Z1. As we know, the standard normal distribution is symmetric, so we can find the percentage for either Z1 or Z2 and subtract it from 0.5.
The percentage corresponding to Z = -1 is approximately 0.1587 (found in the Z-table).
Therefore, the percentage corresponding to Z = 1.03 is also approximately 0.1587.
To calculate the percentage between these Z-values:
Percentage = 2 * (0.1587) = 0.3174
Converting this percentage to a percentage format gives us 31.74%.
So, the correct answer is: c) 34%.
To solve this problem, we used the concept of standardizing values using Z-scores and referencing the Z-table to find the corresponding percentages.
you can play around with Z table stuff at
https://davidmlane.com/hyperstat/z_table.html