If a distribution is mound-shaped and symmetric, what percent of homes will have a monthly bill of more than $115? The mean monthly bill is $125 and has a standard deviation of $10.
Well, $115 is exactly one SD below the mean, so that's handy.
A useful guideline is that, moving left to right, over 2SD covers the first 2%; 2-to-1 SD is another 14%, bringing you up to 16%, and 1-to-0 SD covers another 33%, bringing you to the mean at 50% (not quite, because I've omitted the decimals, but the 1/3 and 1/6 rules are easy to remember.)
So about 16% will be below 1SD below the mean, which is about 16% of homes having a monthly bill of _less_ that $115.
If you want more accuracy, Google yourself a z-score table and read off the exact percentile.
To find the percent of homes that will have a monthly bill of more than $115 in a mound-shaped and symmetric distribution, we can use the concept of standard deviation.
First, let's calculate the z-score for $115. The z-score measures the number of standard deviations above or below the mean a particular value is.
The formula for calculating the z-score is:
z = (x - μ) / σ
Where:
x = the value we want to find the z-score for
μ = the mean of the distribution
σ = the standard deviation of the distribution
In this case, x = $115, μ = $125, and σ = $10.
Plugging these values into the formula, we get:
z = (115 - 125) / 10
z = -10 / 10
z = -1
Now that we have the z-score, we can use a standard normal distribution table (also known as the Z-table) to find the proportion of values that are above the z-score of -1. The standard normal distribution table provides the area to the left of the z-score, so we'll need to subtract that value from 1 to get the area to the right (i.e., the proportion of values above the z-score).
Looking up -1 in the standard normal distribution table, we find that the area to the left of -1 is approximately 0.1587. So the area to the right (above -1) is 1 - 0.1587 = 0.8413.
Therefore, approximately 84.13% of homes will have a monthly bill of more than $115 in this scenario.