Electricity bills in a certain city have mean $120.08. Assume the bills are normally distributed with standard deviation $14.40. Find the value that separates the lower 52% of the bills from the rest.
To find the value that separates the lower 52% of the bills from the rest, we need to use the concept of z-scores and the standard normal distribution.
1. Start by finding the z-score corresponding to the lower 52% of the bills. This can be done using the standard normal distribution table or a calculator. The lower 52% corresponds to the cumulative area of 0.52.
2. Subtracting this cumulative area from 1 gives us the upper cumulative area of 0.48. This is because the cumulative area of the entire normal distribution is always equal to 1.
3. Next, we need to find the z-score that corresponds to this upper cumulative area of 0.48. In other words, we're looking for the z-score that represents the proportion of values above the separating value. The inverse standard normal distribution function (also known as the z-table or calculator) can be used to find this z-score.
4. Once we have the z-score, we can use it to find the actual value that separates the lower 52% of the bills from the rest. We do this by using the formula:
Value = Mean + (z-score × Standard Deviation)
Let's carry out these steps using the given information:
1. Using the standard normal distribution table, the z-score corresponding to a cumulative area of 0.48 is approximately 0.06.
2. Now we can calculate the value that separates the lower 52% of the bills from the rest:
Value = Mean + (z-score × Standard Deviation)
= $120.08 + (0.06 × $14.40)
Calculating this equation, we find:
Value ≈ $120.08 + ($0.86)
Value ≈ $120.94
Therefore, the value that separates the lower 52% of the bills from the rest is approximately $120.94.
To find the value that separates the lower 52% of the bills from the rest, we need to find the z-score corresponding to the 52th percentile.
The z-score can be calculated using the formula:
z = (x - μ) / σ
Where:
x is the value we want to find (the value separating the lower 52% from the rest)
μ is the mean ($120.08)
σ is the standard deviation ($14.40)
To find the z-score corresponding to the 52th percentile, we need to find the z-value from the standard normal distribution table.
The z-value can be found using the inverse standard normal distribution function or by referring to the standard normal distribution table.
The 52th percentile corresponds to a z-value of approximately 0.062.
Now, let's solve for x using the z-score formula:
0.062 = (x - 120.08) / 14.40
Rearranging the equation to solve for x:
0.062 * 14.40 = x - 120.08
0.874 = x - 120.08
x = 0.874 + 120.08
x ≈ 120.954
Therefore, the value that separates the lower 52% of the bills from the rest is approximately $120.954.
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 (.52) and its Z score. Insert in above equation and solve for score.