in 2000 the average charge of tax preparation was $95. assuming a normal distribution and a standard deviation of $10. What proportion of tax preparation fees were more than $95 ?
exactly half.
It doesn't matter what the std is. In a normal distribution, 1/2 of the population is above or below the mean.
To find the proportion of tax preparation fees that were more than $95, we need to calculate the area under the normal distribution curve to the right of $95.
Given that the mean is $95 and the standard deviation is $10, we can use a z-score to calculate the proportion.
The z-score formula is: z = (x - μ) / σ
Where:
x is the value we want to find the proportion for (in this case, $95)
μ is the mean ($95)
σ is the standard deviation ($10)
Plugging in the values:
z = (95 - 95) / 10
z = 0 / 10
z = 0
The z-score is 0, which means that the value $95 corresponds to the mean of the distribution.
To find the proportion of tax preparation fees greater than $95, we need to find the area under the normal curve to the right of this z-score of 0.
The z-table provides the proportion or area under the curve for different z-scores. Since the z-score of 0 is in the center of the curve, we can assume that the proportion to the right is 0.5.
So, the proportion of tax preparation fees that were more than $95 is 0.5 or 50%.
To find the proportion of tax preparation fees that were more than $95, we need to calculate the area under the normal distribution curve that corresponds to fees that are greater than $95.
Here's how we can calculate it step-by-step:
Step 1: Convert the problem into a standard normal distribution.
In order to work with the standard normal distribution, we will convert the given information into the standard form by using the formula:
Z = (X - μ) / σ
Where:
Z is the standard score (the number of standard deviations away from the mean).
X is the value of the variable (in this case, the tax preparation fee).
μ is the mean (average charge of tax preparation).
σ is the standard deviation.
Plugging in the given values:
X = $95
μ = $95
σ = $10
Z = ($95 - $95) / $10
Z = 0
So, we have converted the original problem into a standard normal distribution problem with a mean (μ) of 0 and a standard deviation (σ) of 1.
Step 2: Find the proportion of fees greater than $95.
We want to find the area under the standard normal curve to the right of Z = 0, which represents the proportion of fees that are greater than $95.
To do this, we need to calculate the cumulative probability of the standard normal distribution, which is often denoted as P(Z > 0) or 1 - P(Z ≤ 0).
Using a standard normal distribution table, we can look up the cumulative probability for Z = 0, which is 0.5000. Subtracting this value from 1 gives us the probability of Z being greater than 0:
P(Z > 0) = 1 - P(Z ≤ 0) = 1 - 0.5000 = 0.5000
So, the proportion of tax preparation fees that were more than $95 is 0.5000, or 50%.
Therefore, we can conclude that approximately 50% of the tax preparation fees were more than $95.