in 2000, the averages change 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.
as always in a normal distribution, 50% are above the mean, regardless of std.
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 is greater than $95.
To do this, we can use the standard normal distribution table or calculate the z-score and use a z-table to find the proportion.
Here's how you can calculate the z-score and find the proportion:
Step 1: Calculate the z-score
The z-score formula is given by: z = (x - μ) / σ
Where:
- x is the given value (in this case $95)
- μ is the mean (average) of the distribution (in this case $95)
- σ is the standard deviation (in this case $10)
Using the formula, we can calculate the z-score as follows:
z = ($95 - $95) / $10 = 0
Step 2: Use the z-table to find the proportion
The z-table provides the cumulative probabilities for different z-scores. We want to find the proportion of values greater than the z-score of 0.
Looking up the z-score of 0 in the z-table, we can find that the proportion to the left of 0 is 0.5000. However, since we're interested in values greater than 0, we need to subtract this proportion from 1.
So, the proportion of tax preparation fees that were more than $95 is 1 - 0.5000 = 0.5000 or 50%.
Therefore, 50% of tax preparation fees were more than $95 in the year 2000.