In 2000,the average charge of tax preparation was $95.Assuming a normal distibution and standard deviation of $10.What proportion of tax preparation fees were more than $95?
See process in later post.
Of course, you can shorten the process if you remember that the mean, mode and median are the same value in a normal distribution.
To find the proportion of tax preparation fees that were more than $95, we need to calculate the z-score for that amount and then find the corresponding area under the normal distribution curve.
The z-score formula is given by:
z = (x - µ) / σ
Where:
x = the value we are interested in (in this case, $95)
µ = the mean of the distribution (given as $95)
σ = the standard deviation of the distribution (given as $10)
Calculating the z-score:
z = ($95 - $95) / $10 = 0 / $10 = 0
Now, we need to find the area under the normal distribution curve to the right of this z-score. We can consult a standard normal distribution table or use a calculator to find this area.
Using a standard normal distribution table, we can look up the value for a z-score of 0. The table gives us the proportion of the area to the left of the z-score. Since we want the area to the right of the z-score, we subtract the left area from 1.
Hence, the proportion of tax preparation fees that were more than $95 can be found as 1 - Left Area.
Since the z-score is 0, the left area is 0.5000 (50% on the standard normal distribution table).
Therefore, the proportion of tax preparation fees that were more than $95 is:
1 - 0.5000 = 0.5000 or 50%.