In 2020, the average charge of tax preparation was $95. Assuming a normal distribution and a standard deviation of $10, use the approximate areas beneath the normal curve, as discussed in this section, to answer the following questions.
What proportion of tax preparation fees were between $75 and $115?
A) 0.84
B) 0.34
C) 0.955
D) 0.75
This will be within two standard deviations of the mean. To show this, use z-scores.
Formula is this:
z = (x - mean)/sd
Therefore:
z = (75 - 95)/10 = ?
z = (115 - 95)/10 = ?
I'll let you finish the calculations. Remember that one z-score is below the mean and one is above the mean.
I hope this will help get you started.
To find the proportion of tax preparation fees between $75 and $115, we need to calculate the z-scores for each of these values and then find the area under the normal curve between these two z-scores.
First, let's find the z-score for $75. We can use the formula:
z = (x - μ) / σ
where x is the value, μ is the mean, and σ is the standard deviation.
z = (75 - 95) / 10
z = -20 / 10
z = -2
Next, let's find the z-score for $115:
z = (115 - 95) / 10
z = 20 / 10
z = 2
Now, we need to find the area under the normal curve between these two z-scores. We can use a standard normal distribution table or a calculator to find the corresponding probabilities.
Using a standard normal distribution table, we find that the area to the left of z = -2 is approximately 0.0228, and the area to the left of z = 2 is approximately 0.9772.
To find the area between these z-scores, we subtract the smaller area from the larger area:
0.9772 - 0.0228 = 0.9544
So, the proportion of tax preparation fees between $75 and $115 is approximately 0.9544.
Therefore, the answer is C) 0.955.
To find the proportion of tax preparation fees between $75 and $115, we can start by standardizing the values using the Z-score formula:
Z = (X - μ) / σ
Where:
X is the value we are interested in (in this case, $75 and $115)
μ is the mean of the distribution (in this case, $95)
σ is the standard deviation of the distribution (in this case, $10)
For $75:
Z1 = ($75 - $95) / $10 = -2
For $115:
Z2 = ($115 - $95) / $10 = 2
Now, we need to find the area under the normal curve between these two Z-scores. Since the values are symmetrically distributed around the mean in a standard normal distribution, we can find the area between these Z-scores by subtracting the area to the left of Z1 from the area to the left of Z2.
Using a standard normal distribution table or a calculator, you can find the area to the left of Z1 = -2, which is approximately 0.0228.
You can also find the area to the left of Z2 = 2, which is approximately 0.9772.
Now, subtracting the smaller area from the larger area:
0.9772 - 0.0228 = 0.9544
Therefore, approximately 0.9544 (or 95.44%) of tax preparation fees were between $75 and $115.
Since none of the given options perfectly match the result, we can approximate:
C) 0.955