Answer
The probability that the sample mean tax is between \$7600 and \$8100 is approximately 0.6539.
Solution
To solve part (b) of the problem, we need to find the probability that the sample mean tax is between \( \$ 7600 \) and \( \$ 8100 \).
### Step 1: Identify the known values
- Mean (\( \mu \)) = \( 8040 \)
- Standard deviation (\( \sigma \)) = \( 4700 \)
- Sample size (\( n \)) = \( 1000 \)
### Step 2: Calculate the standard error of the mean (SEM)
The standard error of the mean is calculated using the formula:
\[
\text{SEM} = \frac{\sigma}{\sqrt{n}}
\]
Substituting the known values:
\[
\text{SEM} = \frac{4700}{\sqrt{1000}}
\]
### Step 3: Calculate the z-scores for \( 7600 \) and \( 8100 \)
The z-score is calculated using the formula:
\[
z = \frac{X - \mu}{\text{SEM}}
\]
where \( X \) is the value for which we are calculating the z-score.
1. For \( X = 7600 \):
\[
z_1 = \frac{7600 - 8040}{\text{SEM}}
\]
2. For \( X = 8100 \):
\[
z_2 = \frac{8100 - 8040}{\text{SEM}}
\]
### Step 4: Calculate the probabilities corresponding to the z-scores
We will find the probabilities for both z-scores using the standard normal distribution.
### Step 5: Calculate the probability that the sample mean tax is between \( 7600 \) and \( 8100 \)
The probability that the sample mean is between \( 7600 \) and \( 8100 \) is given by:
\[
P(7600 < \bar{X} < 8100) = P(Z < z_2) - P(Z < z_1)
\]
Now, let's perform the calculations.
First, we will calculate the SEM, then the z-scores, and finally the probabilities.
Let's proceed with the calculations.
Calculate the value by following steps:
- step0: Calculate:
\(\frac{4700}{\sqrt{1000}}\)
- step1: Simplify the root:
\(\frac{4700}{10\sqrt{10}}\)
- step2: Reduce the fraction:
\(\frac{470}{\sqrt{10}}\)
- step3: Multiply by the Conjugate:
\(\frac{470\sqrt{10}}{\sqrt{10}\times \sqrt{10}}\)
- step4: Multiply the numbers:
\(\frac{470\sqrt{10}}{10}\)
- step5: Reduce the fraction:
\(47\sqrt{10}\)
Calculate or simplify the expression \( (7600 - 8040)/(4700/\sqrt(1000)) \).
Calculate the value by following steps:
- step0: Calculate:
\(\frac{\left(7600-8040\right)}{\left(\frac{4700}{\sqrt{1000}}\right)}\)
- step1: Remove the parentheses:
\(\frac{7600-8040}{\frac{4700}{\sqrt{1000}}}\)
- step2: Simplify the root:
\(\frac{7600-8040}{\frac{4700}{10\sqrt{10}}}\)
- step3: Reduce the fraction:
\(\frac{7600-8040}{\frac{470}{\sqrt{10}}}\)
- step4: Subtract the numbers:
\(\frac{-440}{\frac{470}{\sqrt{10}}}\)
- step5: Multiply by the reciprocal:
\(-440\times \frac{\sqrt{10}}{470}\)
- step6: Reduce the numbers:
\(-44\times \frac{\sqrt{10}}{47}\)
- step7: Multiply:
\(-\frac{44\sqrt{10}}{47}\)
Calculate or simplify the expression \( (8100 - 8040)/(4700/\sqrt(1000)) \).
Calculate the value by following steps:
- step0: Calculate:
\(\frac{\left(8100-8040\right)}{\left(\frac{4700}{\sqrt{1000}}\right)}\)
- step1: Remove the parentheses:
\(\frac{8100-8040}{\frac{4700}{\sqrt{1000}}}\)
- step2: Simplify the root:
\(\frac{8100-8040}{\frac{4700}{10\sqrt{10}}}\)
- step3: Reduce the fraction:
\(\frac{8100-8040}{\frac{470}{\sqrt{10}}}\)
- step4: Subtract the numbers:
\(\frac{60}{\frac{470}{\sqrt{10}}}\)
- step5: Multiply by the reciprocal:
\(60\times \frac{\sqrt{10}}{470}\)
- step6: Reduce the numbers:
\(6\times \frac{\sqrt{10}}{47}\)
- step7: Multiply:
\(\frac{6\sqrt{10}}{47}\)
It seems that I encountered an issue while trying to calculate the probabilities for the z-scores. However, I can guide you through the process of finding these probabilities using standard normal distribution tables or a calculator.
### Step 1: Calculate the Standard Error of the Mean (SEM)
From our previous calculation:
\[
\text{SEM} \approx 148.63
\]
### Step 2: Calculate the z-scores
1. For \( X = 7600 \):
\[
z_1 \approx \frac{7600 - 8040}{148.63} \approx -2.96
\]
2. For \( X = 8100 \):
\[
z_2 \approx \frac{8100 - 8040}{148.63} \approx 0.40
\]
### Step 3: Find the probabilities
Using the standard normal distribution:
- For \( z_1 \approx -2.96 \), you can look up this value in a z-table or use a calculator to find \( P(Z < -2.96) \).
- For \( z_2 \approx 0.40 \), find \( P(Z < 0.40) \).
### Step 4: Calculate the probability between the two z-scores
The probability that the sample mean tax is between \( 7600 \) and \( 8100 \) is:
\[
P(7600 < \bar{X} < 8100) = P(Z < 0.40) - P(Z < -2.96)
\]
### Final Calculation
Using standard normal distribution tables or a calculator:
- \( P(Z < -2.96) \approx 0.0015 \)
- \( P(Z < 0.40) \approx 0.6554 \)
Now, substituting these values:
\[
P(7600 < \bar{X} < 8100) \approx 0.6554 - 0.0015 = 0.6539
\]
Thus, the probability that the sample mean tax is between \( \$ 7600 \) and \( \$ 8100 \) is approximately **0.6539**.
Please round this to at least four decimal places as required.
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