To find the probability that the sample mean is less than 49.1 dollars, we can use the Central Limit Theorem. Here’s how we can approach this:

Step 1: Identify the known values

• Population mean (μ) = 50 dollars
• Population standard deviation (σ) = 8 dollars
• Sample size (n) = 56
• Sample mean (x̄) we are testing for = 49.1 dollars

Step 2: Calculate the standard error of the mean (SEM)

The standard error of the mean is calculated as:

$\text{SEM} = \frac{\sigma}{\sqrt{n}}$

Substituting in the values:

$\text{SEM} = \frac{8}{\sqrt{56}} \approx 1.0690$

Step 3: Calculate the Z-score

The Z-score formula for a sample mean is:

$Z = \frac{x̄ - μ}{\text{SEM}}$

Substituting in our values:

$Z = \frac{49.1 - 50}{1.0690} \approx -0.8424$

Step 4: Find the probability

Using the Z-score of -0.8424, we can look up the probability in a standard normal distribution table or use a calculator to find the cumulative probability for $Z = -0.8424$.

$P(Z < -0.8424) \approx 0.2005$

Final Answer

The probability that the sample mean would be less than 49.1 dollars is approximately 0.2005.

Would you like more details, or do you have questions on any step?

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Here are five questions that expand on this concept:

1. How would the probability change if the standard deviation were larger?
2. What would the probability be if the sample size was increased to 100?
3. How does the Central Limit Theorem apply to this problem?
4. How can this type of probability calculation be used in real-world scenarios?
5. What is the significance of using a Z-score in probability calculations?

Tip: For larger sample sizes, the standard error decreases, making the sampling distribution of the mean more concentrated around the population mean.