This problem involves using the Central Limit Theorem to calculate the probability related to the sample mean.

Given dataThis problem involves using the Central Limit Theorem to calculate the probability related to the sample mean.

Given

• Population mean ($\mu$): 2.24
• Population standard deviation ($\sigma$): 1.2
• Sample size ($n$): 80

We are asked to find the probability that the sample mean number of TV sets is greater than 2.

Step 1: Calculate the standard error (SE) of the mean.

The standard error is given by: $SE = \frac{\sigma}{\sqrt{n}}$ Substitute the given values: $SE = \frac{1.2}{\sqrt{80}} \approx 0.1342$

Step 2: Compute the z-score.

The z-score for the sample mean greater than 2 is calculated using the formula: $z = \frac{\bar{x} - \mu}{SE}$ where:

• $\bar{x}$ is the sample mean (2 in this case).
• $\mu$ is the population mean (2.24).

Substitute the values: $z = \frac{2 - 2.24}{0.1342} \approx -1.788$

Step 3: Find the probability corresponding to the z-score.

Using the standard normal distribution table, the probability associated with a z-score of $-1.788$ is approximately 0.0369.

Step 4: Calculate the final probability.

Since we are asked for the probability that the sample mean is greater than 2, we need to subtract the value from 1: $P(\bar{x} > 2) = 1 - 0.0369 = 0.9631$

Final Answer:

The probability that the sample mean number of TV sets is greater than 2 is approximately 0.9631.

Let me know if you'd like further details!

Follow-up Questions:

1. How does increasing the sample size affect the standard error?
2. What is the probability that the sample mean is less than 2?
3. What is the probability that the sample mean is between 2 and 2.5?
4. If the population standard deviation were smaller, how would the probability change?
5. What is the z-score formula and how is it used in statistics?

Tip:

Always check if your problem involves a sample or a population to apply the correct formulas, especially when using standard deviation or standard error!