Final answer:
To calculate probabilities around the sample mean TV sets, we utilize the Central Limit Theorem, look up Z-scores, and use the standard normal distribution. The 20th percentile of the sample mean and the unusualness of certain sample means are determined similarly. For individual households, variation expectations differ from sample means.
Step-by-step explanation:
Considering the given population mean of 2.24 TV sets with a standard deviation of 1.2, we can use the Central Limit Theorem to calculate the probabilities for a sample mean of 80 households.
The sampling distribution of the sample mean will have a standard error (SE), which is the standard deviation of the sample means. This is calculated as the population standard deviation divided by the square root of the sample size.
SE = 1.2 / √80 = 1.2 / 8.944 = 0.134
The probability that the sample mean number of TV sets is greater than 2 can be found using the Z-score formula:
Z = (X - μ) / SE = (2 - 2.24) / 0.134 ≈ -1.791 Since we are looking for the area to the right, we lookup the corresponding area for the Z-score in the standard normal distribution table or use a calculator. The probability of the sample mean being greater than 2 is P(Z > -1.791), which should be looked up or calculated.
For the probability that the sample mean number of TV sets is between 2.5 and 3, you would find the Z-scores for each value and then find the probability between them.
To find the 20th percentile of the sample mean, you would look for the Z-score that corresponds to the 0.20 area to the left in the standard normal distribution.
It would be unusual for the sample mean to be less than 2 if the probability of such an occurrence is very low (generally below 0.05).
It wouldn't be unusual for an individual household to have fewer than 2 TV sets since we have to consider the individual variation rather than the sample mean variation.