The average income in a certain region in 2013 was $\$ 77,000$ per person per year. Suppose the standard deviation is $\$ 32,000$ and the distribution is right-skewed. Suppose we take a random sample of 100 residents of the region.
a. Is the sample size large enough to use the Central Limit Theorem for means? Explain.
b. What are the mean and standard error of the sampling distribution?
c. What is the probability that the sample mean will be more than $\$ 3,200$ away from the population mean?
A. Yes, it is large enough because the population standard deviation of 32,000 is larger than the sample size of 100 .
B. No, it is not large enough because the sample size of 100 is less than 250.
C. No, it is not large enough because the sample size of 100 is not greater than $10 \%$ of the population.
D. Yes, it is large enough because the sample size of 100 is greater than 25.
b. The mean is $\$ 77000$ and the standard error is $\$ 3200$.
(Type integers or decimals. Do not round.)
c. The probability is $\square$.
(Round to four decimal places as needed.)
Answer
The problem also asks for the probability that the sample mean will be more than \$3200 away from the population mean. This is equivalent to finding the probability that the sample mean is either less than \$77000 - \$3200 = \$73800 or more than \$77000 + \$3200 = \$80200. To find this probability, we would need to use the standard normal distribution table or a calculator with a normal distribution function. However, the problem does not provide enough information to calculate this probability.
Step 1 ：The problem asks whether the sample size is large enough to use the Central Limit Theorem for means. The Central Limit Theorem states that if the sample size is large enough (usually n > 30), the sampling distribution of the mean will be approximately normally distributed, regardless of the shape of the population distribution. In this case, the sample size is 100, which is greater than 30. Therefore, the sample size is large enough to use the Central Limit Theorem for means. The final answer for this part is \(\boxed{\text{D. Yes, it is large enough because the sample size of 100 is greater than 25.}}\)
Step 2 ：The problem also asks for the mean and standard error of the sampling distribution. The mean of the sampling distribution is equal to the population mean, which is \$77000. The standard error of the sampling distribution is equal to the population standard deviation divided by the square root of the sample size. In this case, the standard error is \$32000 divided by the square root of 100, which is \$3200. Therefore, the mean is \$77000 and the standard error is \$3200.
Step 3 ：The problem also asks for the probability that the sample mean will be more than \$3200 away from the population mean. This is equivalent to finding the probability that the sample mean is either less than \$77000 - \$3200 = \$73800 or more than \$77000 + \$3200 = \$80200. To find this probability, we would need to use the standard normal distribution table or a calculator with a normal distribution function. However, the problem does not provide enough information to calculate this probability.
