Question

Random samples of size 81 are taken from an infinite population whose mean and standard deviation are 200 and 18, respectively. The distribution of the population is unknown. The mean and the standard error of the mean area. 200 and 18b. 81 and 18c. 9 and 2d. 200 and 2

Answer 1

Answer:

d. 200 and 2

Step-by-step explanation:

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

In this problem, we have that:

Mean = 200

Standard deviation = 18

Sample size: 81

Standard error: $s = \frac{18}{\sqrt{81}} = 2$

So the correct answer is:

d. 200 and 2

Answer 2

Answer:

(D) 200 and 2

Step-by-step explanation:

From the data given, the mean is 200

Standard error is obtained by dividing the population standard deviation by square root of the sample size.

population standard deviation = 18

sample size = 81

Standard error = population standard deviation ÷ sqrt (sample size) = 18 ÷ sqrt(81) = 18 ÷ 9 = 2

Mean = 200

Standard error = 2