Question

For samples of the specified size from the population described, find the mean and standard deviation of the sample mean x-bar.

The mean and the standard deviation of the sampled population are, respectively, 182.1 and 29.4. n = 36

	μx-bar = 29.4 and σx-bar = 4.9
	μx-bar = 356.9 and σx-bar = 1.0
	μx-bar = 182.1 and σx-bar = 4.9
	μx-bar = 4.9 and σx-bar = 182.1

Answer 1

Let X be the random variable with the population parameter mean = 182.1 and standard deviation = 29.4.

Now, n = 36.

Now the distribution of is .

Therefore, the distribution of Mean X bar is μx-bar = 182.1 and σx-bar = 4.9.