There are 120 students at Francis Howell North High School. Mr. Willott conducts a poll that asks 20 randomly selected students how many hours of sleep they got last night. They find that the mean amount of sleep is 6.7 hours. The population standard deviation is σ = 2 hours. Can we estimate the standard deviation of the sampling distribution of means by using σ divided by the square root of n (2/sqrt(20) = 0.557 hours)?

Yes, we estimate the standard deviation of the sampling distribution of means by using σ divided by the square root of n and this can be determined by finding the mean and standard deviation for the distribution of the sample means.

Given :

The mean for the distribution of the sample means is given by:

The standard deviation for the distribution of the sample means is given by:

Yes, we estimate the standard deviation of the sampling distribution of means by using σ divided by the square root of n.

For more information, refer to the link given below:

link

Yes

Step-by-step explanation:

The central limit theorem says that If a random variable X from a population has mean u and finite variance σ² , then the sampling distribution of the sample mean X~ approaches a normal distribution with mean u and variance σ²/n as the sample size n approaches infinity.

It is interesting to note that we have neither assumed that the distribution of X is continuous nor we have said anything about the shape of the distribution , whereas the limiting distribution of X is continuous and normal. Thus the distribution of the sample means  regardless of the shape of the population having a finite variance , is approximately normal with mean u and variance σ²/n .

Therefore

(standard deviation/ √sample size)²= variance / n

2/ √20

= 2/4.472

=0.44721

The sampling distribution of X` is therefore approximately normal with mean ux=u and σx =σ/n