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Sagot :
Answer:
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
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 :
- 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.
The mean for the distribution of the sample means is given by:
[tex]\mu_x = \mu[/tex]
[tex]\mu_x = 6.7[/tex]
The standard deviation for the distribution of the sample means is given by:
[tex]\rm \sigma_x = \dfrac{\sigma}{\sqrt{n} }[/tex]
[tex]\sigma_x = \dfrac{2}{\sqrt{20} }[/tex]
[tex]\sigma_x = 0.44721[/tex]
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:
https://brainly.com/question/17716064
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