Suppose a random sample of 80 measurements is selected from a population with a mean of 25 and a variance of 200. Select the pair that is the mean and standard error of x.
a) [25, 2.081]
b) [25, 1.981
c) [25, 1.681]
d) [25, 1.581]
e) [80. 1.681]
f) None of the above

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean [tex]\mu[/tex] and standard deviation(which is the square root of the variance) [tex]\sigma[/tex], the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation, which is also standard error, [tex]s = \frac{\sigma}{\sqrt{n}}[/tex].

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

In this question:

[tex]\sigma = \sqrt{200}, n = 80[/tex]

So the standard error is:

[tex]s = \frac{\sqrt{200}}{\sqrt{80}} = 1.581[/tex]

By the Central Limit Theorem, the mean is the same, so 25.

The correct answer is:

d) [25, 1.581]

Suppose a random sample of 80 measurements is selected from a population with a mean of 25 and a variance of 200. Select the pair that is the mean and standard error of x. a) [25, 2.081] b) [25, 1.981 c) [25, 1.681] d) [25, 1.581] e) [80. 1.681] f) None of the above

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Suppose a random sample of 80 measurements is selected from a population with a mean of 25 and a variance of 200. Select the pair that is the mean and standard error of x.
a) [25, 2.081]
b) [25, 1.981
c) [25, 1.681]
d) [25, 1.581]
e) [80. 1.681]
f) None of the above