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xenn [34]
3 years ago
10

A population has a mean of 75 and a standard deviation of 8. A random sample of 800 is selected. The expected value of LaTeX: \b

ar{x}x ¯ is
a.8

b.75

c.800

d.None of these alternatives is correct.
Mathematics
2 answers:
GREYUIT [131]3 years ago
8 0

Answer:

\bar X = \frac{\sum_{i=1}^n X_i}{n}

And if we find the expected value for this statistic we got:

E(\bar X) = E(\frac{\sum_{i=1}^n X_i}{n})= \frac{\sum_{i=1}^n E(X_i)}{n}

And assuming that each observation have the same expected value \mu we got:

E(\bar X) = \frac{n \mu}{n}=\mu

So then the expected value for the sample mean would be:

b.75

Step-by-step explanation:

For this case we have the information about the population:

\mu = 75 represent the mean

\sigma = 8 represent the deviation

We select a sample size of n = 800, a very large sample indeed. We know that the sample mean is given by:

\bar X = \frac{\sum_{i=1}^n X_i}{n}

And if we find the expected value for this statistic we got:

E(\bar X) = E(\frac{\sum_{i=1}^n X_i}{n})= \frac{\sum_{i=1}^n E(X_i)}{n}

And assuming that each observation have the same expected value \mu we got:

E(\bar X) = \frac{n \mu}{n}=\mu

So then the expected value for the sample mean would be:

b.75

Blababa [14]3 years ago
5 0

Answer:

b.75

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:

Mean of the population is 75.

By the Central Limit Theorem,

The mean of the sample, \bar{x}, is expected to be also 75.

So the correct answer is:

b.75

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