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expeople1 [14]
3 years ago
5

A simple random sample of size nequals10 is obtained from a population with muequals63 and sigmaequals18. ​(a) What must be true

regarding the distribution of the population in order to use the normal model to compute probabilities involving the sample​ mean? Assuming that this condition is​ true, describe the sampling distribution of x overbar.
Mathematics
1 answer:
umka2103 [35]3 years ago
3 0

Answer:

The sample size is smaller than 30, so we need to assume that the underlying population is normally distributed.

The  sampling distribution of x overbar will be approximately normally distributed with mean 63 and standard deviation 5.69.

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

The sample size is smaller than 30, so we need to assume that the underlying population is normally distributed.

If it is:

\mu = 63, \sigma = 18, n = 10, s = \frac{18}{\sqrt{10}} = 5.69

The  sampling distribution of x overbar will be approximately normally distributed with mean 63 and standard deviation 5.69.

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Step-by-step explanation:

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Your expression can be simplified as follows:

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<em>Additional comment</em>

If you think of an exponent as signifying repeated multiplication, the rules of exponents may be easier to remember. The exponent tells you how many times the base is a factor in the product.

Consider multiplication:

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Consider division:

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This may help you see that a positive exponent in the denominator is equivalent to a negative exponent in the numerator (and vice versa).

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