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The mean height of women in a country​ (ages 20minus​29) is 64.2 inches. A random sample of 75 women in this age group is selected. What is the probability that the mean height for the sample is greater than 65 ​inches? Assume sigmaequals2.84. The probability that the mean height for the sample is greater than 65 inches is nothing.

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Answer:


z=(65-64.2)/((2.84)/(√(75))) = 2.440

And we can find the probability using the complement rule and with the normal standard table like this:


P(Z>2.440) =1-P(Z<2.440) = 1-0.993 =0.007

The probability that the mean height for the sample is greater than 65 inches is 0.007

Explanation:

Let X the random variable that represent the women heights of a population, and we know the following parameters


\mu=64.2 and
\sigma=2.84

We are interested on this probability


P(X>65)

Since the sample size selected is 75>30 we can use the centrel limit theorem and the appropiate formula to use would be the z score given by:


z=(x-\mu)/((\sigma)/(√(n)))

If we find the z score for 65 inches we got:


z=(65-64.2)/((2.84)/(√(75))) = 2.440

And we can find the probability using the complement rule and with the normal standard table like this:


P(Z>2.440) =1-P(Z<2.440) = 1-0.993 =0.007

The probability that the mean height for the sample is greater than 65 inches is 0.007

User Bruno Guarita
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