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Suppose x has a distribution with a mean of 80 and a standard deviation of 3. Random samples of size n = 36 are drawn. (a) Describe the x distribution and compute the mean and standard deviation of the distribution. x has distribution with mean μx = and standard deviation σx = . (b) Find the z value corresponding to x = 81. z = (c) Find P(x < 81). (Round your answer to four decimal places.) P(x < 81) = (d) Would it be unusual for a random sample of size 36 from the x distribution to have a sample mean less than 81? Explain. Yes, it would be unusual because less than 5% of all such samples have means less than 81. No, it would not be unusual because more than 5% of all such samples have means less than 81. No, it would not be unusual because less than 5% of all such samples have means less than 81. Yes, it would be unusual because more than 5% of all such samples have means less than 81.

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Answer:

a)
\mu_(\bar X)=80


\sigma_(\bar X)=(3)/(√(36))

b)
z=(81 -80)/((3)/(√(36)))=2

c)
P(\bar X <81)=P((\bar X -\mu)/((\sigma)/(√(n)))<2)=P(z<2)=0.977

d) No, it would not be unusual because more than 5% of all such samples have means less than 81

Explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".

Part a

Let X the random variable that represent a population, and for this case we know the distribution for X is given by:


X \sim N(80,3)

Where
\mu=80 and
\sigma=3

And let
\bar X represent the sample mean, the distribution for the sample mean is given by:


\bar X \sim N(\mu,(\sigma)/(√(n)))

On this case
\bar X \sim N(80,(3)/(√(36)))


\mu_(\bar X)=80


\sigma_(\bar X)=(3)/(√(36))

Part b

In order to find the z score we can use this formula:


z=(\bar X -\mu_(\bar X))/((\sigma)/(√(n)))

If we replace on the before formula we got:


z=(81 -80)/((3)/(√(36)))=2

Part c

We want this probability:


P(\bar X <81)=P((\bar X -\mu)/((\sigma)/(√(n)))<2)=P(z<2)=0.977

Part d

If we see the probability obtained on part c we have that more than 90% of the samples have samples means less than 81. So the best answer is:

No, it would not be unusual because more than 5% of all such samples have means less than 81

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