Question

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.

Answer 1

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