Question

Suppose x has a distribution with a mean of 75 and a standard deviation of 45. Random samples of size n = 81 are drawn.

(a) Describe the x distribution and compute the mean and standard deviation of the distribution.


(b) Find the z value corresponding to x = 90.


(c) Find P(x < 90).


(d) Would it be unusual for a random sample of size 81 from the x distribution to have a sample mean less than 90? Explain.

Answer 1

a)
`X \sim N(\mu=75,\sigma=45)`

`\bar X \sim N(75,(45)/(√(81)))`

b)
`z=(90-75)/((45)/(√(81)))=3`

c)
`P(\bar X <90) = P(Z<3)=0.99865`

d) No. it would not be unusual because more than 5% of all such samples hav means less than 90.

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".

(a) Describe the x distribution and compute the mean and standard deviation of the distribution.

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

`X \sim N(\mu=75,\sigma=45)`

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(75,(45)/(√(81)))`

(b) Find the z value corresponding to
`\bar X = 90`.

The z score on this case is given by this formula:

`z=(\bar x-\mu)/((\sigma)/(√(n)))`

And if we replace we got:

`z=(90-75)/((45)/(√(81)))=3`

(c) Find
`P(\bar X < 90)`.

For this case we can use a table or excel to find the probability required:

`P(\bar X <90) = P(Z<3)=0.99865`

(d) Would it be unusual for a random sample of size 81 from the x distribution to have a sample mean less than 90? Explain.

For this case the best conclusion is:

No. it would not be unusual because more than 5% of all such samples hav means less than 90.