Question

A simple random sample of size nequals=8181 is obtained from a population with mu equals 77μ=77 and sigma equals 27σ=27. ​(a) Describe the sampling distribution of x overbarx. ​(b) What is Upper P (x overbar greater than 81.5 )P x>81.5​? ​(c) What is Upper P (x overbar less than or equals 69.5 )P x≤69.5​? ​(d) What is Upper P (73.4 less than x overbar less than 84.05 )P 73.4

Answer 1

a)
`P(\bar X>81.5)=1-0.933=0.067`

b)
`P(\bar X<69.5)=0.0062`

c)
`P(73.4<\bar X<84.05)=0.8755`

Explanation:

1) 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".

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=77,\sigma=27)`

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(77,(27)/(√(81)))`

Part a

We want this probability:

`P(\bar X>81.5)=1-P(\bar X<81.5)`

The best way to solve this problem is using the normal standard distribution and the z score given by:

`z=(x-\mu)/((\sigma)/(√(n)))`

If we apply this formula to our probability we got this:

`P(\bar X >81.5)=1-P(Z<(81.5-77)/((27)/(√(81))))=1-P(Z<1.5)`

`P(\bar X>81.5)=1-0.933=0.067`

Part b

We want this probability:

`P(\bar X\leq 69.5)`

If we apply the formula for the z score to our probability we got this:

`P(\bar X \leq 69.5)=P(Z\leq (69.5-77)/((27)/(√(81))))=P(Z<-2.5)`

`P(\bar X\leq 69.5)=0.0062`

Part c

We are interested on this probability

`P(73.4<\bar X<84.05)`

If we apply the Z score formula to our probability we got this:

`P(73.4<\bar X<84.05)=P((73.4-\mu)/((\sigma)/(√(n)))<(X-\mu)/((\sigma)/(√(n)))<(84.05-\mu)/((\sigma)/(√(n))))`

`=P((73.4-77)/((27)/(√(81)))<Z<(84.05-77)/((27)/(√(81))))=P(-1.2<z<2.35)`

And we can find this probability on this way:

`P(-1.2<z<2.35)=P(z<2.35)-P(z<-1.2)`

And in order to find these probabilities we can find tables for the normal standard distribution, excel or a calculator.

`P(-1.2<z<2.35)=P(z<2.35)-P(z<-1.2)=0.9906-0.1151=0.8755`