Question

A simple random sample of size nequals81 is obtained from a population with mu equals 83 and sigma equals 27. ​(a) Describe the sampling distribution of x overbar. ​(b) What is Upper P (x overbar greater than 89 )​? ​(c) What is Upper P (x overbar less than or equals 75.65 )​? ​(d) What is Upper P (79.4 less than x overbar less than 89.3 )​?

Answer 1

a)
`\bar X \sim N (\mu, (\sigma)/(√(n)))`

With:

`\mu_(\bar X)= 83`

`\sigma_(\bar X)=(27)/(√(81))= 3`

b)
`z= (89-83)/((27)/(√(81)))= 2`

`P(Z>2) = 1-P(Z<2)= 1-0.97725= 0.02275`

c)
`z= (75.65-83)/((27)/(√(81)))= -2.45`

`P(Z<-2.45) =0.0071`

d)
`z= (89.3-83)/((27)/(√(81)))= 2.1`

`z= (79.4-83)/((27)/(√(81)))= -1.2`

`P(-1.2<Z<2.1)= P(Z<2.1) -P(z<-1.2) = 0.982-0.115= 0.867`

Explanation:

For this case we know the following propoertis for the random variable X

`\mu = 83, \sigma = 27`

We select a sample size of n = 81

Part a

Since the sample size is large enough we can use the central limit distribution and the distribution for the sampel mean on this case would be:

`\bar X \sim N (\mu, (\sigma)/(√(n)))`

With:

`\mu_(\bar X)= 83`

`\sigma_(\bar X)=(27)/(√(81))= 3`

Part b

We want this probability:

`P(\bar X>89)`

We can use the z score formula given by:

`z = (\bar X -\mu)/((\sigma)/(√(n)))`

And if we find the z score for 89 we got:

`z= (89-83)/((27)/(√(81)))= 2`

`P(Z>2) = 1-P(Z<2)= 1-0.97725= 0.02275`

Part c

`P(\bar X<75.65)`

We can use the z score formula given by:

`z = (\bar X -\mu)/((\sigma)/(√(n)))`

And if we find the z score for 75.65 we got:

`z= (75.65-83)/((27)/(√(81)))= -2.45`

`P(Z<-2.45) =0.0071`

Part d

We want this probability:

`P(79.4 < \bar X < 89.3)`

We find the z scores:

`z= (89.3-83)/((27)/(√(81)))= 2.1`

`z= (79.4-83)/((27)/(√(81)))= -1.2`

`P(-1.2<Z<2.1)= P(Z<2.1) -P(z<-1.2) = 0.982-0.115= 0.867`