Question

Suppose a simple random sample of size nequals64 is obtained from a population with mu equals 88 and sigma equals 8. ​(a) Describe the sampling distribution of x overbar. ​(b) What is Upper P (x overbar greater than 89.7 )​? ​(c) What is Upper P (x overbar less than or equals 85.7 )​? ​(d) What is Upper P (87.35 less than x overbar less than 90.5 )​?

Answer 1

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

With:

`\mu_(\bar X)= 88`

`\sigma_(\bar X)= 8`

b)
`z=(89.7-88)/((8)/(√(64)))= 1.7`

`P(Z>1.7) = 1-P(Z<1.7) =1-0.955=0.0446`

c)
`z =(85.7-88)/((8)/(√(64)))= -2.3`

`P(Z<-2.3) = 0.0107`

d)
`z =(87.35-88)/((8)/(√(64)))= -0.65`

`z =(90.5-88)/((8)/(√(64)))= 2.5`

`P(-0.65<z<2.5)=P(Z<2.5)-P(Z<-0.65) =0.994-0.258 = 0.736`

Explanation:

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

`\mu = 88, \sigma = 8`

We select a sample size of n = 64

Part a

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

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

With:

`\mu_(\bar X)= 88`

`\sigma_(\bar X)= 8`

Part b

We want this probability:

`P(\bar X>89.7)`

We can use the z score formula given by:

`z = (\bar X -\mu)/((\sigma)/(√(n)))`

And if we find the z score for 89.7 we got:

`z=(89.7-88)/((8)/(√(64)))= 1.7`

`P(Z>1.7) = 1-P(Z<1.7) =1-0.955=0.0446`

Part c

`P(\bar X<85.7)`

We can use the z score formula given by:

`z = (\bar X -\mu)/((\sigma)/(√(n)))`

And if we find the z score for 85.7 we got:

`z =(85.7-88)/((8)/(√(64)))= -2.3`

`P(Z<-2.3) = 0.0107`

Part d

We want this probability:

`P(87.35 <\bar X< 90.5)`

We find the z scores:

`z =(87.35-88)/((8)/(√(64)))= -0.65`

`z =(90.5-88)/((8)/(√(64)))= 2.5`

`P(-0.65<z<2.5)=P(Z<2.5)-P(Z<-0.65) =0.994-0.258 = 0.736`