Question

A random sample of size n1 = 16 is selected from a normal population with a mean of 75 and a standard deviation of 9. A second random sample of size n2 = 8 is taken from another normal population with mean 69 and standard deviation 10. Let X1 and X 2 be the two sample means. Find:

a. The probability that X1 - X2 exceeds 4.
b. The probability that 4.3 <= X1 - X2 <= 5.0.

Answer 1

a)
`P( \bar X_1 - \bar X_2 >4)`

And we can use the z score given by:

`z = (D -\mu_D)/(\sigma_D)`

And replacing we got:

`P( \bar X_1 - \bar X_2 >4)= P(z >(4-6)/(4.191)) =P(z>-0.477) = 0.683`

b)
`P(4.3 \leq \bar X_1 - \bar X_2 \leq 5)`

And we can use the z score given by:

`z = (D -\mu_D)/(\sigma_D)`

And replacing we got:

`P(4.3 \leq \bar X_1 - \bar X_2 \leq 5)= P((4.3-6)/(4.191)<z <(5-6)/(4.191)) =P(-0.406< z<-0.239) = P(z<-0.239) -P(z<-0.406) =0.406-0.342=0.0632`

Explanation:

For this case we have the following info given:

`X_1 \sim N(\mu_1 = 75, \sigma_1 = 9)`

`X_2 \sim N(\mu_2 = 69, \sigma_2 = 10)`And we are interested in the probabilities associated to
`\bar X_1 -\bar X_2`

For this case the distribution of
`\bar X_1 -\bar X_2` is given by:

`\bar X_1 -\bar X_2 \sim N (\mu_1 -\mu_2 , \sqrt{(\sigma^2_1)/(n_1) +(\sigma^2_2)/(n_2)}`

And the parameters are:

`\mu_D = 75-69 = 6`

`\sigma_D = \sqrt{(9^2)/(16) +(10^2)/(8)} =4.191`

Part a

For this case we want this probability:

`P( \bar X_1 - \bar X_2 >4)`

And we can use the z score given by:

`z = (D -\mu_D)/(\sigma_D)`

And replacing we got:

`P( \bar X_1 - \bar X_2 >4)= P(z >(4-6)/(4.191)) =P(z>-0.477) = 0.683`

Part b

`P(4.3 \leq \bar X_1 - \bar X_2 \leq 5)`

And we can use the z score given by:

`z = (D -\mu_D)/(\sigma_D)`

And replacing we got:

`P(4.3 \leq \bar X_1 - \bar X_2 \leq 5)= P((4.3-6)/(4.191)<z <(5-6)/(4.191)) =P(-0.406< z<-0.239) = P(z<-0.239) -P(z<-0.406) =0.406-0.342=0.0632`