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.

[tex] P(4.3 \leq \bar X_1 - \bar X_2 \leq 5)= P(\frac{4.3-6}{4.191}<z <\frac{5-6}{4.191}) =P(-0.406< z<-0.239) = P(z<-0.239) -P(z<-0.406) =0.406-0.342=0.0632 [/tex]

Step-by-step explanation:

For this case we have the following info given:

[tex] X_1 \sim N(\mu_1 = 75, \sigma_1 = 9)[/tex]

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

For this case the distribution of [tex] \bar X_1 -\bar X_2 [/tex] is given by:

[tex] \bar X_1 -\bar X_2 \sim N (\mu_1 -\mu_2 , \sqrt{\frac{\sigma^2_1}{n_1} +\frac{\sigma^2_2}{n_2}}[/tex]

And the parameters are:

[tex] \mu_D = 75-69 = 6[/tex]

[tex]\sigma_D = \sqrt{\frac{9^2}{16} +\frac{10^2}{8}} =4.191[/tex]

Part a

For this case we want this probability:

[tex] P( \bar X_1 - \bar X_2 >4)[/tex]

And we can use the z score given by:

[tex] z = \frac{D -\mu_D}{\sigma_D}[/tex]

And replacing we got:

[tex] P( \bar X_1 - \bar X_2 >4)= P(z >\frac{4-6}{4.191}) =P(z>-0.477) = 0.683[/tex]

Part b

[tex] P(4.3 \leq \bar X_1 - \bar X_2 \leq 5)[/tex]

And we can use the z score given by:

[tex] z = \frac{D -\mu_D}{\sigma_D}[/tex]

And replacing we got:

[tex] P(4.3 \leq \bar X_1 - \bar X_2 \leq 5)= P(\frac{4.3-6}{4.191}<z <\frac{5-6}{4.191}) =P(-0.406< z<-0.239) = P(z<-0.239) -P(z<-0.406) =0.406-0.342=0.0632 [/tex]