Answer:
Mean = 1.570
Variance = 0.311
Step-by-step explanation:
The tear of the left knee has a width of less than 3 mm, so it has a probability of success P(xl) of 0.9.
The tear of the right knee has a width of 3 to 6 mm, so the probability of success P(xr) is 0.67.
There are 3 outcomes possible:
1) Both surgeries failed
2) One surgery succeed (can be the right or the left knee) and the other fail
3) Both surgeries succeed
We have this sampling distribution:
P(0) = P(L=0;R=0) = (1-0.9)*(1-0.67)=0.1*0.33=0.033
P(1) = P(L=1;R=0) + P(L=0;R=1) = 0.9*(1-0.67) + (1-0.9)*0.67 = 0.297 + 0.067 = 0.364
P(2) = P(L=1;R=1) = 0.9*0.67 = 0.603
We can calculate the mean as
[tex]\bar{X}=\sum_{X=0}^{2} P(X)*X\\\\\bar{X}=0.033*0+0.364*1+0.603*2=1.57[/tex]
And the variance as
[tex]var=\sum_{X=0}^{2} P(X)*(X-\bar{X})^{2} \\var=0.033*(0-1.57)^{2}+0.364*(1-1.57)^{2}+0.603*(2-1.57)^{2}\\var=0.033*2.465+0.364*0.325+0.603*0.185\\var=0.311[/tex]
