Answer:
Mean = 1.57
Variance=0.31
Step-by-step explanation:
To calculate the mean and the variance of the number of successful surgeries (X), we first have to enumerate the possible outcomes:
1) Both surgeries are successful (X=2).
[tex]P(e_1)=0.90*0.67=0.603[/tex]
2) Left knee unsuccessful and right knee successful (X=1).
[tex]P(e_2)=(1-0.9)*0.67=0.1*0.67=0.067[/tex]
3) Right knee unsuccessful and left knee successful (X=1).
[tex]P(e_3)=0.90*(1-0.67)=0.9*0.33=0.297[/tex]
4) Both surgeries are unsuccessful (X=0).
[tex]P(e_4)=(1-0.90)*(1-0.67)=0.1*0.33=0.033[/tex]
Then, the mean can be calculated as the expected value:
[tex]M=\sum p_iX_i=0.603*2+0.067*1+0.297*1+0.033*0\\\\M=1.206+0.067+0.297+0\\\\M=1.57[/tex]
The variance can be calculated as:
[tex]V=\sum p_i(X_i-\bar{X})^2\\\\V=0.603(2-1.57)^2+(0.067+0.297)*(1-1.57)^2+0.033*(0-1.57)^2\\\\V=0.603*0.1849+0.364*0.3249+0.033*2.4649\\\\V=0.1115+0.1183+0.0813\\\\V=0.3111[/tex]