Question

An article in Knee Surgery, Sports Traumatology, Arthroscopy, "Arthroscopic meniscal repair with an absorbable screw: results and surgical technique," (2005, Vol. 13, pp. 273-279) cites a success rate more than 90% for meniscal tears with a rim width of less than 3 mm, but only a 67% success rate for tears of 3-6 mm. If you are unlucky enough to suffer a meniscal tear of less than 3 mm on your left knee, and one of width 3-6 mm on your right knee, what are the mean and variance of the number of successful surgeries?

Answer 1

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).

$P(e_1)=0.90*0.67=0.603$

2) Left knee unsuccessful and right knee successful (X=1).

$P(e_2)=(1-0.9)*0.67=0.1*0.67=0.067$

3) Right knee unsuccessful and left knee successful (X=1).

$P(e_3)=0.90*(1-0.67)=0.9*0.33=0.297$

4) Both surgeries are unsuccessful (X=0).

$P(e_4)=(1-0.90)*(1-0.67)=0.1*0.33=0.033$

Then, the mean can be calculated as the expected value:

$M=\sum p_iX_i=0.603*2+0.067*1+0.297*1+0.033*0\\\\M=1.206+0.067+0.297+0\\\\M=1.57$

The variance can be calculated as:

$V=\sum p_i(X_i-\bar{X})^2\\\\V=0.603(2-1.57)^2+(0.067+0.297)*(1-1.57)^2+0.033*(0-1.57)^2\\\\V=0.603*0.1849+0.364*0.3249+0.033*2.4649\\\\V=0.1115+0.1183+0.0813\\\\V=0.3111$