a. C42=2!2!4!=6
b.m (125,120)=(125+120)/2=122.5
m(125,130)=(125+130)/2=127.5
m(125,110)=(125+110)/2=117.5
m(120,130)=(120+130)/2=125
m(120,110)=(120+110)/2=115
m(130,110)=(130+110)/2=120
Frequency
F(122.5)=F(127.5)=F(117.5)=F(125)=F(115)=F(120)=1
Probability
P(x)=F(x)/∑F(x)
P(122.5)=P(127.5)=P(117.5)=P(125)=P(115)=P(120)=1/6
E(x)=∑Px=1/6(122.5+127.5+117.5+125+115+120)=121.25
σ2=∑Px2−(∑Px)2=1/6(15006.25+16256.25+13806.25+15625+13225+14400)−14701.56=18.23
c.
