Question 1053131
No. of successes (x) ======0= 1== 2= 3== 4 
Probability of successes f(x) 0.1 0.4 0.3 0.1 0.1 
E(x)=x(p(x))
=0+0.4+0.6+0.3+0.4=1.7  ANSWER
V(x)= sum p(x) (x-mean)^2
-1.7^2*0+-0.7^2*0.4+0.3^2*0.3+1.3^2*0.1+2.3^2*0.1
=0+0.196+0.027+0.169+0.529=0.921
SD=sqrt(0.921)=0.960  ANSWER
mean +/- 2 sd=2.1+/-1.92 or 1.92; (-0.22,3.62) or (0.3.62) to conform with reality
Using Chebyshev"s Theorem, at least 3/4 of the data lie within this interval.  Here, 9 do, with only 4 successes being the outlier.  This is consistent.