Question 1064324
Binomial distribution with n=13 and p=0.4. Want probability p <2.  I think this is what you want.  The normal approximation is met given the values stated.
np=5.2; n(1-p)=7.8
variance is np(1-p)=13*0.4*0.6=3.12
sqrt (variance)=sd=1.767
want p (value is less than 2, when mean is 5.2 and sd is 1.767)
z=(0-5.2)/1.767=-2.943
z=(2-5.2)/1.767=-1.811
probability is -2.943 < z < -1.811=0.0334

Compare with using binomial distribution
13C1(0.4)^1(0.6)^12=0.0113
13C0(0.6^13)=0.0113
Total is 0.0226