p=0.04, q=1-p=0.96, n=12.
(p+q)12=p12+12C1p11q+12C2p10q2+...+12C2p2q10+12C1pq11+q12.
The terms are p(x=12)+p(x=11)+p(x=10)+...+p(x=2)+p(x=1)+p(x=0).
P(x=4)=12C4p4q8=495×0.044×0.968=0.000914145 approx.
P(x≤4)=∑12Crprq12-r for 0≤r≤4.
For normal distribution, μ=np=0.48, σ=√(np(1-p))=√(0.48×0.96)=√0.4608=0.68 (approx). The normal distribution is continuous while the binomial distribution is discrete, so we need to use N(4.5)-N(3.5) to approximate P(x=4).
The Z-scores for these would be Z1=(4.5-0.48)/0.68), Z2=(3.5-0.48)/0.68. N(Z=Z1)-N(Z=Z2).