Find the binominal probability P(x=4) where n=12 and P = 0.04

Question

Also set up without solving the binominal probability P(x is at most 4) using probability-notation.   How would you find the normal approximation fo the binominal probability P(x=4) in part A?  How you would calculate the mean and standard deviation in the formula for the normal approximation to the binomial, and show the final formula you would use without going through all the calculations.

Answer 1

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)=∑¹²C_rp^rq^12-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).