=0.2311
(B) p(x is at most 4)=P (x,=4)=p(0)+p(1)+p(2)+p(3)+p(4)
=c(12,0)(0.30)^0(1-0.30)^2(12-0)+c(12,1)(0.30)^1(1-0.30)^(12-1)
+c(12,2)(0.30)^2(1-0.30)^(12-2)+c(12,3)(0.30)^3(1-0.30)^(12-3)+c(12,4)(0.30)^4(1-0.30)^(12-4)
(C) u=np=12*0.3=3.6
12*0.3*0.7=2.52
(2.52)=1.5875
to find p(4), we applies the continuity correction factor and find p(3.5<x<4.5). this because using the normal distribution p(x=4) will be 0.
z=(3.4-3.6)/1.5875=-0.0630and z=(4.5-3.6)/1.5875=0.5669
P(4)= P(3.5<x<4.5)=P(-0.063<z<0.5669)=0.2397
Answer the following:
(A) Find the binomial probability P(x = 4), where n = 12 and p = 0.30.
(B) Set up, without solving, the binomial probability P(x is at most 4) using probability notation.
(C) How would you find the normal approximation to the binomial probability P(x = 4) in part A? Please show how you would calculate µ and σ in the formula for the normal approximation to the binomial, and show the final formula you would use without going through all the calculations.
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