Question 615019
(A) Find the binomial probability P(x = 4), where n = 12 and p = 0.40.
P(x=4) = 12C4*0.4^4*0.6^8 = 0.2128
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(B) Set up, without solving, the binomial probability P(x is at most 4) using probability notation.
P(0<= x <=4) = 12C0*0.4^0*0.6^12 + 12C1*0.4*0.6^11+...+12C4*0.4^4*0.6^8
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(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 &#963; in the formula for the normal approximation to the binomial, and show the final formula you would use without going through all the calculations.
u = np = 0.4*12 = 4.8
s = sqrt(npq) = sqrt(4.8*0.6) = 1.6971
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Normal approx: P(3.5< x < 4.5)
z(3.5) = (3.5-4.8)/1.6971
z(4.5) = (4.5-4.8)/1.6871
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P(z(3.5)<= z <z(4.5)) is the approximation.
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cheers,
Stan H.
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