# SOLUTION: Please help me answer the following: (A) Find the binomial probability P(x = 4), where n = 12 and p = 0.40. (B) Set up, without solving, the binomial probability P(x is at most 4

Algebra ->  Algebra  -> Probability-and-statistics -> SOLUTION: Please help me answer the following: (A) Find the binomial probability P(x = 4), where n = 12 and p = 0.40. (B) Set up, without solving, the binomial probability P(x is at most 4      Log On

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 Question 615019: Please help me answer the following: (A) Find the binomial probability P(x = 4), where n = 12 and p = 0.40. (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. Answer by stanbon(57307)   (Show Source): You can put this solution on YOUR website!(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 ------------------------------------------------- (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 =============================================================================== (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. u = np = 0.4*12 = 4.8 s = sqrt(npq) = sqrt(4.8*0.6) = 1.6971 ----- 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 --- P(z(3.5)<= z ================================================= cheers, Stan H. =================