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Unformatted text preview: n=20, so the population mean np μ = =20*0.52 = 10.4; the standard deviation is = σ (1 ) 20*0.52*0.48 np p= = 2.234. Using the normal approximation P(X B 5) P(X ≤ ≈ N 5.5); but the standardized value of X ≤ N = 5.5 is Z = 5.5 10.4 2.19 2.234= Thus P(X B 5) P(Z 2.19) = 0.0143 ≤ ≈ ≤ (b) P(X B 10) P(X ≤ ≈ N 10.5) = P(Z 0.04) = 0.5160 ≤ ≤ (c) P(X B 15) P(X ≤ ≈ N 15.5) = P(Z 2.28) = 0.9887 ≤ ≤ There is no binomial table for n= 20, p =0.52, so we calculate the exact probability using the binomial probability formula P(X 15) = 1 – P(X >15) = ≤ 20 20 16 20 (0.52) (0.48) x x x Cx= ∑ =0.9904 (using Excel ) We see that the approximation is correct to 2 decimal place....
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 Fall '09
 Darkoh
 Normal Distribution, Standard Deviation, Probability theory, Binomial distribution, µ

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