as5-083-soln

as5-083-soln - n=20 so the population mean np μ = =20*0.52...

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Stat 1181/Darkoh Fall 2008 Assignment 5 – Solutions Supplement V-4 #s 6, 10, 18, 21, 25, 26 Chapter 5 #5.5 ( ) xP x μ = = 0*0.1+1*0.2+…+5*0.05=1.27 2 2 2 ( ) x P x σ = - = (0 2 *0.1 + 1 2 *0.2 + 3 2 *0.45 + … + 52*.05) - 1.27 2 = 1.6771 So = 2 1.6771 1.27 = = #5.7 Divide each frequency by 200 to convert it into proportion (or probability of x); then apply the formula for the mean and standard deviation of x: ( ) xP x = = 0*14/200+1*31/200+2*47/200+…+8*2/200 =2.9 2 2 2 ( ) x P x = - = (0 2 *14/200+1 2 *31/200 +2 2 *47/200+… + 8 2 *2/200) – 2.9 2 =3.14 So = 2 3.14 1.77 = = 5.24
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5.27 5.31 To use the normal approximation, calculate the mean and standard deviation of the binomial and equate them to the mean and stardard deviation of the normal distribution. (a) Le X B = number in a random sample of 20 that will agree. Then p = P(success) = 0.52,
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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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This note was uploaded on 02/20/2009 for the course STATS 1181 taught by Professor Darkoh during the Fall '09 term at Langara.

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as5-083-soln - n=20 so the population mean np μ = =20*0.52...

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