Roderick_April_Unit_7_Project_MM207-05

# Roderick_April_Unit_7_Project_MM207-05 - 7.5#4 a since the...

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7.5 #4 a. since the sample mean is below 30 we can not say anything about x distribution of sample means. b. Since the x distribution is given to be normal, the x distribution also will be normal even though the sample size is much less than 30. The mean is 72 2 16 8 = The x distribution is normal with the mean of 72 and standard deviation of 2 x μ = μ =72 = = 16 / 8 \ x σ 2 z= 2 72 73 2 72 68 - - x = -2 z 0.5 P ) 5 . 0 2 ( ) 73 68 ( - = z P x =P (z ) 2 ( ) 5 . 0 - - z P 0.5199-0.0228=0.4971 #6 a. 68 = = μ μ x / 3 = x σ = 18 0.71 z 8414 . 0 0793 . 0 9207 . 0 ) 41 . 1 ( ) 41 . 1 ( 41 . 1 41 . 1 ( ) 69 67 ( 41 . 1 41 . 1 71 . 0 68 69 71 . 0 68 67 = - = - - = - = = - = - - Pz z P z P P z z b. 1 9 / 3 68 = = = = x x σ μ μ z 6826 . 0 1587 . 0 8413 . 0 ) 00 . 1 ( ) 00 . 1 ( 00 . 1 00 . 1 ( ) 69 67 ( 00 . 1 00 . 1 1 68 69 1 68 67 = - = - - = - = = - = - - - Pz z P z P P z z c. no, it is not higher, because you only have a sample of 9 in part b and a all the population in part a. #10 a. It is normal probability distribution, yes it is approximately normal, the mean is 16 and sd is approximately 2 37 . 0 30 / 2 16 = = = = x x σ μ μ

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z 4965 . 0 5000 . 0 9965 . 0 ) 0 ( ) 70 . 2 ( ) 70 . 2 0 ( 16 16 ( 70 . 2 0 37 . 0 16 17 37 . 0 16 16 = - = - = = = = - - Pz z P z P P z z c. 0035 . 0 70 . 2 37 . 0 16 15 = - = - less than 15 days 7.6 #4 a. ) 89 . 0 )( 11 . 0 ( 316 76 . 34 ) 11 . 0 )( 316 = = = = σ μ μ μ np 9364 . 30 = σ 562 . 5 = σ 56 . 5 = σ z=x- σ μ / z=(39.5)-34.76/4.74 z=(4.74)/5.56 z=0.85 P(39.5 ) x =P(0.85 z ) P(0.85 ) 85 . 0 ( 1 - = z P z 1-P(z>0.85)=1-0.1977 1-0.1977=0.8023 b. n=316, p=0.11, q=0.89
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