solutions_HMWK_4

# solutions_HMWK_4 - 9i99ISDS-361A Homework 4. - Solutions...

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9i99ISDS-361A Homework 4. - Solutions Normal Distribution – Sampling & Central Limit Theorem Panayiotis Skordi 1. A normal population has a mean of 60 and a standard deviation of 12. You select a random sample of 9. Compute the probability the sample mean is: a. Greater than 63. b. Less than 56. c. Between 56 and 63. ANSWERS X= Population variable = X Sample mean variable Mean of X = 60 Mean of = X 60 SD of X= 12 SD of 4 9 12 = = X from n σ where is the population SD Sample size = 9 = n sample size a. Greater than 63 ) 75 . 0 ( ) 4 60 63 ( ) 63 ( = - = Z p Z p X p Standard Normal P(z>0.75) 0.25 0.5 0.75 1 1.25 1.5 1.75 2 2.25 2.5 2.75 3 3.25 –0.25 –0.5 –0.75 –1 –1.25 –1.5 –1.75 –2 –2.25 –2.5 –2.75 –3 –3.25 z So probability is 0.2266 , found by 0.5000 - 0.2734. 1

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Less than 56 ) 1 ( ) 4 60 56 ( ) 56 ( - < = - < = < Z p Z p X p Standard Normal P(z<-1) 0.25 0.5 0.75 1 1.25 1.5 1.75 2 2.25 2.5 2.75 3 3.25 –0.25 –0.5 –0.75 –1 –1.25 –1.5 –1.75 –2 –2.25 –2.5 –2.75 –3 –3.25 z So the probability is 0.1587 , found by 0.5000 - 0.3413 c. Between 56 and 63 ) 75 . 0 1 ( ) 4 60 63 4 60 56 ( ) 63 56 ( < < - = - < < - = < < Z p Z p X p Standard Normal P(-1<z<0.75) 0.25 0.5 0.75 1 1.25 1.5 1.75 2 2.25 2.5 2.75 3 3.25 –0.25 –0.5 –0.75 –1 –1.25 –1.5 –1.75 –2 –2.25 –2.5 –2.75 –3 –3.25 z 0.6147 , found by 0.3413 + 0.2734. 2
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## This note was uploaded on 05/06/2008 for the course ISDS 361A taught by Professor Skordi during the Fall '07 term at CSU Fullerton.

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solutions_HMWK_4 - 9i99ISDS-361A Homework 4. - Solutions...

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