solutions_HMWK_3

solutions_HMWK_3 - ISDS-361A Homework 3 - Solutions....

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ISDS-361A Homework 3 - Solutions. Normal Distribution Panayiotis Skordi 1. A bank has determined that the monthly balances of the saving accounts of its customers are normally distributed with an average balance of $1,200 and a standard deviation of $250. a. What proportions of customers have monthly balances less than $1,000? b. What proportions of customers have monthly balances more than $1,125? c. What proportions of customers have monthly balances between $950 and $1,075? ANSWER a. What proportions of customers have monthly balances less than $1,000? Let X = monthly balances. Mean=1,200 SD=250 ) 8 . 0 ( ) 250 1200 1000 250 200 , 1 ( ) 1000 ( - < = - < - = < Z p X p X p Standard Normal p(z<-0.8) 0.25 0.5 0.75 1 1.25 1.5 1.75 2 2.25 2.5 2.75 3 –0.25 –0.5 –0.75 –1 –1.25 –1.5 –1.75 –2 –2.25 –2.5 –2.75 –3 z 1
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Standard Normal p(z>0.8) By symmetry, equivalent to 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 z ) 8 . 0 ( ) 8 . 0 ( = - < Z p Z p Standard Normal 0.5 - p(0<z<0.8) is 0.2881 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 z 2119 . 0 2881 . 0 5 . 0 ) 8 . 0 0 ( 5 . 0 ) 8 . 0 ( = - = < < - = Z p Z p So answer is 0.2119 b. What proportions of customers have monthly balances more than $1,125? Let x = monthly balances. ) 3 . 0 ( ) 250 1200 125 , 1 250 200 , 1 ( ) 125 , 1 ( - = - - = Z p X p X p 2
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Standard Normal p(-0.3<z<infinity) 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 z 6179 . 0 1179 . 0 5 . 0 ) 3 . 0 0 ( 5 . 0 ) 3 . 0 ( = + = < < + = - Z p Z p So answer is 0.6179 c. What proportions of customers have monthly balances between $950 and $1,075? ) 50 . 0 0 . 1 ( ) 250 200 , 1 075 , 1 250 200 , 1 950 ( ) 075 , 1 950 ( - < < - = - < < - = < < Z p Z p X p Standard Normal p(-1<z<-0.5) 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 z Standard Normal p(0.5<z<1.0) By symmetry - same as 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 z ) 0 . 1 5 . 0 ( ) 50 . 0 0 . 1 ( < < = - < < - Z p Z p 3
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Standard Normal 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 z 1498 . 0 1915 . 0 3413 . 0 ) 5 . 0 0 . 0 ( ) 0 . 1 0 ( ) 0 . 1 5 . 0 ( = - = < < - < < = < < Z p Z p Z p So answer is 0.1498 4
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2. The life expectancy of refrigerators is normally distributed with a mean of 20 years and a standard deviation of 30 months . a.
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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_3 - ISDS-361A Homework 3 - Solutions....

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