practice1 - Section 6.1 6. (a) X1 29.7(employee 1)...

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Section 6.1 6. (a) 1 X 2 X X 29.7(employee 1) 33.6(employee 2) 31.65 29.7(employee 1) 30.2(employee 3) 29.95 29.7(employee 1) 33.6(employee 4) 31.65 29.7(employee 1) 25.8(employee 5) 27.75 29.7(employee 1) 29.7(employee 6) 29.70 33.6(employee 2) 30.2(employee 3) 31.90 33.6(employee 2) 33.6(employee 4) 33.60 33.6(employee 2) 25.8(employee 5) 29.70 33.6(employee 2) 29.7(employee 6) 31.65 30.2(employee 3) 33.6(employee 4) 31.9 30.2(employee 3) 25.8(employee 5) 28 30.2(employee 3) 29.7(employee 6) 29.95 33.6(employee 4) 25.8(employee 5) 29.7 33.6(employee 4) 29.7(employee 6) 31.65 25.8(employee 5) 29.7(employee 6) 27.75 By above table, we can obtain a new table as shown below. X 27.75 28 29.7 29.95 31.65 31.9 33.6 ( ) X p X 2/15 1/15 1/15 2/15 4/15 2/15 1/15 (b) 1 X 2 X X ( ) X p X 29.7 (Office 1) 33.6 (Office 1) 31.65 1/3 30.2 (Office 2) 33.6 (Office 2) 31.9 1/3 25.8 (Office 3) 29.7 (Office 3) 27.75 1/3 (c) ( ) EX of (a) 21 1 27.75 28 33.6 30.43 15 15 15 ⎛⎞ =⋅ + + + = ⎜⎟ ⎝⎠ L ( ) of (b) 11 1 31.65 31.9 27.75 30.43 33 3 + +⋅ = and the population mean salary 30.43 μ = Section 6.2 16. () 1 11 10 11 1.12 0.8686 2/ 5 <= < = < = PX PZ 1 11 10 11 1.23 0.8906 2/ 6 < = < = Second solution: This question, is not very clear. Most of you did it as above and they got full credit. Those that asked for help I told them to find one probability for the average of the two days as follows: The average of the first day is denoted with X and for the second day as Y . Let 2 X Y W + = . We know that 44 ~1 0 , , 0 , 56 XN YN so:
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() 11 1 10 10 10 22 2 2 1 1 4 1
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This note was uploaded on 01/23/2009 for the course STAT 319 taught by Professor Smith during the Fall '08 term at Pennsylvania State University, University Park.

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practice1 - Section 6.1 6. (a) X1 29.7(employee 1)...

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