I already fill the table for question 1. i hope someone expert can help me find answer for questions 2.
(a.i.) frequency distribution of X and Y
X | frequency |
1 | 4 |
2 | 3 |
2 | 5 |
Y | frequency |
1 | 6 |
2 | 6 |
(a.ii.) joint probability distribution of X and Y
X | |||||
1 | 2 | 3 | total | ||
Y | 1 | 0.1667 | 0.0833 | 0.2500 | 0.5000 |
2 | 0.1667 | 0.1667 | 0.1667 | 0.5000 | |
total | 0.3333 | 0.2500 | 0.4167 | 1.0000 |
(b) P(X=2|Y=1)=P(X=2 and Y=1)/P(Y=1)=0.0833/0.5=0.1667 and P(X=2)=0.25
here P(X=2) is greater than P(X=2|Y=1)
(c)Yes, number of pages are related to language used in the book
E(X)=sum(x*P(x))=1*0.3333+2*0.2500+3*0.4167=2.0834
E(Y)=sum(y*P(y))=1*0.5+2*0.5=1.5
E(XY)=sum(x*y*P(xy))=1*1*0.1667+1*2*0.0833+1*3*0.25+2*1*0.1667+2*2*0.1667+2*3*0.1667=3.0837
E(X)*E(Y)=2.0834*1.5=3.1251 which is not equal to E(XY)=3.0837, so X and Y are not independent and conclude that number of pages are related to language used in the book
following information has been generated when x is in ascending order for answering the above question
Y | x | X |
1 | 20 | 1 |
2 | 22 | 1 |
1 | 28 | 1 |
2 | 32 | 1 |
2 | 39 | 2 |
2 | 42 | 2 |
1 | 45 | 2 |
2 | 48 | 3 |
1 | 52 | 3 |
2 | 71 | 3 |
1 | 72 | 3 |
1 | 72 | 3 |
|
45.25 | |
20%of |
9.05 | |
|
36.2 | |
|
||
54.3 |
I already fill the table for question 1. i hope someone expert can help me find answer for questions 2.
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