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Question
Two dice are thrown simultaneously 25 times. The following price of observation are obtained.
(2, 3) (2, 5) (5, 5) (4, 5) (6, 4) (3, 2) (5, 2) (4, 1) (2, 5) (6, 1) (3, 1) (3, 3) (4, 3) (4, 5) (2, 5) (3, 4) (2, 5) (3, 4) (2, 5) (4, 3) (5, 2) (4, 5) (4, 3) (2, 3) (4, 1)
Prepare a bivariate frequency distribution table for the above data. Also, obtain the marginal distributions.
Solution
Let X = Observation on 1st die
Y = Observation on 2nd die
Now, minimum value of X is 1 and maximum value is 6.
Also, minimum value of Y is 1 and maximum value is 6.
Bivariate frequency distribution can be prepared by taking X as row and Y as column.
Bivariate frequency distribution is as follows:
Y/X | 1 | 2 | 3 | 4 | 5 | 6 | Total (fy) |
1 | – | – | I | II | – | I | 4 |
2 | – | – | I | – | II | – | 3 |
3 | – | II | I | III | – | – | 6 |
4 | – | – | II | – | – | I | 3 |
5 | – | IIII | – | III | I | – | 9 |
6 | – | – | – | – | – | – | 0 |
Total (fx) | 0 | 7 | 5 | 8 | 3 | 2 | 25 |
Marginal frequency distribution of X:
X | 1 | 2 | 3 | 4 | 5 | 6 | Total |
Frequency | 0 | 7 | 5 | 8 | 3 | 2 | 25 |
Marginal frequency distribution of Y:
Y | 1 | 2 | 3 | 4 | 5 | 6 | Total |
Frequency | 4 | 3 | 6 | 3 | 9 | 0 | 25 |
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