5 The numbers of daily morning operations, \(X\), and daily afternoon operations, \(Y\), in an operating theatre of a small private hospital can be modelled by the following bivariate probability distribution.
| \multirow{2}{*}{} | Number of morning operations ( \(\boldsymbol { X }\) ) | |
| | 2 | 3 | 4 | 5 | 6 | \(\mathbf { P } ( \boldsymbol { Y } = \boldsymbol { y } )\) |
| \multirow{3}{*}{Number of afternoon operations ( \(\boldsymbol { Y }\) )} | 3 | 0.00 | 0.05 | 0.20 | 0.20 | 0.05 | 0.50 |
| 4 | 0.00 | 0.15 | 0.10 | 0.05 | 0.00 | 0.30 |
| 5 | 0.05 | 0.05 | 0.10 | 0.00 | 0.00 | 0.20 |
| \(\mathrm { P } ( \boldsymbol { X } = \boldsymbol { x } )\) | 0.05 | 0.25 | 0.40 | 0.25 | 0.05 | 1.00 |
- State why \(\mathrm { E } ( X ) = 4\) and show that \(\operatorname { Var } ( X ) = 0.9\).
- Given that
$$\mathrm { E } ( Y ) = 3.7 , \operatorname { Var } ( Y ) = 0.61 \text { and } \mathrm { E } ( X Y ) = 14.4$$
calculate values for \(\operatorname { Cov } ( X , Y )\) and \(\rho _ { X Y }\).
- Calculate values for the mean and the variance of:
- \(T = X + Y\);
- \(\quad D = X - Y\).
[0pt]
[4 marks]