6 The table shows the probability distribution for the number of weekday (Monday to Friday) morning newspapers, \(X\), purchased by the Reed household per week.
| \(\boldsymbol { x }\) | 0 | 1 | 2 | 3 | 4 | 5 |
| \(\mathbf { P } ( \boldsymbol { X } = \boldsymbol { x } )\) | 0.16 | 0.15 | 0.25 | 0.25 | 0.15 | 0.04 |
- Find values for \(\mathrm { E } ( X )\) and \(\operatorname { Var } ( X )\).
- The number of weekday (Monday to Friday) evening newspapers, \(Y\), purchased by the same household per week is such that
$$\mathrm { E } ( Y ) = 2.0 , \quad \operatorname { Var } ( Y ) = 1.5 \quad \text { and } \quad \operatorname { Cov } ( X , Y ) = - 0.43$$
Find values for the mean and variance of:
- \(S = X + Y\);
- \(\quad D = X - Y\).
- The total cost per week, \(L\), of the Reed household's weekday morning and evening newspapers may be assumed to be normally distributed with a mean of \(\pounds 2.31\) and a standard deviation of \(\pounds 0.89\).
The total cost per week, \(M\), of the household's weekend (Saturday and Sunday) newspapers may be assumed to be independent of \(L\) and normally distributed with a mean of \(\pounds 2.04\) and a standard deviation of \(\pounds 0.43\).
Determine the probability that the total cost per week of the Reed household's newspapers is more than \(\pounds 5\).