91影视

Consider the random variable $T=X+Y$. The values of $\mathrm{T}$is to compute the probability that $T$takes each of these values for the probability distribution.

According to the information, the possible values of$D$ are:
$$\begin{gathered} 1-2=-1 \\ 1鈭�4=鈭�3 \\ 2鈭�2=0 \\ 2鈭�4=鈭�2 \end{gathered}$$
Similarly,
$\begin{array}{r}5鈭�2=3\\ 5鈭�4=1\end{array}$
Determine the probabilities by:
$$\begin{gathered} P(D=-3)=P(X=1)脳P(Y=4) \\ =0.2(0.3) \\ =0.06 \end{gathered}$$
$$\begin{gathered} P(D=-2)=P(X=2)脳P(Y=4) \\ =0.5(0.3) \\ =0.15 \end{gathered}$$
$$\begin{gathered} P(D=-1)=P(X=1)脳P(Y=2) \\ =0.2(0.7) \\ =0.14 \end{gathered}$$
$$\begin{gathered} P(D=0)=P(X=2)脳P(Y=2) \\ =0.5(0.7) \\ =0.35 \end{gathered}$$
Also,
$$\begin{gathered} P(D=1)=P(X=5)脳P(Y=4) \\ =0.3(0.3) \\ =0.09 \end{gathered}$$
$$\begin{gathered} P(D=3)=P(X=5)脳P(Y=2) \\ =0.3(0.7) \\ =0.21 \end{gathered}$$

The random variable is $D=X-Y$. The mean of $D$is equal to ${渭}_X-{渭}_Y$.

The mean of$\mathrm{T}$ be calculated as:
$$\begin{gathered} \begin{array}{c}{渭}_D=鈭�x脳P\left(x\right)\end{array} \\ =鈭�3\left(0.06\right)+(鈭�2)\left(0.15\right)+鈥�.+3\left(0.21\right) \\ =0.1 \end{gathered}$$

Then,

$$\begin{gathered} {渭}_X鈭�{渭}_Y=2.7+2.6 \\ =0.1 \end{gathered}$$

Confirm the variance of $D$is equal to ${蟽}_X^2+{蟽}_Y^2$.

Determine the variance of D by:
$$\begin{gathered} \begin{array}{c}{蟽}_D^2=\sqrt{鈭�x^2脳P\left(x\right)鈭�{\left(鈭�x脳P\left(x\right)\right)}^2}\end{array} \\ =\sqrt{\left(鈭�3^2脳0.06+鈭�2^2脳0.15+鈥�..+3^2脳0.21\right)鈭�(0.1{)}^2} \\ =3.25 \end{gathered}$$

Then,

$$\begin{gathered} {蟽}_X^2+{蟽}_Y^2={1.55}^2+{0.917}^2 \\ =3.2433 \end{gathered}$$