Answer:
The probability that X and Y are both positive and that their sum is less or equal to 1 0.64.
Step-by-step explanation:
It is provided that the random variables X and Y follows a standard normal distribution.
That is, [tex]X,Y\sim N(0, 1)[/tex]
It is also provided that the variables X and Y are statistically independent of each other.
Compute the probability that X and Y are both positive and that their sum is less or equal to 1 as follows:
The mean and standard deviation of X + Y are:
[tex]E(X+Y)=E(X)+E(Y)=0+0=0\\\\SD(X+Y)=\sqrt{V(X)+V(Y)+2Cov(X,Y)}=\sqrt{1+1+0}=\sqrt{2}[/tex]
The probability is:
[tex]P(X+Y\leq 1)=P(X+Y<1-0.50)\ [\text{Apply continuity correction}]\\[/tex]
[tex]=P(X+Y<0.50)\\\\=P(\frac{(X+Y)-E(X+Y)}{SD(X+Y)}<\frac{0.50-0}{\sqrt{2}})\\\\=P(Z<0.354)\\\\=0.63683\\\\\approx 0.64[/tex]
*Use the z-table.
Thus, the probability that X and Y are both positive and that their sum is less or equal to 1 0.64.
