Question

In: Math

(a) Suppose that Marks Man is shooting at a target, where the center of the target...

(a) Suppose that Marks Man is shooting at a target, where the center of the target is at (0, 0) in the
plane. Let fX,Y (x, y) be the joint PDF of his shot. Assume that X and Y are independent random variables,
each distributed as N (0, 1). What is P(X ≥ 0, Y ≥ 0)? (Write your answer as a decimal).

(b) Now suppose Joe Schmo is shooting at the same target, and let fX,Y (x, y) be the joint PDF of his
shot. Assume that X and Y are independent random variables, each distributed as N (−1, 4) (He tends to
miss down and to the left, with a higher variance of his shots.) What is P(X ≥ 1, Y ≥ 0)? (Write your
answer as a decimal).

Solutions

Expert Solution

(a) X,Y are independent random each distributed as N(0,1)-Standard normal distribution

therefore,

fX,Y(x,y) = fX(x)fY(y)

P(X0,Y0) = P(X0)P(Y0)

For Standard normal distributions, (P(X0)=P(Y0) = 0.5

P(X0,Y0) = P(X0)P(Y0)= 0.5 x 0.5=0.25

P(X0,Y0) = 0.25

(b) X,Y are independent random each distributed as N(-1,4) i.e mean = -1 and variance =4; standard deviation =2

therefore,

fX,Y(x,y) = fX(x)fY(y)

P(X1,Y0) = P(X1)P(Y0)

P(X1) =1 - P(X<1)

Z-score for X=1 = (1-mean)/Standard deviation = (1-(-1))/2 = 2/2 =1

From standard normal tables, P(Z<1) = 0.8413

P(X<1) =P(Z<1) = 0.8413

P(X1) =1 - P(X<1) = 1-0.8413=0.1587

P(Y0) =1 - P(Y<0)

Z-score for Y=0 = (1-mean)/Standard deviation = (0-(-1))/2 = 1/2 =0.5

From standard normal tables, P(Z<0.5) = 0.6915

P(Y<0) =P(Z<0.5) =0.6915

P(Y0) =1 - P(Y<0) = 1-0.6915=0.3085

P(X1,Y0) = P(X1)P(Y0) = 0.1587 x 0.3085 = 0.04895895

P(X1,Y0) = 0.04895895


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