Seventy-five percent of claims have a normal distribution with a mean of 3,000 and a variance of 1,000,000. The remaining 25% have a normal distribution with a mean of 4,000 and a variance of 1,000,000. Determine the probability that a randomly selected claim exceeds 5,000.

p = -4, q = -3

Step-by-step explanation:

y = -2x +4  ... (1)   perpendicular bisector of AB, slope = -2

slope of AB = 1/2

Line AB pass (8,3): (y-3) / (x-8) = 1/2

AB equation: y-3 = 1/2(x-8)      y = 1/2x - 1  ... (2)

(2)-(1): 5/2 x = 5      x = 2

y = 0       (2,0) intercept of bisector and AB, it is midpoint of A (8,3) and (p,q)

(8+p)/2 = 2

p = -4

(3+q)/2 = 0

q = -3