The positional relationship between the straight line ax+by+0.5=0 and the circle X 2+Y 2 = 2 is intersecting.

Because x ~ n (100, δ 2), p (x greater than 120)=a, P(80 less than x less than or equal to 100)=b, so a+b= 1.

The center of the circle x 2+y 2 = 2 is (0,0), and the distance from the straight line ax+by+0.5=0 is 0.5/√√√ A 2+B 2 √.

Mean inequality: If both A and B are positive numbers, then √ ((A2+B2)/2) ≥ (A+B)/2 ≥√ AB ≥ 2/(1/a+1/b) (If and only if a=b, the equal sign holds. )

It is known that √ ((A 2+B 2)/2) ≥ (A+B)/2 =1/4.

Because obviously a≠b

So it is simplified to √ √ A 2+B 2 √ > √ 2/4.

So 0.5/√ √ A 2+B 2 √ < √ 2.

So the straight line ax+by+0.5=0 intersects with the circle X 2+Y 2 = 2.

Hope to adopt!