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!