4. Correct. The answer is false

Let X_A* and X_B* denote the mean lifetimes of A and B respectively. Then

m_{XA*- XB*}=m_XA*-m_XB*=1500-1300=200 hr.

and

The standardized variable for the difference in means is

Z=[(X_A^*-X_B^*)-(m_XA*-XB*)]/s_XA*-XB*=[(X_A^*-X_B^*)-200]/20 and is nearly normally distributed.

a) The difference 160 hours in standard units = (160-200)/20 = -2.

Required probability = (area under the normal curve to the right of z=-2) =0.5+0.4772=0.9772

b) The difference 250 hours in standard units = (250-200)/20=2.50.

Required probability = (area under normal curve to the right of z=2.50) = 0.5-0.4938=0.0062