4. Incorrect. The answer is false not true.

 

Let XA* and XB* denote the mean lifetimes of A and B respectively. Then

mXA*- XB*=mXA*-mXB*=1500-1300=200 hr.

and

The standardized variable for the difference in means is

Z=[(XA*-XB*)-(mXA*-XB*)]/sXA*-XB*=[(XA*-XB*)-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