**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_{}