Verified answer

I think the answer is D. The mean is 80, so technically the sample mean should be the same as the population mean. The only thing that changes with population size is the variance.

a) If X denotes the score, and Y=(X-80)/20, then P[seventy 8<X<80 two]=P[(seventy 8-80)/20<Y<(80 two-80)/20] =P[-0.a million<Y<0.a million]=section between -0.a million and 0.a million under the widespread known curve. you could look this up from the traditional hazard table. b) If X denotes the mean, then X follows a classic distribution with mean=80 and classic deviation=20/sqrt(4)=10. permit Y=(X-80)/10 P[seventy 8<X<80 two]=P[(seventy 8-80)/10<Y<(80 two-80)/10] =P[-0.2<Y<0.2]=section between -0.2 and 0.2 under stanadrd known curve. observe that this section would be extra beneficial than the section in (a) because of the fact the era (-0.2,0.2) is larger and thoroughly covers(-0.a million,0.a million) c) If X denotes the mean, then X follows a classic distribution with mean=80 and classic deviation=20/sqrt(25)=5. permit Y=(X-80)/4 P[seventy 8<X<80 two]=P[(seventy 8-80)/4<Y<(80 two-80)/4] =P[-0.5<Y<0.5]=section between -0.5 and 0.5 under widespread known curve. observe that this section would be extra beneficial than the section in (b) because of the fact the era (-05,0.5) is larger and thoroughly covers(-0.2,0.2)