Answer: We know when x∼N(mu,sigma^(2)) then (x-mu)/(sigma)∼N(0,1)let x be scores on statistics finalX⇝N(78,(12)^(2))We have to find P[x < 71] and P[x⩾95]{:[P[x < 71],=P[(x-mu)/(sigma) < (71-mu)/(sigma)]],[,=P[z < (71-78)/(12)],],[,=P[z < -0.5833],],[,=0.2798,:'" from standard ... See the full answer