急 ! F5 about Dispersion 13q3

13) The standard score of Kin Ming in arecital competition is -2.0. If there are 400 participants in the recitalcompetition and their marks are normally distributed, find the number ofparticipants whose marks are higher than Kin Ming.

P(Z<-2.0)= 0.02275 [By using Excel function normsdist(-2)]
the no. of participants whose marks are higher than Kin Ming
=(1-0.02275)*400
=390.89 [~391]
14) In a singing contest tryout, the meanmark of the participants is 7.9 and the standard deviation is 0.64. Suppose themarks of the participants are normally distributed, and the participants withmarks in the lowest 97.5% would be eliminated. For those who were selected.(a) what is the lowest standard score ?
P(Z<K)=0.975 [By using Excel function normsinv(0.975)]
K=1.959964
The lowest standard score is 1.959964 (b) what is the lowest mark ?
(X-7.9)/0.64 = 1.959964
X=0.64*1.959964+7.9
=9.154377