here we use standard normal variate z=(x-mean)/sd
(a) required answer is 0.8904
for x=6.2, z=(6.2-7)/0.5=-1.6
for x=7.8, z=(7.8-7)/0.5=1.6
P(6.2<X<7.8)=P(-1.6<Z<1.6)=P(Z<1.6)-P(Z<-1.6)=0.9452-0.0548=0.8904
P(Z<1.6)=0.9452( using ms-excel=normsdist(1.6))
P(Z<-1.6)=0.0548 ( using ms-excel=normsdist(-1.6))
(b) The normal distribution is a continuous distribution and hence the probability of getting exactly 7 is zero.
(c) answer is 0.7257
for x=7.3, z=(7.3-7)/0.5=0.6
P(X<7.3)=P(Z<0.6)=0.7257 ( using ms-excel=normsdist(0.6))
(d) answer is 0.0227
for x=8, z=(8-7)/0.5=2
P(cup will overflow)=P(X>8)=1-P(X<8)=1-P(Z<2)=1-0.9773=0.0227
(e) answer is 8.165
here we find x such that P(X>x)=0.01,
for this first we find z such that P(Z>z)=0.01
or, P(Z<z)=1-P(Z>z)=1-0.01=0.99
and z=2.33 and corresponding x=7+2.33*0.5=8.165
(g) answer is 0.9999
here z=(x--mean)/(sd/sqrt(n))=(7.4-7)/(0.5/sqrt(25))=4
P(X-<7.4)=P(Z<4)=0.9999
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