Given that z is a standard normal random variable, use the Excel to compute the following probabilities.
a) P(z > 0.5)
b) P(z ≤ −1)
c) P(1≤ Z ≤ 1.5)
d) P(0.5 ≤ z ≤ 1.25)
e) P(0 < z < 2.5)
Z~Normal(0,1)
a) P(z > 0.5)=1-p(z<0.5)
=1-0.6914
=0.3085
#value of z is obtained from Excell command is =NORM.S.DIST(B5,1)
b) P(z ≤ −1)=0.1587
c) P(1≤ Z ≤ 1.5)=P( Z ≤ 1.5)-) P(Z ≤ 1)
=0.9332-0.8413=0.0918
# P(1≤ Z ≤ 1.5)=0.0918
d) P(0.5 ≤ z ≤ 1.25)= P( z ≤ 1.25)- P( z ≤ 0.5)
=0.8944-0.6915=0.2029
# P(0.5 ≤ z ≤ 1.25)=0.2029
e) P(0 < z < 2.5)=p(z<2.5)-p(z<0)
=0.9938-0.5=0.4938
#P(0 < z < 2.5)=0.4938
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