Question #216530

Given that z is a standard normal random variable, compute the following

probabilities.

a) P (Z < 0.49)

b) P (0.52 < Z)

c) P (- 1.98 ≤ Z ≤ .49)

d) P (.52 ≤ Z ≤ 1.22)

e) P (- 1.75 ≤ Z ≤ - 1.04)


Expert's answer

a)

P(Z<0.49)=0.687933P(Z<0.49)=0.687933

b)

P(Z>0.52)=0.301532P(Z>0.52)=0.301532


P(Z<0.52)=0.698468P(Z<0.52)=0.698468

c)


P(−1.98<Z<0.49)=P(Z<0.49)−P(Z≤−1.98)P(-1.98<Z<0.49)=P(Z<0.49)-P(Z\leq-1.98)

≈0.6879331−0.0238518≈0.664081\approx0.6879331-0.0238518\approx0.664081

d)


P(0.52≤Z≤1.22)=P(Z≤1.22)−P(Z<0.52)P(0.52\leq Z\leq 1.22)=P(Z\leq1.22)-P(Z<0.52)

≈0.8887678−0.6984682≈0.190300\approx0.8887678-0.6984682\approx0.190300

e)


P(−1.75≤Z≤−1.04)=P(Z≤−1.04)−P(Z<−1.75)P(-1.75\leq Z\leq -1.04)=P(Z\leq-1.04)-P(Z<-1.75)

≈0.1491700−0.0400592≈0.109112\approx0.1491700-0.0400592\approx0.109112


LATEST TUTORIALS
APPROVED BY CLIENTS