Answer to Question #275807 in Statistics and Probability for Gon

Question #275807

Assume z is a normal random variable with mean 0 and standard deviation 1.

1
Expert's answer
2021-12-07T13:44:15-0500

ZN(0,1)Z\sim N(0,1)

a)


P(1.55<Z<1.86)P(-1.55<Z<1.86)

=P(Z<1.86)P(Z1.55)=P(Z<1.86)-P(Z\leq -1.55)

0.96855720.0605708\approx0.9685572-0.0605708

0.907986\approx0.907986

b)


P(X<1.55 or x>1.86)P(X<-1.55\ or\ x>1.86)

=1P(1.55<X<1.86)=1-P(-1.55<X<1.86)

10.907986\approx1-0.907986

0.092014\approx0.092014

c)


P(Z>z)=0.015P(Z>z)=0.015

1P(Zz)=0.0151-P(Z\leq z)=0.015

P(Zz)=0.985P(Z\leq z)=0.985

z2.1701z\approx2.1701


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