Question

For standadrd normal random variable Z, find (i)

p(0 < Z < 1.35), (ii) p(-1.04 < Z < 1.45), (iii) p(-1.40

< Z < -0.45), (iv) p(1.17 < Z < 1.45), (v) p( Z < 1.45), (vi) p(1.0 < Z < 3.45)

Answer 1

Note: Z~N(0,1), so cdf of normal distribution= P(Z<z) = P(Z<=z) = pnorm(z) (in R studio)

i) P(0<Z<1.35) = P(Z<1.35) - P(Z<=0) = 0.911492- 0.5 = 0.411492

ii) P(-1.04<Z<1.45) = P(Z<1.45)- P(Z<= -1.04) = 0.9264707-0.14917 = 0.7773007

iii) P(-1.40< Z<-0.45) = P(Z< -0.45) - P(Z<=-1.4) = 0.3263552 - 0.08075666 = 0.245599

iv) P(1.17<Z<1.45) = P(Z<1.45) - P(Z<=1.17) = 0.9264707 -0.8789995 = 0.0474712

v) P(Z<1.45) = 0.9264707

vi) P(1<Z<3.45) = P(Z<3.45)- P(Z<=1) = 0.9997197 - 0.8413447 = 0.158375

R Output:

> pnorm(1.35)
[1] 0.911492
> pnorm(0)
[1] 0.5
> pnorm(1.45)
[1] 0.9264707
> pnorm(-1.04)
[1] 0.14917
> pnorm(-0.45)
[1] 0.3263552
> pnorm(-1.4)
[1] 0.08075666
> pnorm(1.17)
[1] 0.8789995
> pnorm(3.45)
[1] 0.9997197
> pnorm(1)
[1] 0.8413447