Question

Variable Z is distributed standard normal: Z ~ N(0; 1), so using the statistical table of the Standard Normal Distribution provided, find the values of the following probabilities:

(A) P[Z ≤ 1.27] =

(B) P[Z ≥ 0.41] =

(C) P[Z ≤ 1:96] =

(D) P[-1 ≤ Z ≤ 2] =

(E) P(Z ≤ 1.87)

(F) P(Z ≥ - 0.53). Hint: P(Z ≤ - a)=P(Z ≥ a), where a ∈ ?, and a≥0.

(G) P(Z ≤ - 0.06)

(H) P(1.12 ≤ Z ≤ 2.38). Hint: P(a ≤ Z ≤ b) = P(Z ≤ b) - P(Z ≤ a), where a, b ∈ ?.

(I) Find the zvalue such that P(Z ≤ zvalue)=0.975.

Answer 1

a)

P(Z <= 1.27) = 0.8980

b)

P(Z >= 0.41) = 1 - P(Z < 0.41)

= 1 - 0.6591

= 0.3409

c)

P(Z <= 1.96) = 0.9750

d)

P(-1 <= Z <= 2) = P(Z <=2 ) - P(Z <= -1)

= P(Z <= 2) - ( 1 - P(Z <= 1) )

= 0.9772 - ( 1- 0.8413)

= 0.8185

e)

P(Z <= 1.87) = 0.9693

f)

P(Z >= -0.53) = P(Z <= 0.53)

= 0.7019

g)

P(Z <= -0.06) = 1 - P(Z <= 0.06)

= 1 - 0.5239

= 0.4761

h)

P(1.12 <= Z <= 2.38) = P(Z <= 2.38)- P(Z <= 1.12)

= 0.9913 - 0.8686

= 0.1227

I )

We have to calculate z such that P(Z <= z) = 0.9750

From Z table, z-score for the probability of 0.975 is 1.96

z = 1.96