Question

Find the value of the standard normal random variable zz , called z0z0 such that:

(a) P(z≤z0)=0.9999P(z≤z0)=0.9999
z0=z0=

(b) P(−z0≤z≤z0)=0.922P(−z0≤z≤z0)=0.922
z0=z0=

(c) P(−z0≤z≤z0)=0.3954P(−z0≤z≤z0)=0.3954
z0=z0=

(d) P(z≥z0)=0.4497P(z≥z0)=0.4497
z0=z0=

(e) P(−z0≤z≤0)=0.3225P(−z0≤z≤0)=0.3225
z0=z0=

(f) P(−1.66≤z≤z0)=0.5474P(−1.66≤z≤z0)=0.5474
z0=z0=

Answer 1

solution=

a. P (z ≤ z0) = 0.9999=P(z ≥ z0) = 1-0.9999 = 0.0001 = P(z ≤ -z0) = 0.0001
From tables z0 = -3.72

b. P (-z0 ≤ z ≤ z0) = .922 = P (z ≤ z0)- P (z ≤ -z0) = P (z ≤ z0)- P (z ≥ z0) =
P (z ≤ z0)-(1- P (z ≤ z0))
P (z ≤ z0) = (0.922+1)/2=0.961
From tables z0 = 1.76

c. P (-z0 ≤ z ≤ z0) = 0.3954
the procedure is the same that exercise b P (z ≤ z0) = (0.3954 +1)/2=0.6977
From tables the nearest value is z0 = 0.52

d. P (z ≥ z0) = 0.4497 = 1- P (z ≤ z0)
P (z ≤ z0) = 0.4497
From tables value z0 = 0.15

e. P (-z0 ≤ z ≤ 0) = 0.3225= P (z ≤ 0) - P (z ≤ -z0) = P (z ≤ 0) - P (z ≥ z0) =
P (z ≤ 0) - (1- P (z ≤ z0))
P (z ≤ z0) = 0.3225 + 1 - P (z ≤ 0)= 0.3225 + 1 - 0.5 = 0.8225
From tables z0 = 0.92

f. P (-1.66 ≤ z ≤ z0) = 0.5474 = P (z ≤ z0) - P (z ≤ -1.66) = P (z ≤ z0) - P (z ≥ 1.66) =
P (z ≤ z0) - (1- P (z ≤ 1.66))
P (z ≤ z0) = 0.5474 + 1 - P(z ≤ 1.66) = 0.5474 + 1 - 0.9515 = 0.5959
From tables the nearest value is z0 = 0.24