Question

In each case, determine the value of the constant c that makes the probability statement correct. (Round your answers to two decimal places.)

(a)    Φ(c) = 0.9850

(b)    P(0 ≤ Z ≤ c) = 0.2939

(c)    P(c ≤ Z) = 0.1251

(d)    P(−c ≤ Z ≤ c) = 0.6528

(e)    P(c ≤ |Z|) = 0.0128

Answer 1

Part a

φ(c) = 0.9850

c = φ’(0.9850)

c = 2.17009

(by using z-table)

c = 2.17

Part b

P(0≤Z≤c) = 0.2939

P(0≤Z≤c) = P(Z<c) – P(Z<0)

P(Z<c) – P(Z<0) = 0.2939

P(Z<c) – 0.5 = 0.2939

P(Z<c) = 0.2939 + 0.5 = 0.7939

c = 0.820028

(by using z-table)

c = 0.82

Part c

P(c≤Z) = 0.1251

P(c≤Z) = P(Z≥c) = P(Z>c) = 1 – P(Z<c) = 0.1251

P(Z<c) = 1 – 0.1251 = 0.8749

P(Z<c) = 0.8749

c = 1.149864

(by using z-table)

c = 1.15

Part d

P(-c≤Z≤c) = 0.6528

P(Z<-c) + P(Z>c) = 1 – 0.6528

P(Z<-c) + P(Z>c) = 0.3472

P(Z<-c) = 0.3772/2 = 0.1736

[For normal distribution, P(Z<-c) = P(Z>c)]

-c = -0.94003

(by using z-table)

c = 0.94

Part e

P(c≤|Z|) = 0.0128

P(-c<Z<c) = 0.0128

P(Z<-c) + P(Z>c) = 1 – 0.0128 = 0.9872

P(Z<-c) + P(Z>c) = 0.9872

2*P(Z<-c) = 0.9872

P(Z<-c) = 0.4936

[For normal distribution, P(Z<-c) = P(Z>c)]

-c = -0.01604

(by using z-table)

c = 0.02