Question

Find the value of the standard normal random variable ?, called ?0 such that:

(a)  ?(?≤?0)=0.8651
?0=

(b)  ?(−?0≤?≤?0)=0.2858
?0=

(c)  ?(−?0≤?≤?0)=0.6074
?0=

(d)  ?(?≥?0)=0.0888
?0=

(e)  ?(−?0≤?≤0)=0.3981
?0=

(f)  ?(−1.24≤?≤?0)=0.4718
?0=

Answer 1

a)

P(Z <= z0) = 0.8651

From Z table, z-score for the probability of 0.8651 is 1.10

z0 = 1.10

b)

P ( -z0 <= Z <= z0) = 0.2858

P(Z <= z0) - P(Z <= -z0) = 0.2858

P(Z <= z0) - ( 1 - P(Z < z0) ) = 0.2858

P(Z <= z0) - 1 +  P(Z < z0) = 0.2858

2 P(Z < z0) = 1.2858

P(Z < z0) = 0.6429

From Z table, z-score for the probability of 0.6429 is 0.37

z0 = 0.37

c)

P ( -z0 <= Z <= z0) = 0.6074

P(Z <= z0) - P(Z <= -z0) = 0.6074

P(Z <= z0) - ( 1 - P(Z < z0) ) = 0.6074

P(Z <= z0) - 1 +  P(Z < z0) = 0.6074

2 P(Z < z0) = 1.6074

P(Z < z0) = 0.8037

From Z table, z-score for the probability of 0.8037 is 0.85

z0 = 0.85

d)

P(Z >= z0) = 0.0888

P(Z <= z0) = 1 - 0.0888

P(Z <= z0) = 0.9112

From Z table, z-score for the probability of 0.9112 is 1.35

z0 = 1.35

e)

P ( -z0 <= Z <= 0) = 0.6981

P(Z <= 0) - P(Z <= -z0) = 0.3981

P(Z <= 0) - ( 1 - P(Z < z0) ) = 0.3981

P(Z <= 0) - 1 +  P(Z < z0) = 0.3981

P(Z < z0) = 1.3981 + P(Z <= 0)

P(Z < z0) = 1.3981 - 0.5

P(Z < z0) = 0.8981

From Z table, z-score for the probability of 0.8981 is 1.27

z0 = 1.27

f)

P(-1.24 <= Z <= z0) = 0.4718

P(Z <= z0) - P(Z < -1.24) = 0.4718

P(Z <= z0) - 0.1075 = 0.4718

P(Z <= z0) = 0.5793

From Z table , z-score for the probability of 0.5793 is 0.20

z0 = 0.20