Question

Let z denote the standard normal random variable. Find the value of z0z0 such that:

(a)  P(z≤z0)=0.89P(z≤z0)=0.89
z0=z0=

(b)  P(−z0≤z≤z0)=0.039P(−z0≤z≤z0)=0.039
z0=z0=

(c)  P(−z0≤z≤z0)=0.1112P(−z0≤z≤z0)=0.1112
z0=z0=

(d)  P(z≥z0)=0.0497P(z≥z0)=0.0497
z0=z0=

(e)  P(−z0≤z≤0)=0.4874P(−z0≤z≤0)=0.4874
z0=z0=

(f)  P(−2.03≤z≤z0)=0.5540P(−2.03≤z≤z0)=0.5540
z0=z0=

Answer 1

We have used standard normal distribution table (Z table) to calculate these values.

(a)

P(z≤z0)=0.89

=> z0 = 1.2265

(b)

P(−z0≤z≤z0)=0.039

-=> P(0 ≤ z≤ z0) = 0.039/2 = 0.0195

=> P(z ≤ z0) - P(z ≤ 0) = 0.0195

=> P(z ≤ z0) -  0.5 = 0.0195

=> P(z ≤ z0) = 0.5195

=> z0 = 0.0489

(c)

P(−z0≤z≤z0)=0.1112

-=> P(0 ≤ z≤ z0) = 0.1112/2 = 0.0556

=> P(z ≤ z0) - P(z ≤ 0) = 0.0556

=> P(z ≤ z0) -  0.5 = 0.0556

=> P(z ≤ z0) = 0.5556

=> z0 = 0.1398

(d)

P(z≥z0)=0.0497

z0 = 1.6478

(e)

P(−z0≤z≤0)=0.4874

-=> P(0 ≤ z≤ z0) = 0.4874/2 = 0.2437

=> P(z ≤ z0) - P(z ≤ 0) = 0.2437

=> P(z ≤ z0) -  0.5 = 0.2437

=> P(z ≤ z0) = 0.7437

=> z0 = 0.6548

(f)

P(−2.03≤z≤z0)=0.5540

=> P(z ≤ z0) - P(z ≤ -2.03) = 0.5540

=>  P(z ≤ z0) - 0.0212 = 0.5540

=>  P(z ≤ z0) = 0.5540 + 0.0212 = 0.5752