Find a value of the standard normal random variable z, call it z0, such that

a.P(zz0)=0.2090b.P(zz0)=0.7090c.P(-z0z<z0)=0.8472d.P(-z0z<z0)=0.1664e.P(z0zz0)=0.4798f.P(-1<z<z0)

Short Answer

Expert verified

a.z0=-0.81b.z0=0.55c.z0=1.43d.z0=0.21e.z0=-2.05f.P-1<z<z0=Φz0-0.1587

Step by step solution

01

Given information

z is a standard normal variable which is calledz0

02

Finding the value of z0 when P(z≤z0)=0.2090

a.Pzz0=0.2090Φz0=0.2090z0=Φ-10.2090z0=-0.81

Therefore, thez0 score is -0.81.

03

Finding the value of z0 when P(z≤z0)=0.7090

b.P(zz0)=0.7090Φz0=0.7090z0=Φ-10.7090z0=0.55

Therefore, thez0 score is 0.55.

04

Finding the value z0 of when P(- z0≤z< z0)=0.8472

c.P-z0z<z0=0.84722P0z<z0=0.8472P0z<z0=0.4236Pz<z0-Pz<0=0.4236Pz<z0-0.50=0.4236Φz0=0.9236z0=Φ-10.9236z0=1.43

Therefore, thez0 score is 1.43.

05

Finding the value of z0 when P(- z0≤z< z0)=0.1664

d.P-z0z<z0=0.16642P0z<z0=0.1664P0z<z0=0.0832Pz<z0-Pz<0=0.0832Pz<z0-0.50=0.832Pz<z0=0.5832Φz0=0.5832z0=Φ-10.5832z0=0.21

Therefore, thez0 score is 0.21.

06

Finding the value of z0 when P(-1<z<z0)

e.Pz0z<0=0.4798Pz<0)-P(z<z0=0.47980.50-PPz<z0=0.4798Pz<0=0.020Φz0=0.0202z0=-2.05

Therefore, thez0 score is -2.05.

07

Finding the value of z0 when P(-1<z<z0)

f.P-1<z<z0=Pz<z0-Pz-1=Φz0-Φ-1=Φz0-Φ-1=Φz0-0.1587

Therefore, thez0 score is -0.1587.

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