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Chapter 4: Q166SE (page 281)

Find az-score, call itz0, such that

a.P(z≤z0)=.5080

b.P(z≥z0)=.5517

c.P(z≥z0)=.1492

d.P(z0≤z≤.59)=.4773

Short Answer

Expert verified

a.z0=0.02

b.z0=-0.129

c. z0=1.039

d. z0=-0.689

Step by step solution

01

Given information

z-score is denoted by.z0

02

CalculateP(z≤z0)=.5080

P(z≤z0)=.5080Φ(z0)=.5080z0=Φ−1(.5080)z0=0.02

Hence.z0=0.02

03

CalculateP(z≥z0)=.5517

P(z≥z0)=.55171−P(z≤z0)=.55171−Φ(z0)=.5517Φ(z0)=1−.5517Φ(z0)=0.4483z0=Φ−1(0.4483)z0=-0.129

Hencez0=-0.129.

04

CalculateP(z≥z0)=.1492

P(z≥z0)=.14921−P(z≤z0)=.14921−Φ(z0)=.1492Φ(z0)=1−.1492Φ(z0)=0.8508z0=Φ−1(0.8508)z0=1.039

Hence.z0=1.039

05

CalculateP(z0≤z≤.59)=.4773

P(z0≤z≤.59)=.4773P(z≤.59)−P(z≤z0)=.4773P(z≤z0)=P(z≤.59)−.4773Φ(z0)=0.7224−0.4773Φ(z0)=0.2451z0=Φ−1(0.2451)z0=-0.689

Hence z0=-0.689.

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Most popular questions from this chapter

Industrial filling process. The characteristics of an industrialfilling process in which an expensive liquid is injectedinto a container were investigated in the Journal of QualityTechnology(July 1999). The quantity injected per containeris approximately normally distributed with mean 10

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