Question: Find the following probabilities for the standard normal random variable z:

a.P(z>1.46)b.P(z<-1.56)c.P(.67z<2.41)d.P(-1.96z-.33)e.P(Z0)f.P(-2.33<z<1.50)

Short Answer

Expert verified

Answer

A random variable is a mathematical expression of a statistical study's result.

Step by step solution

01

(a) The data is given below

The calculation is given below:

P(z>1.46)=1-P(z1.46)=1-0.9279=0.0721

02

(b) The data is given below

The calculation is given below:

P(z<-1.56)=0.0594

03

(c) The data is given below

The calculation is given below:


P(.67z<2.41)=P(.67z<2.41)-P(z<.0.67)=0.99920-0.7486=0.2434

04

(d) The data is given below

The calculation is given below:

P(-1.96z-.33)=P(-1.96Z<0)-P(-0.33Z<0)=0.4750-0.1293=0.3457

05

(e) The data is given below

The calculation is given below:

P(Z0)=1-P(z<0)=1-0.5=.5

06

(f) The data is given below

The calculation is given below:

P(-2.33<z<1.50)=P(-2.33<z<0)+P(0<Z<-1.50)

=0.4901+0.4332=0.9233

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Most popular questions from this chapter

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Mean

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91.28 92.83 89.35 91.90 82.85 94.83 89.83 89.00 84.62

86.96 88.32 91.17 83.86 89.74 92.24 92.59 84.21 89.36

90.96 92.85 89.39 89.82 89.91 92.16 88.67 89.35 86.51

89.04 91.82 93.02 88.32 88.76 89.26 90.36 87.16 91.74

86.12 92.10 83.33 87.61 88.20 92.78 86.35 93.84 91.20

93.44 86.77 83.77 93.19 81.79

Descriptive statistics(Quantitative data)

Statistic

Content

Nbr.of Observation

50

Minimum

81.79

Maximum

94.83

1st Quartile

87.2725

Median

89.375

3rd Quartile

91.88

Mean

89.2906

Variance(n-1)

10.1343

Standard deviation(n-1)

3.1834

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