Question

1. On the standard normal curve, find the following values of z. a. the value of z representing the 75th percentile or upper

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Answer #1

Solution(1a)

Given in the question

P-value =0.75

From Z table we found z value

So z value = 0.675

Solution(1b)

P-value =0.15

So from z table we found z value

Zvalue = -1.04

Solution(1c)

P-value =0.75

And z value is 0.675

Solution(2)

Z =1.21

This is left tailed test so p-value from z table is

P(Z<=1.21)= 0.8869

Solution(3)

Z =2.48

And this is right tailed test

So P(Z>2.48)= 0.0066

Solution(4)

P(-1.30<Z<1.75) = P(Z<1.75)-P(Z<-1.30)

= 0.9599- 0.0968 = 0.8631

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