**Let Z be a standard normal random variable and calculate the following probabilities, drawing pictures whenever…**

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## Question “Let Z be a standard normal random variable and calculate the following probabilities, drawing pictures whenever…”

Let *Z* be a standard normal random variable and

calculate the following probabilities, drawing pictures whenever

appropriate. (Do this on paper. Your instructor may ask you to turn

in this work.)

## Answer

A *Z* score indicates how far an element has deviated from its mean. *Z*-scores can be either positive or negative. A positive score indicates that the score is higher than the mean value. A negative score indicates that the score is lower than the mean. *Z*_score can help you determine the normal probabilities.

This problem deals with the concept of finding normal area probabilities using known * and z* scores.

Table of areas under normal distributions gives probabilities. You can also use the Excel function $math_tag_0 to calculate the probability.

The table shows the probabilities for the P\left( {Z < z} \right). form

(a)

The $math_tag_2 shaded area is in the normal curve

The area to the right of z = 2.74

The area to the right of $math_tag_2

The area between $math_tag_2 and $math_tag_3

\begin{array}{c}\\P\left( {0 < z < 2.74} \right) = 0.9969 - 0.5000\\\\ = 0.4969\\\end{array}(b)

The $math_tag_2 shaded area is in the normal curve

The area to the right of $math_tag_8

The area to the right of z = 0

The area between $math_tag_2 and $math_tag_3

\begin{array}{c}\\P\left( {0 < z < 1} \right) = 0.8413 - 0.5000\\\\ = 0.3413\\\end{array}(c)

The $math_tag_12 shaded area is in the normal curve

The area to the right of $math_tag_12

The area to the right of z = 0

The area between $math_tag_12 and $math_tag_12

\begin{array}{c}\\P\left( { - 2.40 < z < 0} \right) = P\left( {z < 0} \right) - P\left( {z < - 2.40} \right)\\\\ = 0.5000 - 0.0082\\\\ = 0.4918\\\end{array}(d)

The $math_tag_12 shaded area is in the normal curve

The area to the right of $math_tag_12

The area to the right of $math_tag_19

The area between $math_tag_12 and $math_tag_12

\begin{array}{c}\\P\left( { - 2.40 < z < 2.40} \right) = P\left( {z < 2.40} \right) - P\left( {z < - 2.40} \right)\\\\ = 0.9918 - 0.0082\\\\ = 0.9836\\\end{array}(e)

The $math_tag_22 shaded area is in the normal curve

The area to the right of $math_tag_22

This is it.

P\left( {z \le 1.63} \right) = 0.9484(f)

The $math_tag_25 shaded area is in the normal curve

The area to the right $math_tag_25

This is it.

\begin{array}{c}\\P\left( { - 1.74 \le z} \right) = P\left( {z \ge - 1.74} \right)\\\\ = 1 - P\left( {z < - 1.74} \right)\\\\ = 1 - 0.0409\\\\ = 0.9591\\\end{array}(g)

The $math_tag_28 shaded area is in the normal curve

The area to the right of $math_tag_28

The area to the right of $math_tag_30

The area between $math_tag_28 and $math_tag_28

\begin{array}{c}\\P\left( { - 1.40 < z < 2.00} \right) = P\left( {z < 2.00} \right) - P\left( {z < - 1.40} \right)\\\\ = 0.9772 - 0.0808\\\\ = 0.8964\\\end{array}(h)

The $math_tag_22 shaded area is in the normal curve

The area to the right of $math_tag_22

The area to the right of $math_tag_35

The area between $math_tag_22 and $math_tag_22

\begin{array}{c}\\P\left( {1.63 < z < 2.50} \right) = P\left( {z < 2.50} \right) - P\left( {z < 1.63} \right)\\\\ = 0.9938 - 0.9484\\\\ = 0.0453\\\end{array}(i)

The $math_tag_38 shaded area is in the normal curve

The area right of $math_tag_39

This is it.

\begin{array}{c}\\P\left( {1.4 \le z} \right) = P\left( {z \ge 1.4} \right)\\\\ = 1 - P\left( {z < 1.4} \right)\\\\ = 1 - 0.9192\\\\ = 0.0808\\\end{array}(j)

The $math_tag_41 shaded area is in the normal curve

The area to the right of $math_tag_41

The area to the right of $math_tag_35

The area between $math_tag_41 and $math_tag_41

\begin{array}{c}\\P\left( { - 2.50 < z < 2.50} \right) = P\left( {z < 2.50} \right) - P\left( {z < - 2.50} \right)\\\\ = 0.9938 - 0.0062\\\\ = 0.9876\\\end{array}Ans Part a

The area between z = 0Part B is therefore

The area between z = 0Part C is therefore

The area between z = - 2.40Part D is therefore

The area between z = - 2.40Part E is therefore

The area to the left is 1.63.

Part f

The area to the right is z = - 1.74Part G.

The area between z = - 1.40Part H is therefore.

The area between z = 1.63Part I is therefore

The area to the right is z = 1.40Part J.

The area between $math_tag_41 and $math_tag_41 is therefore

## Conclusion

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