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
Above is the solution for “Let Z be a standard normal random variable and calculate the following probabilities, drawing pictures whenever…“. We hope that you find a good answer and gain the knowledge about this topic of math.