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Given that z is a standard normal random variable, compute the following probabilities

– $ P (z  \space \leq \space – \space 1.0 )$

– $ P (z  \space \geq \space – \space 1 )$

– $ P (z  \space \geq \space – \space 1.5 )$

– $ P ( – \space 2.5  \space \geq \space  \space z )$

– $ P (- \space 3  \space < \space z \space \geq \space \space 0 )$

The main objective of this question is to find the probabilities for the given expressions given the z score, which is a standard random variable.

This question uses the concept of z-score. The standard normal z-table is the abbreviation for the z-table. Standard Normal models are used in hypothesis testing as well as the differences between two means.  $100 \space % $ of an area under a distribution of normal curve is represented by a value of one hundred percent or $ 1 $. The z-table tells us how much of the curve is below a given point. The z-score is calculated as:

\[ \space z \space = \frac{ score \space – \space mean }{ standard deviation} \]

Expert Answer

We have to compute the probabilities.

a) From the z-table, we know that the value of $ – \space 1 $ is:

\[ \space = \space 0.1587 \]

So:

\[ \space P (z  \space \leq \space – \space 1.0 )  \space = \space 0.1587 \]

b) Given that:

\[ \space P (z  \space \geq \space – \space 1 ) \]

Thus:

\[ \space = \space 1 \space – \space P (z  \space \leq \space – \space 1 ) \]

We know that:

\[ \space P (z  \space \leq \space – \space 1.0 )  \space = \space 0.1587 \]

So:

\[ \space = \space 1 \space – \space 0.1587 \]

\[ \space = \space 0.8413 \]

c) Given that:

\[ \space P (z  \space \geq \space – \space 1.5 ) \]

So:

\[ \space = \space 1 \space – \space P(z \space \leq \space – \space 1.5 \]

\[ \space = \space 1 \space – \space 0.0668 \]

\[ \space = \space 0.9332 \]

d) Given that:

\[ \space P ( – \space 2.5  \space \geq \space  \space z ) \]

So:

\[ \space P(z \space \geq \space – \space 2.5) \]

\[ \space 1 \space – \space P(z \space \leq \space – \space 2.5) \]

\[ \space = \space 1 \space – \space 0.0062 \]

\[ \space = \space 0.9938 \]

e) Given that:

\[ \space P (- \space 3  \space < \space z \space \geq \space \space 0 ) \]

So:

\[ \space P(z \space \leq \space 0) \space – \space P(z \leq \space – \space 3) \]

\[ \space 0.5000 \space – \space 0.0013 \]

\[ \space = \space 0.4987 \]

Numerical Answer

The probability for the $ P (z  \space \leq \space – \space 1.0 )$  is:

\[ \space = \space 0.1587 \]

The probability for the $ P (z  \space \geq \space – \space 1 ) $  is:

\[ \space = \space 0.8413 \]

The probability for the $  P (z  \space \geq \space – \space 1.5 )$  is:

\[ \space = \space 0.9332 \]

The probability for the $   P ( – \space 2.5  \space \geq \space  \space z )$  is:

\[ \space = \space 0.9938 \]

The probability for the $ P (- \space 3  \space < \space z \space \geq \space \space 0 )$  is:

\[ \space = \space 0.4987 \]

Example

Find the probability for  $ z $ which is a standard random variable.

\[ \space  P (z  \space \leq \space – \space 2.0 ) \]

We have to compute the probabilities. From the z-table, we know that the value of $ – \space 2 $ is:

\[ \space = \space 0.228 \]

So:

\[ \space P (z  \space \leq \space – \space 1.0 )  \space = \space 0.228 \]

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