Find the following probability for the standard normal random variable z

– $ 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:

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

Video Transcript

For this question, there is a standard normal of distribution here for standard normal distribution, the mean value which is denoted by mood that should be 0 and the standard deviation, which is not by sigma. That should be 1, so i can define the random variable for the standard, normal distribution. It has the mean value 0 and the standard deviation is 1, so we're going to find the values by using this mean and the standard division. So, for part a what i have to do, i need to just find the c is greater than 1.32. So what that means. So in this case, i'm going to use the normal c d f function, so the lower boundary is 1.32 and the upper boundary there is no in upper ondara. We just graph the normal distribution here. This is the mean value where this is 0 and this is 1.32, so we have to get the are of this 1, so there is no upper boundary here that goes to positive infinity. So i'm going to put a very big number here and the mean value is a standard equation is 1. Let'S get the answer. Just pressed second and distribution, there's normal c d f over boundary, which is 1.2, the upper boundary is 1. This is second 99 and the mean value is 0. Standard equation is 1. Let'S get the answer, so we have to run 3 decimal places and these would be 0.093. This is the answer for part a what about for part b. It says what about the probability of c is less than negative, 1.99 n. So what that means? If i just use the same graph here so negative 1.99 is here so we have to get this blue shaded region. So when we look at the lower boundary that goes to negative infinity again, i'm going to use the normal cth f function and the lower boundary that goes to negative infinite. I'M going to put a very small number and the upper boundary is negative 1.99 and the mean value is your standard. Deviation is 1, let's get the answer just pre second and distribution, or this is quite second distribution, normal c d. F over battery negative 1 point: this is 99 and the upper boundary is 1.99 and the mean value is a standard equation is 1, so the answer would be 0.023, that's and what about 4 parts? So there is an interval here which is 0.62 less than or equal to. This is z and less than or equal to 2.30 2 point again, i'm going to use the same function, which is normal c d f. The lover boundary is 0.62. Upper boundary is 2.20 and 1. Let'S get answered, second distribution, normal chelower, boundary .62 upper boundary is 2.32 and it is and 1 at so the answer would be a 0.257. That'S it. Okay, let's keep solving with the choice here so again, there's an interval here, which is between negative 1.82 less than or equal to z. Less than this is so. These are continues variable so so less than or equal or less than doesn't matter anything for the result. So again, i'm going to use the normal c d f and the lower boundary would be negative. 1.82 upper boundary is negative 0.58 that is 0 and 1. So the answer would be a second servation known as c d fate, 1.82, a propounder is negative, or this is negative 0.58, and this is 0 and 1, so the answer would be 0.247 rat and the next 1. Let'S say this was e, which is greater than 0 point, so that z is equal to 0. Is the middle point for this distributive for this graph? So, under the curve, the total probity is 1 or the total area is 1, so t is greater than 0, which is 1 half that is equal to 0.5 tit, and the next 1 is about again an interval. The lower boundary is negative, 2.2, .63 z and 1.4 8 point. So again, i'm going to use the normal c d f over boundary, negative 2.63, the upper boundary is 1.48 and the mean value is our standard. Division is 1, so just per second distribution, normal. The lover boundary negative 2.63 and the preponderis 1.48. This is 0 and 1 and let's get the answer here. This is 0.92 and 6 that i and the next 1, which is g, so the probability of z, is greater than or equal to negative. 2.20. 1 point again: i'm going to use the normalcy function, the lower boundary negative 2.21, the upper boundary- there is no input boundary here and the mean value is your standard. Devision is 1. So, let's get the answer here from this is second distribution normal covering 2.21. The upper boundary is 10 and 1, so the answer would be 0.98 and 6 great and the last 1. So the last 1 says the probability of z is less than this is 2.21. Okay and again i'm going to use the normal c d. F lover: boundary that is negative infinity, so the preponderous 01 second distribution, normal the f over boundary negative 1. This is second and 99, and upper boundary is 2.21. That is 0 and 1. So the answer is 0.986. That'S it! So we got the answer for.

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