normal distribution - area under the curve [closed]

-1

The area (in percentage) under standard normal distribution curve of random variable Z within limits from -3 to 3 is ____.

Please provide some hints on how to solve this problem.

asked Dec 21 '18 at 9:26

swapnil

335

closed as off-topic by Kavi Rama Murthy, Did, Davide Giraudo, Brahadeesh, Adrian Keister Dec 21 '18 at 14:45

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If this question can be reworded to fit the rules in the help center, please edit the question.

add a comment |

-1

The area (in percentage) under standard normal distribution curve of random variable Z within limits from -3 to 3 is ____.

Please provide some hints on how to solve this problem.

asked Dec 21 '18 at 9:26

swapnil

335

closed as off-topic by Kavi Rama Murthy, Did, Davide Giraudo, Brahadeesh, Adrian Keister Dec 21 '18 at 14:45

This question appears to be off-topic. The users who voted to close gave this specific reason:

"This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Kavi Rama Murthy, Did, Davide Giraudo, Brahadeesh, Adrian Keister

If this question can be reworded to fit the rules in the help center, please edit the question.

add a comment |

-1

The area (in percentage) under standard normal distribution curve of random variable Z within limits from -3 to 3 is ____.

Please provide some hints on how to solve this problem.

asked Dec 21 '18 at 9:26

swapnil

335

The area (in percentage) under standard normal distribution curve of random variable Z within limits from -3 to 3 is ____.

Please provide some hints on how to solve this problem.

probability normal-distribution

asked Dec 21 '18 at 9:26

swapnil

335

asked Dec 21 '18 at 9:26

swapnil

335

asked Dec 21 '18 at 9:26

swapnil

335

asked Dec 21 '18 at 9:26

swapnil

335

asked Dec 21 '18 at 9:26

swapnil

335

closed as off-topic by Kavi Rama Murthy, Did, Davide Giraudo, Brahadeesh, Adrian Keister Dec 21 '18 at 14:45

This question appears to be off-topic. The users who voted to close gave this specific reason:

"This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Kavi Rama Murthy, Did, Davide Giraudo, Brahadeesh, Adrian Keister

If this question can be reworded to fit the rules in the help center, please edit the question.

closed as off-topic by Kavi Rama Murthy, Did, Davide Giraudo, Brahadeesh, Adrian Keister Dec 21 '18 at 14:45

This question appears to be off-topic. The users who voted to close gave this specific reason:

"This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Kavi Rama Murthy, Did, Davide Giraudo, Brahadeesh, Adrian Keister

If this question can be reworded to fit the rules in the help center, please edit the question.

add a comment |

3 Answers
3

active

oldest

votes

You can look up the probability density for z-values between -3 and +3.In the attached snapshots you can see that the P(0<z<3)=0.4987. Since the standard normal distribution is symmetrical, P(-3<z<0)=0.4987 also. Your question is the probability density in between z-values -3 and +3. In other words, P(-3<z<+3), which is the sum of the two densities I highlighted earlier i.e. P(-3<z<+3)=P(-3<z<0)+P(0<z<3)=0.9974

Density table pic 1

Density table pic 2

edited Dec 21 '18 at 10:09

answered Dec 21 '18 at 9:54

Indula

506

$begingroup$
Sorry about my earlier response.
$endgroup$
– Indula
Dec 21 '18 at 10:10

$begingroup$
One more thing. In a standard normal distribution, almost all the density - that is 99% something - is contained within z-values -3 and +3. So 99% is a rough estimate of the probability for z-values in between -3 and +3
$endgroup$
– Indula
Dec 21 '18 at 10:13

$begingroup$
Thanks for those pics got it.
$endgroup$
– swapnil
Dec 21 '18 at 17:33

add a comment |

Hint: the area satisfying $zle a$ is $Phi(a)$, where $Phi$ is the $N(0,,1)$ cdf. So what's the area satisfying $ale zle b$?

answered Dec 21 '18 at 9:30

J.G.

25.3k22539

$begingroup$
Okay. I got the concept, but how to solve a≤z≤b this part. do I have to use integration from -3 to 3 to f(x) where f(x) is 1/sqrt(2*pi)*e^(-x^2/2).
$endgroup$
– swapnil
Dec 21 '18 at 9:43

$begingroup$
@swapnil That's the integral you need. You'll probably obtain it from a resource that can compute $Phi$. Can you express the value you want in terms of two values of $Phi$?
$endgroup$
– J.G.
Dec 21 '18 at 9:47

$begingroup$
As you mentioned a resource, I saw one graph of standard deviation it took values from -3 to 3 having percentage = 2.14+13.59+34.13+34.13+13.59+2.14 = 99.72
$endgroup$
– swapnil
Dec 21 '18 at 9:54

$begingroup$
but I don't know in exam whether they will provide something like this, so is there any other way of doing it. i.e. calculating graph under the curve from 0 to 3 then I will just double it to get required answer.
$endgroup$
– swapnil
Dec 21 '18 at 9:57

$begingroup$
@swapril You should research what's provided in your exam, but you'll likely receive a table of values of $Phi(z)$ for positive values of $z$, which is enough because of the identity $Phi(-z)=1-Phi(z)$. The case $a=-b$ gives $Phi(b)-Phi(a)=2Phi(b)-1$.
$endgroup$
– J.G.
Dec 21 '18 at 10:32

|
show 1 more comment

It is $$A={1over sqrt {2pi}}int_{-3}^3 e^{-x^2over 2}dx$$which can only be calculated numerically using Q-function as follows $$A=Q(-3)-Q(3)=1-2Q(3)approx 99.74%$$

answered Dec 21 '18 at 10:11

Mostafa Ayaz

15.3k3939

$begingroup$
Thank you I will learn more about Q functions.
$endgroup$
– swapnil
Dec 21 '18 at 17:39

$begingroup$
You're welcome. Good luck!
$endgroup$
– Mostafa Ayaz
Dec 23 '18 at 11:31

add a comment |

3 Answers
3

active

oldest

votes

3 Answers
3

active

oldest

votes

Density table pic 1

Density table pic 2

edited Dec 21 '18 at 10:09

answered Dec 21 '18 at 9:54

Indula

506

$begingroup$
Sorry about my earlier response.
$endgroup$
– Indula
Dec 21 '18 at 10:10

$begingroup$
One more thing. In a standard normal distribution, almost all the density - that is 99% something - is contained within z-values -3 and +3. So 99% is a rough estimate of the probability for z-values in between -3 and +3
$endgroup$
– Indula
Dec 21 '18 at 10:13

$begingroup$
Thanks for those pics got it.
$endgroup$
– swapnil
Dec 21 '18 at 17:33

add a comment |

Density table pic 1

Density table pic 2

edited Dec 21 '18 at 10:09

answered Dec 21 '18 at 9:54

Indula

506

$begingroup$
Sorry about my earlier response.
$endgroup$
– Indula
Dec 21 '18 at 10:10

$begingroup$
One more thing. In a standard normal distribution, almost all the density - that is 99% something - is contained within z-values -3 and +3. So 99% is a rough estimate of the probability for z-values in between -3 and +3
$endgroup$
– Indula
Dec 21 '18 at 10:13

$begingroup$
Thanks for those pics got it.
$endgroup$
– swapnil
Dec 21 '18 at 17:33

add a comment |

Density table pic 1

Density table pic 2

edited Dec 21 '18 at 10:09

answered Dec 21 '18 at 9:54

Indula

506

Density table pic 1

Density table pic 2

edited Dec 21 '18 at 10:09

answered Dec 21 '18 at 9:54

Indula

506

edited Dec 21 '18 at 10:09

answered Dec 21 '18 at 9:54

Indula

506

answered Dec 21 '18 at 9:54

Indula

506

answered Dec 21 '18 at 9:54

Indula

506

$begingroup$
Sorry about my earlier response.
$endgroup$
– Indula
Dec 21 '18 at 10:10

$begingroup$
One more thing. In a standard normal distribution, almost all the density - that is 99% something - is contained within z-values -3 and +3. So 99% is a rough estimate of the probability for z-values in between -3 and +3
$endgroup$
– Indula
Dec 21 '18 at 10:13

$begingroup$
Thanks for those pics got it.
$endgroup$
– swapnil
Dec 21 '18 at 17:33

add a comment |

$begingroup$
Sorry about my earlier response.
$endgroup$
– Indula
Dec 21 '18 at 10:10

$begingroup$
One more thing. In a standard normal distribution, almost all the density - that is 99% something - is contained within z-values -3 and +3. So 99% is a rough estimate of the probability for z-values in between -3 and +3
$endgroup$
– Indula
Dec 21 '18 at 10:13

$begingroup$
Thanks for those pics got it.
$endgroup$
– swapnil
Dec 21 '18 at 17:33

Sorry about my earlier response.

– Indula
Dec 21 '18 at 10:10

One more thing. In a standard normal distribution, almost all the density - that is 99% something - is contained within z-values -3 and +3. So 99% is a rough estimate of the probability for z-values in between -3 and +3

– Indula
Dec 21 '18 at 10:13

Thanks for those pics got it.

– swapnil
Dec 21 '18 at 17:33

add a comment |

Hint: the area satisfying $zle a$ is $Phi(a)$, where $Phi$ is the $N(0,,1)$ cdf. So what's the area satisfying $ale zle b$?

answered Dec 21 '18 at 9:30

J.G.

25.3k22539

$begingroup$
Okay. I got the concept, but how to solve a≤z≤b this part. do I have to use integration from -3 to 3 to f(x) where f(x) is 1/sqrt(2*pi)*e^(-x^2/2).
$endgroup$
– swapnil
Dec 21 '18 at 9:43

$begingroup$
@swapnil That's the integral you need. You'll probably obtain it from a resource that can compute $Phi$. Can you express the value you want in terms of two values of $Phi$?
$endgroup$
– J.G.
Dec 21 '18 at 9:47

$begingroup$
As you mentioned a resource, I saw one graph of standard deviation it took values from -3 to 3 having percentage = 2.14+13.59+34.13+34.13+13.59+2.14 = 99.72
$endgroup$
– swapnil
Dec 21 '18 at 9:54

$begingroup$
but I don't know in exam whether they will provide something like this, so is there any other way of doing it. i.e. calculating graph under the curve from 0 to 3 then I will just double it to get required answer.
$endgroup$
– swapnil
Dec 21 '18 at 9:57

$begingroup$
@swapril You should research what's provided in your exam, but you'll likely receive a table of values of $Phi(z)$ for positive values of $z$, which is enough because of the identity $Phi(-z)=1-Phi(z)$. The case $a=-b$ gives $Phi(b)-Phi(a)=2Phi(b)-1$.
$endgroup$
– J.G.
Dec 21 '18 at 10:32

|
show 1 more comment

Hint: the area satisfying $zle a$ is $Phi(a)$, where $Phi$ is the $N(0,,1)$ cdf. So what's the area satisfying $ale zle b$?

answered Dec 21 '18 at 9:30

J.G.

25.3k22539

$begingroup$
Okay. I got the concept, but how to solve a≤z≤b this part. do I have to use integration from -3 to 3 to f(x) where f(x) is 1/sqrt(2*pi)*e^(-x^2/2).
$endgroup$
– swapnil
Dec 21 '18 at 9:43

$begingroup$
@swapnil That's the integral you need. You'll probably obtain it from a resource that can compute $Phi$. Can you express the value you want in terms of two values of $Phi$?
$endgroup$
– J.G.
Dec 21 '18 at 9:47

$begingroup$
As you mentioned a resource, I saw one graph of standard deviation it took values from -3 to 3 having percentage = 2.14+13.59+34.13+34.13+13.59+2.14 = 99.72
$endgroup$
– swapnil
Dec 21 '18 at 9:54

$begingroup$
but I don't know in exam whether they will provide something like this, so is there any other way of doing it. i.e. calculating graph under the curve from 0 to 3 then I will just double it to get required answer.
$endgroup$
– swapnil
Dec 21 '18 at 9:57

$begingroup$
@swapril You should research what's provided in your exam, but you'll likely receive a table of values of $Phi(z)$ for positive values of $z$, which is enough because of the identity $Phi(-z)=1-Phi(z)$. The case $a=-b$ gives $Phi(b)-Phi(a)=2Phi(b)-1$.
$endgroup$
– J.G.
Dec 21 '18 at 10:32

|
show 1 more comment

Hint: the area satisfying $zle a$ is $Phi(a)$, where $Phi$ is the $N(0,,1)$ cdf. So what's the area satisfying $ale zle b$?

answered Dec 21 '18 at 9:30

J.G.

25.3k22539

Hint: the area satisfying $zle a$ is $Phi(a)$, where $Phi$ is the $N(0,,1)$ cdf. So what's the area satisfying $ale zle b$?

answered Dec 21 '18 at 9:30

J.G.

25.3k22539

answered Dec 21 '18 at 9:30

J.G.

25.3k22539

answered Dec 21 '18 at 9:30

J.G.

25.3k22539

answered Dec 21 '18 at 9:30

J.G.

25.3k22539

$begingroup$
Okay. I got the concept, but how to solve a≤z≤b this part. do I have to use integration from -3 to 3 to f(x) where f(x) is 1/sqrt(2*pi)*e^(-x^2/2).
$endgroup$
– swapnil
Dec 21 '18 at 9:43

$begingroup$
@swapnil That's the integral you need. You'll probably obtain it from a resource that can compute $Phi$. Can you express the value you want in terms of two values of $Phi$?
$endgroup$
– J.G.
Dec 21 '18 at 9:47

$begingroup$
As you mentioned a resource, I saw one graph of standard deviation it took values from -3 to 3 having percentage = 2.14+13.59+34.13+34.13+13.59+2.14 = 99.72
$endgroup$
– swapnil
Dec 21 '18 at 9:54

$begingroup$
but I don't know in exam whether they will provide something like this, so is there any other way of doing it. i.e. calculating graph under the curve from 0 to 3 then I will just double it to get required answer.
$endgroup$
– swapnil
Dec 21 '18 at 9:57

$begingroup$
@swapril You should research what's provided in your exam, but you'll likely receive a table of values of $Phi(z)$ for positive values of $z$, which is enough because of the identity $Phi(-z)=1-Phi(z)$. The case $a=-b$ gives $Phi(b)-Phi(a)=2Phi(b)-1$.
$endgroup$
– J.G.
Dec 21 '18 at 10:32

|
show 1 more comment

$begingroup$
Okay. I got the concept, but how to solve a≤z≤b this part. do I have to use integration from -3 to 3 to f(x) where f(x) is 1/sqrt(2*pi)*e^(-x^2/2).
$endgroup$
– swapnil
Dec 21 '18 at 9:43

$begingroup$
@swapnil That's the integral you need. You'll probably obtain it from a resource that can compute $Phi$. Can you express the value you want in terms of two values of $Phi$?
$endgroup$
– J.G.
Dec 21 '18 at 9:47

$begingroup$
As you mentioned a resource, I saw one graph of standard deviation it took values from -3 to 3 having percentage = 2.14+13.59+34.13+34.13+13.59+2.14 = 99.72
$endgroup$
– swapnil
Dec 21 '18 at 9:54

$begingroup$
but I don't know in exam whether they will provide something like this, so is there any other way of doing it. i.e. calculating graph under the curve from 0 to 3 then I will just double it to get required answer.
$endgroup$
– swapnil
Dec 21 '18 at 9:57

$begingroup$
@swapril You should research what's provided in your exam, but you'll likely receive a table of values of $Phi(z)$ for positive values of $z$, which is enough because of the identity $Phi(-z)=1-Phi(z)$. The case $a=-b$ gives $Phi(b)-Phi(a)=2Phi(b)-1$.
$endgroup$
– J.G.
Dec 21 '18 at 10:32

Okay. I got the concept, but how to solve a≤z≤b this part. do I have to use integration from -3 to 3 to f(x) where f(x) is 1/sqrt(2*pi)*e^(-x^2/2).

– swapnil
Dec 21 '18 at 9:43

@swapnil That's the integral you need. You'll probably obtain it from a resource that can compute $Phi$. Can you express the value you want in terms of two values of $Phi$?

– J.G.
Dec 21 '18 at 9:47

As you mentioned a resource, I saw one graph of standard deviation it took values from -3 to 3 having percentage = 2.14+13.59+34.13+34.13+13.59+2.14 = 99.72

– swapnil
Dec 21 '18 at 9:54

but I don't know in exam whether they will provide something like this, so is there any other way of doing it. i.e. calculating graph under the curve from 0 to 3 then I will just double it to get required answer.

– swapnil
Dec 21 '18 at 9:57

@swapril You should research what's provided in your exam, but you'll likely receive a table of values of $Phi(z)$ for positive values of $z$, which is enough because of the identity $Phi(-z)=1-Phi(z)$. The case $a=-b$ gives $Phi(b)-Phi(a)=2Phi(b)-1$.

– J.G.
Dec 21 '18 at 10:32

|
show 1 more comment

It is $$A={1over sqrt {2pi}}int_{-3}^3 e^{-x^2over 2}dx$$which can only be calculated numerically using Q-function as follows $$A=Q(-3)-Q(3)=1-2Q(3)approx 99.74%$$

answered Dec 21 '18 at 10:11

Mostafa Ayaz

15.3k3939

$begingroup$
Thank you I will learn more about Q functions.
$endgroup$
– swapnil
Dec 21 '18 at 17:39

$begingroup$
You're welcome. Good luck!
$endgroup$
– Mostafa Ayaz
Dec 23 '18 at 11:31

add a comment |

It is $$A={1over sqrt {2pi}}int_{-3}^3 e^{-x^2over 2}dx$$which can only be calculated numerically using Q-function as follows $$A=Q(-3)-Q(3)=1-2Q(3)approx 99.74%$$

answered Dec 21 '18 at 10:11

Mostafa Ayaz

15.3k3939

$begingroup$
Thank you I will learn more about Q functions.
$endgroup$
– swapnil
Dec 21 '18 at 17:39

$begingroup$
You're welcome. Good luck!
$endgroup$
– Mostafa Ayaz
Dec 23 '18 at 11:31

add a comment |

It is $$A={1over sqrt {2pi}}int_{-3}^3 e^{-x^2over 2}dx$$which can only be calculated numerically using Q-function as follows $$A=Q(-3)-Q(3)=1-2Q(3)approx 99.74%$$

answered Dec 21 '18 at 10:11

Mostafa Ayaz

15.3k3939

It is $$A={1over sqrt {2pi}}int_{-3}^3 e^{-x^2over 2}dx$$which can only be calculated numerically using Q-function as follows $$A=Q(-3)-Q(3)=1-2Q(3)approx 99.74%$$

answered Dec 21 '18 at 10:11

Mostafa Ayaz

15.3k3939

answered Dec 21 '18 at 10:11

Mostafa Ayaz

15.3k3939

answered Dec 21 '18 at 10:11

Mostafa Ayaz

15.3k3939

answered Dec 21 '18 at 10:11

Mostafa Ayaz

15.3k3939

$begingroup$
Thank you I will learn more about Q functions.
$endgroup$
– swapnil
Dec 21 '18 at 17:39

$begingroup$
You're welcome. Good luck!
$endgroup$
– Mostafa Ayaz
Dec 23 '18 at 11:31

add a comment |

$begingroup$
Thank you I will learn more about Q functions.
$endgroup$
– swapnil
Dec 21 '18 at 17:39

$begingroup$
You're welcome. Good luck!
$endgroup$
– Mostafa Ayaz
Dec 23 '18 at 11:31

Thank you I will learn more about Q functions.

– swapnil
Dec 21 '18 at 17:39

You're welcome. Good luck!

– Mostafa Ayaz
Dec 23 '18 at 11:31

add a comment |

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