Finding Expected Value of $f(x)= 0.10e^{-x/10}$
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I have the following probability density function:
$$f(x)= cases{0.10e^{-x/10} hspace{0.5cm} x geq 0\
0 hspace{1.9cm}text{otherwise}}$$
and need help answering the following question:
What is $P(Xleq E(X))$? What does this tell you about the distribution of X?
When I tried to complete the above question, I obtained a value of 10 for the expected value and 0.6321 as the probability that the random variable is less than the expected value. But I don't know what this tells me about the distribution, nor if I integrated correctly. Any help would be greatly appreciated.
probability integration
edited Nov 15 at 21:59
David K
51.2k340113
asked Mar 9 '17 at 14:58
Sarah
196
add a comment |
up vote
0
down vote
favorite
I have the following probability density function:
$$f(x)= cases{0.10e^{-x/10} hspace{0.5cm} x geq 0\
0 hspace{1.9cm}text{otherwise}}$$
and need help answering the following question:
What is $P(Xleq E(X))$? What does this tell you about the distribution of X?
When I tried to complete the above question, I obtained a value of 10 for the expected value and 0.6321 as the probability that the random variable is less than the expected value. But I don't know what this tells me about the distribution, nor if I integrated correctly. Any help would be greatly appreciated.
probability integration
edited Nov 15 at 21:59
David K
51.2k340113
asked Mar 9 '17 at 14:58
Sarah
196
Hi. I tried helping with the formatting, please let me know if something went wrong
– mathreadler
Mar 9 '17 at 15:05
1
This tells you that the median is to the left of the mean, thus, the Pearson median skewness, or second skewness coefficient, is positive. en.wikipedia.org/wiki/…
– Did
Mar 9 '17 at 15:05
it's the exponential distribution, it has a long tail on the right, if you look at a graph of the distribution it can take large values on the right, but with low probability - I agree with your calculation, it is 1 - 1/e and the answer 10 is 1 / $lambda$ a standard result - so that is good
– Cato
Mar 9 '17 at 15:19
can you imagine a distribution function for which the expected value and the median are the same? I think this will help you understand the general case.
– Seyhmus Güngören
Mar 9 '17 at 17:17
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I have the following probability density function:
$$f(x)= cases{0.10e^{-x/10} hspace{0.5cm} x geq 0\
0 hspace{1.9cm}text{otherwise}}$$
and need help answering the following question:
What is $P(Xleq E(X))$? What does this tell you about the distribution of X?
When I tried to complete the above question, I obtained a value of 10 for the expected value and 0.6321 as the probability that the random variable is less than the expected value. But I don't know what this tells me about the distribution, nor if I integrated correctly. Any help would be greatly appreciated.
probability integration
edited Nov 15 at 21:59
David K
51.2k340113
asked Mar 9 '17 at 14:58
Sarah
196
I have the following probability density function:
$$f(x)= cases{0.10e^{-x/10} hspace{0.5cm} x geq 0\
0 hspace{1.9cm}text{otherwise}}$$
and need help answering the following question:
What is $P(Xleq E(X))$? What does this tell you about the distribution of X?
When I tried to complete the above question, I obtained a value of 10 for the expected value and 0.6321 as the probability that the random variable is less than the expected value. But I don't know what this tells me about the distribution, nor if I integrated correctly. Any help would be greatly appreciated.
probability integration
probability integration
edited Nov 15 at 21:59
David K
51.2k340113
asked Mar 9 '17 at 14:58
Sarah
196
edited Nov 15 at 21:59
David K
51.2k340113
asked Mar 9 '17 at 14:58
Sarah
196
edited Nov 15 at 21:59
David K
51.2k340113
edited Nov 15 at 21:59
David K
51.2k340113
edited Nov 15 at 21:59
David K
51.2k340113
51.2k340113
asked Mar 9 '17 at 14:58
Sarah
196
asked Mar 9 '17 at 14:58
Sarah
196
asked Mar 9 '17 at 14:58
Sarah
196
196
Hi. I tried helping with the formatting, please let me know if something went wrong
– mathreadler
Mar 9 '17 at 15:05
1
This tells you that the median is to the left of the mean, thus, the Pearson median skewness, or second skewness coefficient, is positive. en.wikipedia.org/wiki/…
– Did
Mar 9 '17 at 15:05
it's the exponential distribution, it has a long tail on the right, if you look at a graph of the distribution it can take large values on the right, but with low probability - I agree with your calculation, it is 1 - 1/e and the answer 10 is 1 / $lambda$ a standard result - so that is good
– Cato
Mar 9 '17 at 15:19
can you imagine a distribution function for which the expected value and the median are the same? I think this will help you understand the general case.
– Seyhmus Güngören
Mar 9 '17 at 17:17
add a comment |
Hi. I tried helping with the formatting, please let me know if something went wrong
– mathreadler
Mar 9 '17 at 15:05
1
This tells you that the median is to the left of the mean, thus, the Pearson median skewness, or second skewness coefficient, is positive. en.wikipedia.org/wiki/…
– Did
Mar 9 '17 at 15:05
it's the exponential distribution, it has a long tail on the right, if you look at a graph of the distribution it can take large values on the right, but with low probability - I agree with your calculation, it is 1 - 1/e and the answer 10 is 1 / $lambda$ a standard result - so that is good
– Cato
Mar 9 '17 at 15:19
can you imagine a distribution function for which the expected value and the median are the same? I think this will help you understand the general case.
– Seyhmus Güngören
Mar 9 '17 at 17:17
Hi. I tried helping with the formatting, please let me know if something went wrong
– mathreadler
Mar 9 '17 at 15:05
Hi. I tried helping with the formatting, please let me know if something went wrong
– mathreadler
Mar 9 '17 at 15:05
1
1
This tells you that the median is to the left of the mean, thus, the Pearson median skewness, or second skewness coefficient, is positive. en.wikipedia.org/wiki/…
– Did
Mar 9 '17 at 15:05
This tells you that the median is to the left of the mean, thus, the Pearson median skewness, or second skewness coefficient, is positive. en.wikipedia.org/wiki/…
– Did
Mar 9 '17 at 15:05
it's the exponential distribution, it has a long tail on the right, if you look at a graph of the distribution it can take large values on the right, but with low probability - I agree with your calculation, it is 1 - 1/e and the answer 10 is 1 / $lambda$ a standard result - so that is good
– Cato
Mar 9 '17 at 15:19
it's the exponential distribution, it has a long tail on the right, if you look at a graph of the distribution it can take large values on the right, but with low probability - I agree with your calculation, it is 1 - 1/e and the answer 10 is 1 / $lambda$ a standard result - so that is good
– Cato
Mar 9 '17 at 15:19
can you imagine a distribution function for which the expected value and the median are the same? I think this will help you understand the general case.
– Seyhmus Güngören
Mar 9 '17 at 17:17
can you imagine a distribution function for which the expected value and the median are the same? I think this will help you understand the general case.
– Seyhmus Güngören
Mar 9 '17 at 17:17
add a comment |
1 Answer
1
active
oldest
votes
up vote
0
down vote
Here's how I would do this.
1) Here is the pdf:
$$f(x)=frac{e^{-frac{x}{10}}}{10}$$
2) Find the mean:
$$E[X] = int_0^{infty} x f(x) = x frac{e^{-frac{x}{10}}}{10} = 10 $$
3) Now, let's find $$P(X leq E[X]) = P(X leq 10)$$
$$ = 1-e^{-frac{x}{10}} = 1-e^{-frac{10}{10}} = 1-e^{-1}= 1-frac{1}{e} = 0.632121$$
4) So what does this tell you about the distribution? It tells you that most events or the mass of the distribution is below or less than the mean.
answered Mar 11 '17 at 17:46
PiE
600410
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Here's how I would do this.
1) Here is the pdf:
$$f(x)=frac{e^{-frac{x}{10}}}{10}$$
2) Find the mean:
$$E[X] = int_0^{infty} x f(x) = x frac{e^{-frac{x}{10}}}{10} = 10 $$
3) Now, let's find $$P(X leq E[X]) = P(X leq 10)$$
$$ = 1-e^{-frac{x}{10}} = 1-e^{-frac{10}{10}} = 1-e^{-1}= 1-frac{1}{e} = 0.632121$$
4) So what does this tell you about the distribution? It tells you that most events or the mass of the distribution is below or less than the mean.
answered Mar 11 '17 at 17:46
PiE
600410
add a comment |
up vote
0
down vote
Here's how I would do this.
1) Here is the pdf:
$$f(x)=frac{e^{-frac{x}{10}}}{10}$$
2) Find the mean:
$$E[X] = int_0^{infty} x f(x) = x frac{e^{-frac{x}{10}}}{10} = 10 $$
3) Now, let's find $$P(X leq E[X]) = P(X leq 10)$$
$$ = 1-e^{-frac{x}{10}} = 1-e^{-frac{10}{10}} = 1-e^{-1}= 1-frac{1}{e} = 0.632121$$
4) So what does this tell you about the distribution? It tells you that most events or the mass of the distribution is below or less than the mean.
answered Mar 11 '17 at 17:46
PiE
600410
add a comment |
up vote
0
down vote
up vote
0
down vote
Here's how I would do this.
1) Here is the pdf:
$$f(x)=frac{e^{-frac{x}{10}}}{10}$$
2) Find the mean:
$$E[X] = int_0^{infty} x f(x) = x frac{e^{-frac{x}{10}}}{10} = 10 $$
3) Now, let's find $$P(X leq E[X]) = P(X leq 10)$$
$$ = 1-e^{-frac{x}{10}} = 1-e^{-frac{10}{10}} = 1-e^{-1}= 1-frac{1}{e} = 0.632121$$
4) So what does this tell you about the distribution? It tells you that most events or the mass of the distribution is below or less than the mean.
answered Mar 11 '17 at 17:46
PiE
600410
Here's how I would do this.
1) Here is the pdf:
$$f(x)=frac{e^{-frac{x}{10}}}{10}$$
2) Find the mean:
$$E[X] = int_0^{infty} x f(x) = x frac{e^{-frac{x}{10}}}{10} = 10 $$
3) Now, let's find $$P(X leq E[X]) = P(X leq 10)$$
$$ = 1-e^{-frac{x}{10}} = 1-e^{-frac{10}{10}} = 1-e^{-1}= 1-frac{1}{e} = 0.632121$$
4) So what does this tell you about the distribution? It tells you that most events or the mass of the distribution is below or less than the mean.
answered Mar 11 '17 at 17:46
PiE
600410
answered Mar 11 '17 at 17:46
PiE
600410
answered Mar 11 '17 at 17:46
PiE
600410
answered Mar 11 '17 at 17:46
PiE
600410
600410
add a comment |
add a comment |
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Hi. I tried helping with the formatting, please let me know if something went wrong
– mathreadler
Mar 9 '17 at 15:05
1
This tells you that the median is to the left of the mean, thus, the Pearson median skewness, or second skewness coefficient, is positive. en.wikipedia.org/wiki/…
– Did
Mar 9 '17 at 15:05
it's the exponential distribution, it has a long tail on the right, if you look at a graph of the distribution it can take large values on the right, but with low probability - I agree with your calculation, it is 1 - 1/e and the answer 10 is 1 / $lambda$ a standard result - so that is good
– Cato
Mar 9 '17 at 15:19
can you imagine a distribution function for which the expected value and the median are the same? I think this will help you understand the general case.
– Seyhmus Güngören
Mar 9 '17 at 17:17