Expected Value Question with Money
$begingroup$
I've been struggling a lot w/ expected value, and was wondering if anyone could help me with this.
I know the equation for expected value is $E(X)$ $=$ $NP$
The question asks:
A 20 dollar bill, two 10 dollar bills, three 5 dollar bills and four 1 dollar bills are placed in a bag. If a bill is chosen at random, what is the expected value for the amount chosen?
I believe the equation would go something like
$E(X) = 20(p) + 10(p) + 5(p) + 1(p)$ but I'm not sure what I would add for P? Theres a total of 10 bills that can be chosen, so would the probability be something like
$E(X) = 20(1/10) + 10(2/10) + 5(3/10) + 1(4/10)$
probability
asked Apr 17 '16 at 23:48
XorXor
357
$endgroup$
add a comment |
$begingroup$
I've been struggling a lot w/ expected value, and was wondering if anyone could help me with this.
I know the equation for expected value is $E(X)$ $=$ $NP$
The question asks:
A 20 dollar bill, two 10 dollar bills, three 5 dollar bills and four 1 dollar bills are placed in a bag. If a bill is chosen at random, what is the expected value for the amount chosen?
I believe the equation would go something like
$E(X) = 20(p) + 10(p) + 5(p) + 1(p)$ but I'm not sure what I would add for P? Theres a total of 10 bills that can be chosen, so would the probability be something like
$E(X) = 20(1/10) + 10(2/10) + 5(3/10) + 1(4/10)$
probability
asked Apr 17 '16 at 23:48
XorXor
357
$endgroup$
$begingroup$
Your last line is the appropriate weighted average!
$endgroup$
– Xoque55
Apr 17 '16 at 23:51
add a comment |
$begingroup$
I've been struggling a lot w/ expected value, and was wondering if anyone could help me with this.
I know the equation for expected value is $E(X)$ $=$ $NP$
The question asks:
A 20 dollar bill, two 10 dollar bills, three 5 dollar bills and four 1 dollar bills are placed in a bag. If a bill is chosen at random, what is the expected value for the amount chosen?
I believe the equation would go something like
$E(X) = 20(p) + 10(p) + 5(p) + 1(p)$ but I'm not sure what I would add for P? Theres a total of 10 bills that can be chosen, so would the probability be something like
$E(X) = 20(1/10) + 10(2/10) + 5(3/10) + 1(4/10)$
probability
asked Apr 17 '16 at 23:48
XorXor
357
$endgroup$
I've been struggling a lot w/ expected value, and was wondering if anyone could help me with this.
I know the equation for expected value is $E(X)$ $=$ $NP$
The question asks:
A 20 dollar bill, two 10 dollar bills, three 5 dollar bills and four 1 dollar bills are placed in a bag. If a bill is chosen at random, what is the expected value for the amount chosen?
I believe the equation would go something like
$E(X) = 20(p) + 10(p) + 5(p) + 1(p)$ but I'm not sure what I would add for P? Theres a total of 10 bills that can be chosen, so would the probability be something like
$E(X) = 20(1/10) + 10(2/10) + 5(3/10) + 1(4/10)$
probability
probability
asked Apr 17 '16 at 23:48
XorXor
357
asked Apr 17 '16 at 23:48
XorXor
357
asked Apr 17 '16 at 23:48
XorXor
357
asked Apr 17 '16 at 23:48
XorXor
357
asked Apr 17 '16 at 23:48
XorXor
357
357
$begingroup$
Your last line is the appropriate weighted average!
$endgroup$
– Xoque55
Apr 17 '16 at 23:51
add a comment |
$begingroup$
Your last line is the appropriate weighted average!
$endgroup$
– Xoque55
Apr 17 '16 at 23:51
$begingroup$
Your last line is the appropriate weighted average!
$endgroup$
– Xoque55
Apr 17 '16 at 23:51
$begingroup$
Your last line is the appropriate weighted average!
$endgroup$
– Xoque55
Apr 17 '16 at 23:51
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
I know the equation for expected value is $E(X) = NP $
That is so for a binomial distribution. However, this is not that. So it is just the wrong equation for this situation.
I believe the equation would go something like
$$E(X) ~=~ 20(1/10) + 10(2/10) + 5(3/10) + 1(4/10)$$
Now you are on the right track. In this instance, that is the correct calculation of the mean, in . (Also known as the weighted average; or the expectation.)
Don't doubt yourself. It appears that you were just trying to apply the wrong shortcut, and that was confusing you, because otherwise you do know what you are about. You have got the basics down.
$Box$
answered Apr 18 '16 at 0:13
Graham KempGraham Kemp
86.5k43479
$endgroup$
$begingroup$
What is the correct equation for these sort of questions? Just so I know which ones to apply to what question =)
$endgroup$
– Xor
Apr 18 '16 at 0:18
1
$begingroup$
$Sigma( p_icdot x_i)$
$endgroup$
– true blue anil
Apr 18 '16 at 7:49
add a comment |
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1 Answer
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1 Answer
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active
oldest
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votes
$begingroup$
I know the equation for expected value is $E(X) = NP $
That is so for a binomial distribution. However, this is not that. So it is just the wrong equation for this situation.
I believe the equation would go something like
$$E(X) ~=~ 20(1/10) + 10(2/10) + 5(3/10) + 1(4/10)$$
Now you are on the right track. In this instance, that is the correct calculation of the mean, in . (Also known as the weighted average; or the expectation.)
Don't doubt yourself. It appears that you were just trying to apply the wrong shortcut, and that was confusing you, because otherwise you do know what you are about. You have got the basics down.
$Box$
answered Apr 18 '16 at 0:13
Graham KempGraham Kemp
86.5k43479
$endgroup$
$begingroup$
What is the correct equation for these sort of questions? Just so I know which ones to apply to what question =)
$endgroup$
– Xor
Apr 18 '16 at 0:18
1
$begingroup$
$Sigma( p_icdot x_i)$
$endgroup$
– true blue anil
Apr 18 '16 at 7:49
add a comment |
$begingroup$
I know the equation for expected value is $E(X) = NP $
That is so for a binomial distribution. However, this is not that. So it is just the wrong equation for this situation.
I believe the equation would go something like
$$E(X) ~=~ 20(1/10) + 10(2/10) + 5(3/10) + 1(4/10)$$
Now you are on the right track. In this instance, that is the correct calculation of the mean, in . (Also known as the weighted average; or the expectation.)
Don't doubt yourself. It appears that you were just trying to apply the wrong shortcut, and that was confusing you, because otherwise you do know what you are about. You have got the basics down.
$Box$
answered Apr 18 '16 at 0:13
Graham KempGraham Kemp
86.5k43479
$endgroup$
$begingroup$
What is the correct equation for these sort of questions? Just so I know which ones to apply to what question =)
$endgroup$
– Xor
Apr 18 '16 at 0:18
1
$begingroup$
$Sigma( p_icdot x_i)$
$endgroup$
– true blue anil
Apr 18 '16 at 7:49
add a comment |
$begingroup$
I know the equation for expected value is $E(X) = NP $
That is so for a binomial distribution. However, this is not that. So it is just the wrong equation for this situation.
I believe the equation would go something like
$$E(X) ~=~ 20(1/10) + 10(2/10) + 5(3/10) + 1(4/10)$$
Now you are on the right track. In this instance, that is the correct calculation of the mean, in . (Also known as the weighted average; or the expectation.)
Don't doubt yourself. It appears that you were just trying to apply the wrong shortcut, and that was confusing you, because otherwise you do know what you are about. You have got the basics down.
$Box$
answered Apr 18 '16 at 0:13
Graham KempGraham Kemp
86.5k43479
$endgroup$
I know the equation for expected value is $E(X) = NP $
That is so for a binomial distribution. However, this is not that. So it is just the wrong equation for this situation.
I believe the equation would go something like
$$E(X) ~=~ 20(1/10) + 10(2/10) + 5(3/10) + 1(4/10)$$
Now you are on the right track. In this instance, that is the correct calculation of the mean, in . (Also known as the weighted average; or the expectation.)
Don't doubt yourself. It appears that you were just trying to apply the wrong shortcut, and that was confusing you, because otherwise you do know what you are about. You have got the basics down.
$Box$
answered Apr 18 '16 at 0:13
Graham KempGraham Kemp
86.5k43479
answered Apr 18 '16 at 0:13
Graham KempGraham Kemp
86.5k43479
answered Apr 18 '16 at 0:13
Graham KempGraham Kemp
86.5k43479
answered Apr 18 '16 at 0:13
Graham KempGraham Kemp
86.5k43479
86.5k43479
$begingroup$
What is the correct equation for these sort of questions? Just so I know which ones to apply to what question =)
$endgroup$
– Xor
Apr 18 '16 at 0:18
1
$begingroup$
$Sigma( p_icdot x_i)$
$endgroup$
– true blue anil
Apr 18 '16 at 7:49
add a comment |
$begingroup$
What is the correct equation for these sort of questions? Just so I know which ones to apply to what question =)
$endgroup$
– Xor
Apr 18 '16 at 0:18
1
$begingroup$
$Sigma( p_icdot x_i)$
$endgroup$
– true blue anil
Apr 18 '16 at 7:49
$begingroup$
What is the correct equation for these sort of questions? Just so I know which ones to apply to what question =)
$endgroup$
– Xor
Apr 18 '16 at 0:18
$begingroup$
What is the correct equation for these sort of questions? Just so I know which ones to apply to what question =)
$endgroup$
– Xor
Apr 18 '16 at 0:18
1
1
$begingroup$
$Sigma( p_icdot x_i)$
$endgroup$
– true blue anil
Apr 18 '16 at 7:49
$begingroup$
$Sigma( p_icdot x_i)$
$endgroup$
– true blue anil
Apr 18 '16 at 7:49
add a comment |
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$begingroup$
Your last line is the appropriate weighted average!
$endgroup$
– Xoque55
Apr 17 '16 at 23:51