3 ms·

If the side length isn't constrained to integers, the expected volume is 39 inches cubed. In a bit more detail: take the integral of x^3 (the volume function)

If the side length isn't constrained to integers, the expected volume is 39 inches cubed. In a bit more detail: take the integral of x^3 (the volume function) from 1 to 5 and you get 5^4/4 - 1^4/4 = 624/4 = 156. Then we divide by the length of the line segment (4) to get 39. If the side lengths have to be integers, then it's just (1 + 8 + 27 + 64 + 125)/5 = 45 inches. Edit: corrected division mistake, thanks to dxbydt Edit 2: and corrected a second mistake thanks to tzs.

dxbydt 13y ago
not 59, 39.
- SatvikBeri 13y ago
  Heh, failed at the simplest step. Thanks for the correction.
drdeca 13y ago
156/4=39
tzs 13y ago
> Edit: corrected division mistake, thanks to dxbydt Well, one of the mistakes: > If the side lengths have to be integers, then it's just (1 + 8 + 27 + 125)/4 = 40.25 inches If I'm not mistaken, 4 is an integer, and so should not have cruelly snubbed when you sent out invitations to your summation of the cubes of the integers from 1 to 5. Sure, 4 is not a cool integer like the others in 1 to 5...he's got a reputation as a square, but that doesn't mean he doesn't have feelings.
- SatvikBeri 13y ago
  Oy. Someone needs to revoke my math license. Thanks for the correction!
  - dxbydt 13y ago
    My, my, 2 mistakes on a silly uniform distribution! You must atone for this sin. Explain what the volume would be if the side came from a normal distribution over [1,5] (ie. mean 3, sigma 2/3) Otherwise I will personally call the MAA director on my speed dial & haul your ass back to stat101 :) But seriously, you write rather well on Quora & I'm sure the crowd would want to know what happens to the volume were you to choose a normal distribution instead of a uniform, so I hope you respond.
    - Cogito 13y ago
      From [0]: Let X be a random variable and g be any function. If X is continuous, then the expectation of g(X) is deﬁned as, E[g(X)] = \int_−∞^∞ g(x)f(x) dx, where f is the probability density function of X. --- For us, with mean 3 and sigma 2/3, f = 3/2 e^(-9/8 (x-3)^2)/sqrt(2 π) We need to integrate numerically; using Wolfram [1] we get an estimate of 30.8099. If we ask for the expected value over the entire distribution we get 31. [0] http://imai.princeton.edu/teaching/files/Expectation.pdf http://imai.princeton.edu/teaching/files/Expectation.pdf [1] http://www.wolframalpha.com/input/?i=integrate+3%2F2+e%5E%28-9%2F8+%28x-3%29%5E2%29%2Fsqrt%282+%CF%80%29*x%5E3%2C+x+%3D+1+..+5 http://www.wolframalpha.com/input/?i=integrate+3%2F2+e%5E%28... [2] http://www.wolframalpha.com/input/?i=expected+value+of+x%5E3+if+x+is+normal+with+mean+3+and+standard+deviation+2%2F3 http://www.wolframalpha.com/input/?i=expected+value+of+x%5E3...

dxbydt 13y ago

not 59, 39.

SatvikBeri 13y ago
Heh, failed at the simplest step. Thanks for the correction.

drdeca 13y ago

156/4=39

tzs 13y ago

> Edit: corrected division mistake, thanks to dxbydt Well, one of the mistakes: > If the side lengths have to be integers, then it's just (1 + 8 + 27 + 125)/4 = 40.25 inches If I'm not mistaken, 4 is an integer, and so should not have cruelly snubbed when you sent out invitations to your summation of the cubes of the integers from 1 to 5. Sure, 4 is not a cool integer like the others in 1 to 5...he's got a reputation as a square, but that doesn't mean he doesn't have feelings.

SatvikBeri 13y ago
Oy. Someone needs to revoke my math license. Thanks for the correction!
- dxbydt 13y ago
  My, my, 2 mistakes on a silly uniform distribution! You must atone for this sin. Explain what the volume would be if the side came from a normal distribution over [1,5] (ie. mean 3, sigma 2/3) Otherwise I will personally call the MAA director on my speed dial & haul your ass back to stat101 :) But seriously, you write rather well on Quora & I'm sure the crowd would want to know what happens to the volume were you to choose a normal distribution instead of a uniform, so I hope you respond.
  - Cogito 13y ago
    From [0]: Let X be a random variable and g be any function. If X is continuous, then the expectation of g(X) is deﬁned as, E[g(X)] = \int_−∞^∞ g(x)f(x) dx, where f is the probability density function of X. --- For us, with mean 3 and sigma 2/3, f = 3/2 e^(-9/8 (x-3)^2)/sqrt(2 π) We need to integrate numerically; using Wolfram [1] we get an estimate of 30.8099. If we ask for the expected value over the entire distribution we get 31. [0] http://imai.princeton.edu/teaching/files/Expectation.pdf http://imai.princeton.edu/teaching/files/Expectation.pdf [1] http://www.wolframalpha.com/input/?i=integrate+3%2F2+e%5E%28-9%2F8+%28x-3%29%5E2%29%2Fsqrt%282+%CF%80%29*x%5E3%2C+x+%3D+1+..+5 http://www.wolframalpha.com/input/?i=integrate+3%2F2+e%5E%28... [2] http://www.wolframalpha.com/input/?i=expected+value+of+x%5E3+if+x+is+normal+with+mean+3+and+standard+deviation+2%2F3 http://www.wolframalpha.com/input/?i=expected+value+of+x%5E3...