Probability density expected values

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Homework Help Overview

The problem involves calculating expected values E(X) and E(X^2) for a probability density function defined for the bearing capacity of soil, which varies between 6 and 15 kips/sq.ft. The function is given as f(x)=1/2.7 * (1- x/15) within the specified range.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • Participants discuss the calculations for E(X) and E(X^2), noting discrepancies between their results and those on the answer sheet. There is confusion regarding the variable "u" in the probability density function and whether it should be "x".

Discussion Status

Some participants express uncertainty about the correctness of their calculations and the answer sheet. There is a suggestion to double-check the work and verify with others in the course, indicating an ongoing exploration of the problem.

Contextual Notes

Participants mention potential errors in the answer sheet and the need for clarification on the definitions and limits of the probability density function. There is a lack of consensus on the correct expected values.

afireinside
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Homework Statement


Bearing capacity of soil varies between 6 and 15 kips/sq.ft. If probability density within this range is given as
f(x)=1/2.7 * (1- x/15), 6 ≤ u ≤ 15
=0 otherwise

Find E(X) and E(X^2)

Homework Equations


E(x) should be ∫x*f(x) dx
E(x^2) should be ∫x^2*f(x) dx


The Attempt at a Solution



E(x) ends up being 9 when I plug it in, but the answer sheet says either 9.55 or 9 * 55 (his decimals can look like multiplication dots)

E(x^2) ends up being 85.5, but the answer sheet says either 230.81 or 230 * 81.

I feel like it should be the * ones, because 9 and 81 are part of them, but I have no clue why, unless my equations are wrong.
 
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f(x)=1/2.7 * (1- x/15), 6 ≤ u ≤ 15
... what is "u" in there?
Do you mean:
$$f(x)=\frac{1}{2.7}\left (1-\frac{x}{15}\right )\; :\; 6\leq x \leq 16$$... then
$$E[X]=\frac{1}{243}\int_6^{15} (90x-6x^2)dx$$... sure enough - unless this is something funny in the course.

Probably you've just made a mistake someplace - or the answer sheet could be wrong.
Double check.
 
Simon Bridge said:
... what is "u" in there?
Do you mean:
$$f(x)=\frac{1}{2.7}\left (1-\frac{x}{15}\right )\; :\; 6\leq x \leq 16$$... then
$$E[X]=\frac{1}{243}\int_6^{15} (90x-6x^2)dx$$... sure enough - unless this is something funny in the course.

Probably you've just made a mistake someplace - or the answer sheet could be wrong.
Double check.

So are you saying the answer should be 9, and the answer sheet is wrong? Then what about E[x^2], is that 81 or is it 85.5 by doing a similar integral just with x^2 instead of x?
 
afireinside said:
So are you saying the answer should be 9, and the answer sheet is wrong? Then what about E[x^2], is that 81 or is it 85.5 by doing a similar integral just with x^2 instead of x?

I get EX = 9 and EX^2 = 85.5.
 
Either the question is not what we think it is or the answer sheet is wrong.
I don't want to be definitive about this because I am not there and not the one doing the course.
You should check with someone else doing the course and ask the person who made the answer sheet.
 

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