Statistics -> Normal distribution problem

[Undyinglova]

Homework Statement

In considering the safety of building, the total force acting on the columns of the building must be examined. This would include the effects of the dead load, D, due to weight of the structure, the live load, L, due to human occupancy, movable furniture and the like, and the wind load, W. Assume that the load effects on the individual columns are statistically independent Normal random variables with:
mean of D (dead load) = 4.2 kips, standard deviation = 0.3 kips
mean of L (live load ) = 6.5 kips, standard deviation = 0.8 kips
mean of W (wind load) = 3.4 kips, standard deviation = 0.7 kips

(a) Determine the mean and standard deviation of the total load acting on a column.
(b) Suppose the strength of the column is also Normal with mean equal to 1.5 times of the total mean load and the coefficient of variation of the strength is 15%. A column fails if the strength is less than the total load. What is the probability of failure of the column?

Homework Equations

The Attempt at a Solution

Part a. Yielded 14.1 kips as the mean, and 1.1045 for the deviation, and is the correct answer.
As for part b, from what i see, the picture for me seems to be 2 normal distributions with their tails overlapping. Any suggestions on how to approach it would be appreciated (hopefully without the use of the normal distribution's equation, I've done it using that and its a double integral of the cumulative distribution density function.. and long enough to be rather annoying).

 

[Dick]

The difference of two normal distributions is also a normal distribution. Just like you combined the three load distributions can't you combine the strength minus total load into a single distribution and look at the negative tail?

 

[Undyinglova]

The difference of two normal distributions is also a normal distribution. Just like you combined the three load distributions can't you combine the strength minus total load into a single distribution and look at the negative tail?

From part a. We know that the mean is 14.1, standard deviation 1.04536.
So for the strength, the mean is 14.1*1.5 = 21.15.
When it says 15% coefficient of variation, i am assuming that the variance is 21.15*0.15 = 3.1725.
So the difference distribution, which also happens to be normal, ends up being 21.15 - 14.1 = 7.05 ( this is the mean for the new normal distributon ).
And the variance of this new distribution is 3.1726 + 1.04536^2 = 4.26538
And hence, the standard deviation is 2.065
7.05/2.065 = 3.414 deviations
If one looks at the z-score table, the value is 0.0003 or 0.03%.
From the solution sheet, the correct answer is 0.018, which is 1.8%.
So yeah.. that doesn't seem to work either :(

 

[Dick]

I've got to confess, I'm not a stats expert. My post was just a suggestion. I do know the difference of normal distributions is normal. In that I have faith. What I looked up said the 'coefficient of variation' meant ratio of the mean to the standard deviation, not the variance. I am not trying to run the numbers on this so you shouldn't put much faith in what I think.

 

[Undyinglova]

I've got to confess, I'm not a stats expert. My post was just a suggestion. I do know the difference of normal distributions is normal. In that I have faith. What I looked up said the 'coefficient of variation' meant ratio of the mean to the standard deviation, not the variance. I am not trying to run the numbers on this so you shouldn't put much faith in what I think.

I've got to confess sir, you're a genius. Thanks for looking up "coefficient of variation" for me. Since it had the word variation, i assumed "related to variance". But clearly it was related to the standard deviation instead, as pointed by you and later confirmed by wiki. After running the numbers again, the new standard deviation computed to 3.34.
7.05/3.34 = 2.11. Looking up the z-scores, it seems the answer indeed is 1.8%.
Thanks a lot Dick.
Problem solved.

 

[Dick]

You're welcome. Thanks especially for saving me the effort of actually having to work it out myself.