Combining normal variables homework question

#1
Hi all,

First post on here, but would really appreciate any help anyone can give as I am stumped on the below question:

The tensile strengths, measured in newtons (N), of a large number of ropes of equal length are independently and normally distributed such that 5% are under 706N and 5% over 1294N.
Four such ropes are randomly selected and joined end-to-end to form a single rope; the strenth of the combined rope is equal to the strength of the weakest of the 4 selected ropes. Derive the probabilities that this combined rope will not break under tensions of 1000N and 900N respectively.
A further 4 ropes are randomly selected and attached between 2 rings, the strength of the arrangeent being the sum of the strengths of the 4 separate ropes. Derive the probabilities that this arrangement will break under tensions of 4000N and 4200N respectively.


Here's where I've got to so far, If we let X be the tensile strength in Newtons of a length of rope, I've worked out that X~N(1000,178.72^2) using simultaneous equations. I've doubled checked this part a couple of times so am fairly sure this bit is correct, although of course please point out if not.

What im not really sure about is what the question is really asking me for when it comes to the specific probabilities. For the first part, I think it would just be the standard normal distribution given above seeing as the combined rope is only as strong as the weakest part. So my assumption would be I'm looking for P(0<X<1000) & P(0<X<900), eg, the rope can take up to 1000/900 newtons of tension, but that doesnt seem to get me the correct answer.

For the second part, if we take Y to be X1 + X2 ... + X4, assuming that X~N(1000,178.72^2) is correct, I make it Y~N(4000, 1227763.35), and that because it is WILL break, the maximum tension for each is one below breaking point so we'd be looking at P(0<Y<3999) & P(0<Y<4299) but again, not correct so I am stuck.

Thank you in advance for any help!
 

katxt

Active Member
#2
X~N(1000,178.72^2)
I haven't checked this.
Four such ropes are randomly selected and joined end-to-end to form a single rope; the strenth of the combined rope is equal to the strength of the weakest of the 4 selected ropes. Derive the probabilities that this combined rope will not break under tensions of 1000N and 900N respectively.
To get you started ... For the combined rope not to break at 1000, then A>1000 and B>1000 and C and D. Find the probability of each and find A and B and C and D not breaking. (Each new rope in the line makes it less likely to hold.)
Y~N(4000, 1227763.35)
Check the variance
 
#3
I haven't checked this.

To get you started ... For the combined rope not to break at 1000, then A>1000 and B>1000 and C and D. Find the probability of each and find A and B and C and D not breaking. (Each new rope in the line makes it less likely to hold.)

Check the variance
Thanks Kat, have got it now with your additional pointer, it was the combination of all ropes making it weaker that I had failed to consider. Appreciate your help :)