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View Full Version : Course 3 #35 -- R and S

3rookie
11-07-2002, 03:30 PM
Could somebody please step by step do this one? I keep coming up with D. Thanks

drctypea
11-07-2002, 03:59 PM
i know this probably wont help much but i backsolved....set up mu as .05 and then just worked through each answer choice to get the correct answer...luckily it was answer a...sorry i cant be of more help

Woody
11-07-2002, 04:02 PM
R = 1-px = qx
So, S = .75qx.
S = 1-px[exp(-k)]
So, .75qx = 1-px[exp(-k)]
Solve for k...

px[exp(-k)]=1-.75qx
exp(-k)=(1-.75qx)/px = (1-.75qx)/(1-qx)
Raising each side to the -1 gives you the reciprocal on the right side of the equal sign. Therefore...

exp(k)=(1-qx)/(1-.75qx)
So...

k = ln[(1-qx)/(1-.75qx)]......A

sleg
11-07-2002, 04:03 PM
I wont detail the solution due to all of the symbols that would be necessary. However, it looks like you have dealt with a negative sign wrong somewhere. That would flip the stuff inside the ln.

3rookie
11-07-2002, 04:08 PM
Thanks Woody

I set up S = 1 - Re^(-k), which now I see is wrong. Hopefully, it doesn't matter.

During the exam, I was very happy to get this one "right". :swear: