Having some troubles with this problem, so any help is appreciated!
From Example 3C of the ASM Exam C manual:
Distribution For Loss Sizes is
Size Pr
1000 .5
1500 .25
2500 .25
It uses the Normal Approximation for Pr(losses<1550) for the average of 100 losses.
The solution states it is 2.5 because the losses are in multiples of 500, and dividing that by the 100 losses and using the midpoint gave 2.5.
My question pertains to the correction used. Why use the multiple of 500? With each not being exactly 500 in difference, what difference would it make if we used, say 20. Is it the highest factor that we should use in determining which distance to use, or am I missing something?
Lets say it was of sizes 2, 3, and 5. Do I then use .005 there for the correction?
Apologies for the long question, I haven't had much experience with the continuity correction out of +/- .5. I'm sure I am overthinking it, I just wanted to get some clarification.
Thanks
From Example 3C of the ASM Exam C manual:
Distribution For Loss Sizes is
Size Pr
1000 .5
1500 .25
2500 .25
It uses the Normal Approximation for Pr(losses<1550) for the average of 100 losses.
The solution states it is 2.5 because the losses are in multiples of 500, and dividing that by the 100 losses and using the midpoint gave 2.5.
My question pertains to the correction used. Why use the multiple of 500? With each not being exactly 500 in difference, what difference would it make if we used, say 20. Is it the highest factor that we should use in determining which distance to use, or am I missing something?
Lets say it was of sizes 2, 3, and 5. Do I then use .005 there for the correction?
Apologies for the long question, I haven't had much experience with the continuity correction out of +/- .5. I'm sure I am overthinking it, I just wanted to get some clarification.
Thanks
Continuity Correction Question for ASM Exam C
