anonymous 5 years ago minerals produced from mines 1, 2, 3 and are independent normal random variables with means 80, 90, 75 pounds, respectively. what is the probability that the combined amount of mineral produced from all three mines exceeds 283 pounds? [Hint: use sum random variable Y = X1 + X2 + X3

1. anonymous

0%. Even if you get 3 of the heaviest mineral, it only gives you 270 lbs (unless you means something else)

2. anonymous

answer is .0351 you may need to use linear combinations

3. anonymous

What would be the constant? To you take out 1 mineral/mine or what?

4. anonymous

I think some info is missing.

5. anonymous

no other information is given sorry

6. anonymous

standard deviations 12, 14, 10 forgot crucial information!

7. anonymous

I got .036 do you know about moment generating functions?

8. anonymous

some what the e^tx thing?

9. anonymous

yes it's the expected value of e^tx. Do you know the moment generating function for a normal random variable?

10. anonymous

no but i can look it up

11. anonymous

i can type it for you... $M(t)=e^{\mu*t+\sigma^{2}t^{2}/2}$

12. anonymous

there's so many variations of formulas i can't retain them in my memory

13. anonymous

that has mu as the mean and sigma as the standard deviation

14. anonymous

i've seen that before but didn't actually had to solve any questions with it so far

15. anonymous

do i use that formula?

16. anonymous

you use it to derive something that is helpful to remember about sum of independent normal vars. Do you want me to derive the result or just tell you what it is?

17. anonymous

you can just tell me what it is lol

18. anonymous

ok haha. The mean of the sum of independent normal vars is the sum of the means, and the variance is the sum of the variances. So we have mean 80+90+75=245 and variance = 144+196+100=440

19. anonymous

then you convert it to standard normal and find the probability of being greater than 283 by looking a a standard normal chart

20. anonymous

do you know how to convert it?

21. anonymous

what is standard normal? and i don't think we can use standard normal chart for this

22. anonymous

standard normal is normal with mean 0 variance 1. Well if you can't use the chart you just integrate. Do you know the pdf for normal variables?

23. anonymous

yeah

24. anonymous

ok so you have the variance and mean, so you know the pdf and you integrate from 283 to infinity

25. anonymous

i use the standard pdf formula?

26. anonymous

yes you can use that. It's kind of a mess though I usually use standard normal

27. anonymous

im looking at the integral now its pretty bad. I think I'd have to do it numerically. You haven't used standard normal in your class before?

28. anonymous

hmm ok i'll have to play around with it for a while

29. anonymous

no we've only used charts for z-score

30. anonymous

everything else we just integrated or didn't use charts

31. anonymous

yeah z score, thats standard normal

32. anonymous

oh haha

33. anonymous

you have mean and standard deviation so use the z score to find probability

34. anonymous

i only knew it as table VI on pg 382 lol

35. anonymous

haha

36. anonymous

using the z = (x-m)/st.dev?

37. anonymous

yeah that's it. stddev is sqrt( sum of the variances of each variable) mean is sum of the means

38. anonymous

oh ok thanks. are you an actuary?

39. anonymous

im a graduate student in math studying modern algebra. I just took a grad probability class last semester though.

40. anonymous

ahh nice

41. anonymous

are you in college?

42. anonymous

no i graduated recently but taking some courses at local college

43. anonymous

44. anonymous

are you taking a prob/stats class now? i noticed a bunch of your questions are from that topic

45. anonymous

probability and stat is worse than multivariable calc for me

46. anonymous

yeah

47. anonymous

i like prob not stats. Did my answers to your other questions show up? Im not sure it worked.

48. anonymous

yeah they did thanks. but haven't worked on them

49. anonymous

but i have a good idea on how to solve them now though

50. anonymous

good. well ill cya around the site I just found it and I think its fun. good luck with your probability.

51. anonymous

thanks!!