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M
 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
M
 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

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.00%. Even if you get 3 of the heaviest mineral, it only gives you 270 lbs (unless you means something else)

M
 5 years ago
Best ResponseYou've already chosen the best response.0answer is .0351 you may need to use linear combinations

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0What would be the constant? To you take out 1 mineral/mine or what?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I think some info is missing.

M
 5 years ago
Best ResponseYou've already chosen the best response.0no other information is given sorry

M
 5 years ago
Best ResponseYou've already chosen the best response.0standard deviations 12, 14, 10 forgot crucial information!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I got .036 do you know about moment generating functions?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yes it's the expected value of e^tx. Do you know the moment generating function for a normal random variable?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i can type it for you... \[M(t)=e^{\mu*t+\sigma^{2}t^{2}/2}\]

M
 5 years ago
Best ResponseYou've already chosen the best response.0there's so many variations of formulas i can't retain them in my memory

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0that has mu as the mean and sigma as the standard deviation

M
 5 years ago
Best ResponseYou've already chosen the best response.0i've seen that before but didn't actually had to solve any questions with it so far

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you 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?

M
 5 years ago
Best ResponseYou've already chosen the best response.0you can just tell me what it is lol

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok 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

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0then you convert it to standard normal and find the probability of being greater than 283 by looking a a standard normal chart

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0do you know how to convert it?

M
 5 years ago
Best ResponseYou've already chosen the best response.0what is standard normal? and i don't think we can use standard normal chart for this

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0standard 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?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok so you have the variance and mean, so you know the pdf and you integrate from 283 to infinity

M
 5 years ago
Best ResponseYou've already chosen the best response.0i use the standard pdf formula?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yes you can use that. It's kind of a mess though I usually use standard normal

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0im 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?

M
 5 years ago
Best ResponseYou've already chosen the best response.0hmm ok i'll have to play around with it for a while

M
 5 years ago
Best ResponseYou've already chosen the best response.0no we've only used charts for zscore

M
 5 years ago
Best ResponseYou've already chosen the best response.0everything else we just integrated or didn't use charts

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yeah z score, thats standard normal

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you have mean and standard deviation so use the z score to find probability

M
 5 years ago
Best ResponseYou've already chosen the best response.0i only knew it as table VI on pg 382 lol

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yeah that's it. stddev is sqrt( sum of the variances of each variable) mean is sum of the means

M
 5 years ago
Best ResponseYou've already chosen the best response.0oh ok thanks. are you an actuary?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0im a graduate student in math studying modern algebra. I just took a grad probability class last semester though.

M
 5 years ago
Best ResponseYou've already chosen the best response.0no i graduated recently but taking some courses at local college

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0are you taking a prob/stats class now? i noticed a bunch of your questions are from that topic

M
 5 years ago
Best ResponseYou've already chosen the best response.0probability and stat is worse than multivariable calc for me

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i like prob not stats. Did my answers to your other questions show up? Im not sure it worked.

M
 5 years ago
Best ResponseYou've already chosen the best response.0yeah they did thanks. but haven't worked on them

M
 5 years ago
Best ResponseYou've already chosen the best response.0but i have a good idea on how to solve them now though

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0good. well ill cya around the site I just found it and I think its fun. good luck with your probability.
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