Recent content by giddy

aha.. ok so i didn't even know how to really solve summation equations, but I looked it up. So Sum(x^2) = 34735178! And its correct... =)

Sorry I am not sure what you mean =S If I do expand (x - 500)^2 it'll be x^2 - 2(500)(x) + 500^2 Right? So where would I get sum of x? How would I expand sum(x^2)

Hi, So this is just part of my problem but its got me stumped for days and I can't ignore it since its popping up too often in my problems. Homework Statement For A sample of 140 bags of flour. The masses of x grams of the contents are summarized by \sum (x - 500) = -266 and \sum...

Sorry, I can be very distracted(Was diagnosed with ADD symptoms but the meds didn't do much) So the second part of what HallsofIvy's post says about the central limit theorem isn't in my book.The part where the sum of the probabilities is mean * n and the SD is sqrt{n * SD}. So that seems to...

yea, I understand. I just wanted to know how to approach the problem. I've tried a few random things. I just tried applying : X \sim N(\mu,\frac{\sigma^2}{n}) for P( X > 200) = 200-2.25/\sqrt{2.25/100} = -1318.33 Which can't be standardized. Also, trying T = probability of total number of...

hey, thanks so much. Seems the problem with the next 2 problems was post #4. But what about the second part of the problem... the next problem in my book is similar and I don't even know how to approach it. A rectangular field is gridded into squares of side 1m. At one time of the year the...

\phi(z) That would be finding the probability after standardizing X to Z so Z ~ N(0,1) there's a table I have in the back of my book where i look up the probability then. Thanks that's correct...but its a little off from the answer: P(Z < -0.980) = 1 - phi(0.980) 1 - 0.8363 = 0.1637 The...

hi.. yes I did mean variance... 1/24 = 0.41666... Comes up close to the same thing... 3.5-3.3/0.2041 = 0.9799 P(Z < 0.9799) = 0.8368.. still not close to 0.155

So from what I understand, the central limit theorem allows us to calculate the probability of the mean of a number of independent observations of the same variable. I probably have not understood something because I can't really solve any of the problems just based on the formula give...

ah, ok I understood quite a bit from that. But there is still just one more problem... f(x) =\frac{1}{4}(1 -\frac{x}{8}) for 0 <= x <= 8, 0 otherwise. Find P(X > 6) So after integration \int_{6}^{\infty }\frac{1}{4}(1 -\frac{x}{8}) dx it turns into (\frac{x}{4} - \frac{x^2}{64})...

Repeat Post!

That makes sense... ok What about situations with a minus infinity and limits that tend to infinity. Like f(x) = \frac{1}{15}(x^2 + 2x) find P(X<1.5) \int_{-\infty}^{1.5} f(x) dx In this case I can see it would tend to zero but what if x were on the denominator? The expression would tend to...

Im sorry I just used to 4x because when I type the equations in my book the latex just gets messed up (Spent about 20 mins on last post) What I really want to understand is how to evaluate an integral when its infinite. I'm going to write integration from a to b as I(a,b) ? Printer cartridge...

Homework Statement So I understand how to evaluate P(4 < X < 6) where the probability density function is f(x)=4x I can't seem to understand how to evaluate P(X>6) I would have to do something like \int^6_\infty 4x ... so how do I evaluate it? Homework Equations The Attempt at a Solution The...

hey your right! So I used a2Var(X) and got 0.129 by making other basic math errors too, and thought it was close enough to the answer --> 0.127 P(V > 0) V ~ N(-10,77) P(V > 0) = P(Z > (0-(-10)/\sqrt{77}) = P(Z > 1.1396) = 1 - \Phi(1.1396) = 1-0.8726 = 0.1274 \cong 0.127 Thanks so...