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Aha, I was dividing the other way around :p

Thanks

Pro tip: include your units when you do division, then you'll always know if you're getting what you want.

100 percent / 16 hours = 6.25 percent / hour ("6.25 percent of the battery is used per hour")
16 hours / 100 percent = 0.16 hours / percent ("It takes 0.16 hours to use 1 percent of the battery")

This is a tip that scales well even when you're doing calculus and more complex things. If the units you get aren't the units you want, then you did something wrong.
 
Should I ask here or in the programming OT for help with some matlab code?

I can't speak to Matlab because my particular discipline is more R-oriented, but my sense is that you generally get better statistical/math programming help from statisticians/mathematicians than from programmers.
 
It's me again ! I'm in another probability dead end ! Though this I only have one question.

Here is the subject: We have a jar which contains 3 ball numbered from 1 to 3. We drawn and put back n ball successively. We assume that all drawn are independent , and Xi is the value (1,2 or 3) of the drawn ball number i.

What is the distribution of Xi, for 1 <= i <= n ?
 
It's me again ! I'm in another probability dead end ! Though this I only have one question.

Here is the subject: We have a jar which contains 3 ball numbered from 1 to 3. We drawn and put back n ball successively. We assume that all drawn are independent , and Xi is the value (1,2 or 3) of the drawn ball number i.

What is the distribution of Xi, for 1 <= i <= n ?

It'd be a uniform distribution, right? At the i-th draw, each ball has an equal chance of 1/3 to be selected. And since each draw is independent from the others, we have this uniform distribution for all n draws.
 
bombing the first test of the semester, I'm remembering that (1) I'm not good at math since HS or (2) I'm not good at school. i need help and I'm hoping you guys will take it easy on me, first time in a math class in 3 years.
its algebra..1.lol :(.
 
A general question I had:

What's the most convenient way to calculate residues of non-rational functions? Say e^(1/x) or cos(sin(1/x)) or something along those lines.

I know that by definition that's the integral of the function over a closed contour encircling whatever singularity there is, but that's usually hard to find.

Do we just take the Laurent expansion in a nbd of the singularity and hope we get clean coefficients?
 
V = 4/3 * pi * r^3

Find r with above, since you know V = 288 * pi.

Differentiating yields...

dV/dt = 4 pi r^2 dr/dt

dr/dt = 2 cm / hr, r = from above.

Answer.

i have solved this on a final exam, haha wanted to get an answer here to check if i was correct

288pi cm^3 /hr

i get kinda weirded out when the solution is in the problem
 
^ Looks fine by me. You get r = 6 (ignore units) and you can write dV/dt = 4 pi r^2 dr/dt = 4/3 pi * r^2 *(3 * dr/dt) and note 3*dr/dt = 6 = r, so that's in fact equal to 4/3 pi r^3 i.e. your volume.
 
^ Looks fine by me. You get r = 6 (ignore units) and you can write dV/dt = 4 pi r^2 dr/dt = 4/3 pi * r^2 *(3 * dr/dt) and note 3*dr/dt = 6 = r, so that's in fact equal to 4/3 pi r^3 i.e. your volume.

thanks so much for checking (:

its a fairly simple problem but i guess maybe the intent to make the answer what it is was to make the student scratch their head for a second
 
it's been a while since I was in a math class so I'm starting low.
Algebra I help, please. graphing linear equations.

I feel like I'm on the right path but when solving for my 3rd ordered paid(I still pick X as my variable to substitute) but no matter the number I pick, it seems to not be a straight line when connecting them. so either my calculations are wrong, I'm picking the wrong numbers, or my graphing is wrong. I have a quiz in a couple hours I'm trying not to fail.

e2zn4nm.jpg
 
it's been a while since I was in a math class so I'm starting low.
Algebra I help, please. graphing linear equations.

I feel like I'm on the right path but when solving for my 3rd ordered paid(I still pick X as my variable to substitute) but no matter the number I pick, it seems to not be a straight line when connecting them. so either my calculations are wrong, I'm picking the wrong numbers, or my graphing is wrong. I have a quiz in a couple hours I'm trying not to fail.

4x + 6 + 3y = 18
3y = 12 - 4x
y = 4 - 4/3x

The graph is definitely a straight line. Pick the points where it intersects the axis ( in this case (0,4) & (3,0)) and draw a straight line between these two and continuing off both ends to infinity.
 
I feel like I'm on the right path but when solving for my 3rd ordered paid(I still pick X as my variable to substitute) but no matter the number I pick, it seems to not be a straight line when connecting them.

It is because you did not consistently use the equation 4x + 3y = 12 (from 4x + 6 + 3y = 18) for each value of x that you chose.

Remember that, unless the problem asks you to find 3 points on the line, you just need 2 points to determine what the line looks like.

When you have the equation in the form of ax + by = c like above, you can pick the x- and y-intercepts (i.e. set one of the variables to 0 and solve for the other) as the two points.
 
4x + 6 + 3y = 18
3y = 12 - 4x
y = 4 - 4/3x

The graph is definitely a straight line. Pick the points where it intersects the axis ( in this case (0,4) & (3,0)) and draw a straight line between these two and continuing off both ends to infinity.

what about when there is a fraction in front of the variable?
(ex: y = 3/4x + 2)

should I set the x variable to 0 and then receive the pair (0,2)?
or do I multiple by the demonitator(4) to cancel it out, then would it turn out to be y=3(4)+2 => y=12+2 => y=14? (4,14)

14 seems very high compared to the others so I'm assuming I'm plugging something in wrong
 
what about when there is a fraction in front of the variable?
(ex: y = 3/4x + 2)

should I set the x variable to 0 and then receive the pair (0,2)?
or do I multiple by the demonitator(4) to cancel it out, then would it turn out to be y=3(4)+2 => y=12+2 => y=14? (4,14)

14 seems very high compared to the others so I'm assuming I'm plugging something in wrong

In you're hypothetical sitation, if you set x=4 you get y=5 (you forgot to divide by the 4). In the original question, set x,y=0 and since the fraction in the question has a demominator=3, set x=3.
 
How do I find the standard deviation and mean of a normal distribution if I am just given a probability at a certain range? I was given the probability of X > 30 = 0.12, and I had to find the probability of X < 5. This was for a bonus on a test, and I tried many different rearrangements and couldn't figure out a way to do it without knowing at least one of the mean or standard deviation.
 
How do I find the standard deviation and mean of a normal distribution if I am just given a probability at a certain range? I was given the probability of X > 30 = 0.12, and I had to find the probability of X < 5. This was for a bonus on a test, and I tried many different rearrangements and couldn't figure out a way to do it without knowing at least one of the mean or standard deviation.

You are correct that you'd need to know one of the two. The density function of a normal distribution is parameterized by mu and sigma. As with any algebraic relationship of three variables, you can recover the third from the first two, but not two of them from one, unless you have a system of equations that allows you to, which in this case you do not.

Here's how I would approach the problem:
How many standard deviations above the mean of a normal distribution is the 88th percent quantile? ~1.18 standard deviations (just did a quick table check). Thus Mu + 1.18sd = 30; given either you could recover the other, and then transform to standard normal.

Let mu=25, then sd = 4.23, and p(X<5) < 0.0002, the table I checked only had quantiles for z-scores down -3.50
Let mu=15, then sd = 12.71, and p(x<5) ~= 0.2514

Did a quick sanity check in case my brain is totally failing me by checking the documentation for R's pnorm (CDF of normal distribution given quantile) and qnorm (recover quantile of value given normal distribution), and both require a mu and a sigma, or they assume standard normal parameters (zero mean unit variance) I think it's somewhat unlikely that your exam intended you to assume this; unit variance would give you a p~=0, zero mean would give you a wide variance.

So yeah unless my brain is totally broken right now, I have no idea what they would intend you to do. It sounds like either they made a mistake or you forgot something (no offence!)

Edit: Gotchaye suggested to me that perhaps the question was referring to a distribution from a previous question on the test, which would make sense. You don't mention if the test had a z-score table and/or computers provided, as well, as this kind of question requires one of the two even if you did have one of the parameters. One useful question type I'd put on a stats test would be if you knew either mu=17.5 or sd=10.59 (the situation where 30 and 5 are equidistant from the mean) in which case they'd be asking you to verify that p(X < mu - alpha) = p(X > mu + alpha) for all alpha in the normal distribution.
 
You are correct that you'd need to know one of the two. The density function of a normal distribution is parameterized by mu and sigma. As with any algebraic relationship of three variables, you can recover the third from the first two, but not two of them from one, unless you have a system of equations that allows you to, which in this case you do not.

Here's how I would approach the problem:
How many standard deviations above the mean of a normal distribution is the 88th percent quantile? ~1.18 standard deviations (just did a quick table check). Thus Mu + 1.18sd = 30; given either you could recover the other, and then transform to standard normal.

Let mu=25, then sd = 4.23, and p(X<5) < 0.0002, the table I checked only had quantiles for z-scores down -3.50
Let mu=15, then sd = 12.71, and p(x<5) ~= 0.2514

Did a quick sanity check in case my brain is totally failing me by checking the documentation for R's pnorm (CDF of normal distribution given quantile) and qnorm (recover quantile of value given normal distribution), and both require a mu and a sigma, or they assume standard normal parameters (zero mean unit variance) I think it's somewhat unlikely that your exam intended you to assume this; unit variance would give you a p~=0, zero mean would give you a wide variance.

So yeah unless my brain is totally broken right now, I have no idea what they would intend you to do. It sounds like either they made a mistake or you forgot something (no offence!)

Edit: Gotchaye suggested to me that perhaps the question was referring to a distribution from a previous question on the test, which would make sense. You don't mention if the test had a z-score table and/or computers provided, as well, as this kind of question requires one of the two even if you did have one of the parameters. One useful question type I'd put on a stats test would be if you knew either mu=17.5 or sd=10.59 (the situation where 30 and 5 are equidistant from the mean) in which case they'd be asking you to verify that p(X < mu - alpha) = p(X > mu + alpha) for all alpha in the normal distribution.

I guess I should've reminded you that this is just for a grade 12 data class, with a 4 lesson long unit on normal distributions.

The things I had available for the test and bonus were a Z-score table and a calculator. The only thing I managed to calculate was the Z-value that was for the 88th percentile, and I believe that was 1.175. I didn't know how to find the probably of X < 5 after that.

I should be getting the test back later today, so I can type up the exact question when I get home from school. Thanks for the reply, and sorry for my delayed response.
 
I guess I should've reminded you that this is just for a grade 12 data class, with a 4 lesson long unit on normal distributions.

The things I had available for the test and bonus were a Z-score table and a calculator. The only thing I managed to calculate was the Z-value that was for the 88th percentile, and I believe that was 1.175. I didn't know how to find the probably of X < 5 after that.

I should be getting the test back later today, so I can type up the exact question when I get home from school. Thanks for the reply, and sorry for my delayed response.

Oh, if your teacher handed out the z-table, then he/she likely meant to ask you to assume that the mean is 0. Otherwise, you and Stump are correct, we would need two information to determine two unknowns.


We figure out the standard deviation from the given, that P(x > 30) = 0.12. You are correct that P(x <= 30) = 0.88 when z = 1.175 approximately.

Hence,

(30 - 0) / sigma = 1.175 => sigma = 25.532 (approx.)

Then, figure out the probability that x < 5.

z = (5 - 0) / 25.532 = 0.196 => P(x < 5) = 0.577 (approx.)


----


(I found out yesterday that I am a full member now, wanted to celebrate it with math-GAF first.)
 
I feel like I'm overthinking this way too much, help:

Imxajgs.png




The answer is
4/3,
but I'm not sure what properties to use to get there anymore. I've been working in circles. I figure l'Hopitals, really.
 
I feel like I'm overthinking this way too much, help:

Imxajgs.png




The answer is
4/3,
but I'm not sure what properties to use to get there anymore. I've been working in circles.

Multiply and divide by sqrt(3x^2 + 3) + sqrt(x^2 + 5).

The numerator becomes 2x^2 - 8, factorization gives: 2 * (x-2)(x+2). Now you can cancel out the x-2 term in the denominator and the rest is trivial.
 
I feel like I'm overthinking this way too much, help:

Imxajgs.png




The answer is
4/3,
but I'm not sure what properties to use to get there anymore. I've been working in circles. I figure l'Hopitals, really.

Take the derivative of both numerator and denominator:

Numerator: (3x^2 - 3)^1/2 - (x^2 + 5)^1/2
Derivative: (1/2)(6x)(3x^2 - 3)^(-1/2) - (1/2)(2x)(x^2 + 5)^(-1/2)
= (3x)(3x^2 - 3)^(-1/2) - x(x^2 + 5)(-1/2)

Denominator: x - 2
Derivative: 1

Divide the two at x = 2:

= 6*9^(-1/2) - 2*9^(-1/2)
= 4*9^(-1/2)
= 4/3
 
Oh, if your teacher handed out the z-table, then he/she likely meant to ask you to assume that the mean is 0. Otherwise, you and Stump are correct, we would need two information to determine two unknowns.


We figure out the standard deviation from the given, that P(x > 30) = 0.12. You are correct that P(x <= 30) = 0.88 when z = 1.175 approximately.

Hence,

(30 - 0) / sigma = 1.175 => sigma = 25.532 (approx.)

Then, figure out the probability that x < 5.

z = (5 - 0) / 25.532 = 0.196 => P(x < 5) = 0.577 (approx.)


----


(I found out yesterday that I am a full member now, wanted to celebrate it with math-GAF first.)

I don't think this is it because the question is actually a word problem. Here it is:

It is know that the lifetime (in years) of a certain brand of TV set is normally distributed. If 12% of the TV's survive past 30 years, determine the probability that a new TV will die within the first 5 years.

So in this word problem, it isn't possible for the mean to be 0 as a TV cannot have a negative lifespan.

Congratulations on the membership.
 
Thanks for the help yesterday, DEAD RABBIT, alatif113; good to see where I went wrong in the second option, and the much simpler alternative of the first response.

Hey all, I'm taking my first ever Calculus class this summer.
Anyways, I wondering how I determine the infinite limit for this problem.

Screen_Shot_2015_06_02_at_7_10_48_PM.png


secx = 1/cosx, that's where you start.

So by this, we know that directly at pi/2, and -pi/2 there are asymptotes in the secx graph, since they're zeroes in the cosx graph and 1/0 is undefined. Not that you're looking for the two sided limit, anyway, just thought I'd throw that in.

But the question asks to approach from the left, so you'd end up with a near insignificant number in the denominator like -0.0000xx. 1 over such a small number is going to be a really big negative, obvious enough, etcetc -- limit is negative infinity.
 
I don't think this is it because the question is actually a word problem. Here it is:

It is know that the lifetime (in years) of a certain brand of TV set is normally distributed. If 12% of the TV's survive past 30 years, determine the probability that a new TV will die within the first 5 years.

So in this word problem, it isn't possible for the mean to be 0 as a TV cannot have a negative lifespan.

Congratulations on the membership.

Thanks. Yeah, I don't think it's possible to solve the problem without the additional information of the mean or the standard deviation.
 
Anyone really good at probability, statistics, and optimization?

Suppose there are 2 events P and Q that each have a payout associated with them. P pays out, on average, P_x of resource X and P_y of resource Y. Q pays out on average Q_x of X and Q_y of Y.

The goal is to figure out the optimal strategy for obtaining some fixed quantity of X and Y in the fewest number of events.

We can write these as linear equations as just say that if we do P p times, and Q q times, then the quantity of X obtained is p*P_x + q*Q_x, and the quantity of Y obtained is p*P_y + q*Q_y. Then we can make inequalities. So we need:

p*P_x + q*Q_x >= N_x
p*P_y + q*Q_y >= N_y

From here it's easy to solve this system of equations for p and q.

But I'm not sure how to interprete the results. Does it mean the optimal strategy is to do P exactly p times, and then do Q until a sufficient quantity is obtained? Because P_x is jsut the expected payout, it's not actually guaranteed on any particular sample, just over the long run.

Is there a better strategy perhaps?
 
Anyone really good at probability, statistics, and optimization?

Suppose there are 2 events P and Q that each have a payout associated with them. P pays out, on average, P_x of resource X and P_y of resource Y. Q pays out on average Q_x of X and Q_y of Y.

The goal is to figure out the optimal strategy for obtaining some fixed quantity of X and Y in the fewest number of events.

We can write these as linear equations as just say that if we do P p times, and Q q times, then the quantity of X obtained is p*P_x + q*Q_x, and the quantity of Y obtained is p*P_y + q*Q_y. Then we can make inequalities. So we need:

p*P_x + q*Q_x >= N_x
p*P_y + q*Q_y >= N_y

From here it's easy to solve this system of equations for p and q.

But I'm not sure how to interprete the results. Does it mean the optimal strategy is to do P exactly p times, and then do Q until a sufficient quantity is obtained? Because P_x is jsut the expected payout, it's not actually guaranteed on any particular sample, just over the long run.

Is there a better strategy perhaps?

You have stumbled onto a harder problem than you think it is. First, your final question, on certainty. You are asking to simultaneously optimize for the values p and q that, on expectation, satisfy your constraints, and optimize for the likelihood of them satisfying the constraints. There are a variety of ways to do this. One is to introduce a penalization term which penalizes uncertainty and is weighted based on to what extent you are willing to trade off certainty for fewer moves. How you weight these terms will determine the solution you get.

Another is to do an algorithmic approach where you choose some of the events and then recalculate your progress relative to your target, reoptimizing on the newly ideal p and q; so for instance if you expect that 3p and 5q will get you there, then maybe you start with q, do say 3 q, and then ask "okay, what do I need to do to get to where I'm trying to get from where I am now"; the new optimization might be 3p and 2q, or it might be some other figures depending on your probabilistic luck on the draws. This would be a heuristic approach and I don't have the time to think through whether this guarantees the minimum possible moves in expectation or if it would just give you a good-enough solution.

Now, your original solution. There are potentially multiple solutions (say for example you can choose between 2p and 1q or 1p and 2q, both satisfy your constraints). You appear to be saying that you are indifferent between those solutions, and want to simply minimize the total number of events regardless of which events they are. Thus you want to minimize p+q s.t. pE[P_x] + qE[Q_x] >= N_x , pE[P_y] + qE[Q_y] >= N_y , p>=0, q>=0. I'm not able to work out the best method to do this, but it visually on first inspection looks to me like a convex optimization problem. Solving it analytically is non-trivial.

However, I think you also over-asked your question by leaving it as general as you did. You're clearly working on something for Final Fantasy Record Keeper, like a way of telling people which levels to play to get the item drops they want. Personally (this is a modified pocket algorithm approach) I would just run a loop over p=1 to N, q=1 to N, if your inequalities hold, record the value of p+q, if it's lower than the previous minimum p+q, record the p and q values and the new minimum p+q value. If your loop runs and you find no combinations of p and q that satisfy the constraint, simply report that it's not feasible for the end user and you have no advice. If you have at least one satisfying combination, report the optimal one. Set N equal to half the maximum feasible number of events you think someone is willing to play to get the outcome they want (certainly no more than say 50). Ta-da.

Literally:
Code:
N=50
bestSum = 99999
bestPQ = (0,0)
for p in 1...N:
	for q in 1...N:
		if p*PX + q*QX >= NX and p*PY + q*QY >= NY and p+q<bestSum:
			bestSum=p+q
			bestPQ = (p,q)
if bestPQ != (0,0):
	print p,q
else:
	print "lol no"
 
You have stumbled onto a harder problem than you think it is. First, your final question, on certainty. You are asking to simultaneously optimize for the values p and q that, on expectation, satisfy your constraints, and optimize for the likelihood of them satisfying the constraints. There are a variety of ways to do this. One is to introduce a penalization term which penalizes uncertainty and is weighted based on to what extent you are willing to trade off certainty for fewer moves. How you weight these terms will determine the solution you get.

Another is to do an algorithmic approach where you choose some of the events and then recalculate your progress relative to your target, reoptimizing on the newly ideal p and q; so for instance if you expect that 3p and 5q will get you there, then maybe you start with q, do say 3 q, and then ask "okay, what do I need to do to get to where I'm trying to get from where I am now"; the new optimization might be 3p and 2q, or it might be some other figures depending on your probabilistic luck on the draws. This would be a heuristic approach and I don't have the time to think through whether this guarantees the minimum possible moves in expectation or if it would just give you a good-enough solution.

Now, your original solution. There are potentially multiple solutions (say for example you can choose between 2p and 1q or 1p and 2q, both satisfy your constraints). You appear to be saying that you are indifferent between those solutions, and want to simply minimize the total number of events regardless of which events they are. Thus you want to minimize p+q s.t. pE[P_x] + qE[Q_x] >= N_x , pE[P_y] + qE[Q_y] >= N_y , p>=0, q>=0. I'm not able to work out the best method to do this, but it visually on first inspection looks to me like a convex optimization problem. Solving it analytically is non-trivial.

However, I think you also over-asked your question by leaving it as general as you did. You're clearly working on something for Final Fantasy Record Keeper, like a way of telling people which levels to play to get the item drops they want. Personally (this is a modified pocket algorithm approach) I would just run a loop over p=1 to N, q=1 to N, if your inequalities hold, record the value of p+q, if it's lower than the previous minimum p+q, record the p and q values and the new minimum p+q value. If your loop runs and you find no combinations of p and q that satisfy the constraint, simply report that it's not feasible for the end user and you have no advice. If you have at least one satisfying combination, report the optimal one. Set N equal to half the maximum feasible number of events you think someone is willing to play to get the outcome they want (certainly no more than say 50). Ta-da.

Literally:
Code:
N=50
bestSum = 99999
bestPQ = (0,0)
for p in 1...N:
	for q in 1...N:
		if p*PX + q*QX >= NX and p*PY + q*QY >= NY and p+q<bestSum:
			bestSum=p+q
			bestPQ = (p,q)
if bestPQ != (0,0):
	print p,q
else:
	print "lol no"

Heh. So since you made it concrete, what I specifically want to do is allow a person to enter up to two types of orbs they want to simultaneously farm. Then it should find all locations where either of the orbs drop, and just tell the user exactly where to go.

I'm open to the possibility of it being an iterative algorithm, so that for example it tells you what to do next, then after you do it, potentially alters its next suggestion based on the outcome.

So you start out and you give it some inputs like "I need 40 of this and 25 of that" and the goal is to suggest where to go to achieve that with the minimum possible stamina usage.
 
Thanks. Yeah, I don't think it's possible to solve the problem without the additional information of the mean or the standard deviation.

Ok, so I asked my teacher and got the solution to the bonus question. It makes sense to me and I don't think there is anything wrong with it. Here it is:

(Excuse the quality of my paint diagram)

So the x-axis is for the word problem, and the z-values are found through conversions. Because of the nature of the word problem, the lowest value on the x-axis is 0. Convert this to a z-value, and you get -3. The mean of the z-curve is 0. The 1.175 represents 88% and is found in the z-table. 1.175 represents 4.175 standard deviations, and because they represent the same area, 30 is also 4.175 standard deviations --> 4.175&#963; = 30, therefore &#963; = 7.186. The mean is 3&#963;, and is therefore 21.557. Now both the mean and the sigma of the x-curve are known, and the probability of x < 5 can easily be found. z = (5 - 21.557) / 7.186 = -2.3, and the probability of -2.3 = 1.07%.
 
On what basis is there a -3 z-score where x=0? There is insufficient information given in the problem to come to that conclusion. If that information was provided, the rest follows, certainly.
 

I don't believe he actually said that, but the lowest value on our z-score table is -2.99, and the highest is 2.99. He also wrote on my page that 0 for x is -3 for z. Looking at the values I can see that 3 isn't supposed to be the highest value, and looking online tells me it is 4. I guess he simplified it because it would make one standard deviation on the standard normal distribution = 1 rather than 1.33.
 
I don't believe he actually said that, but the lowest value on our z-score table is -2.99, and the highest is 2.99. He also wrote on my page that 0 for x is -3 for z. Looking at the values I can see that 3 isn't supposed to be the highest value, and looking one tells me it is 4. I guess he simplified it because it would make one standard deviation on the standard normal distribution = 1 rather than 1.33.

his "simplification" is incorrect, and z-scores have no "highest" (or "lowest") value; that's not a meaningful concept. there is no reason to believe 0 has a z-score of -3. it seems your teacher does not understand where z-scores come from or what the density function of a normal distribution is, or how we derive the p-values that go into the z-score table you use.

if you want i can walk you through the reasoning (this is not a high school topic--this is something that would be taught in a 2nd or 3rd year university stats course at the earliest and would require that you have knowledge of calculus), but an easier way to think about this might be considering the real world:

imagine adult male human height is normally distributed with a mean of 70 inches (5 foot 10) and sd of 2.75 inches. this is approximately correct, by the way. this means that 99.5% of adult males are between 5'2" and 6'6". but there are people in your city taller than 6'6". there's an NBA, right? now look up the tallest and shortest adult men in history. what would be their z-scores?

values outside three standard deviations are improbable, but the likelihood of at least one such improbable case varies depending on the number of units in the population you are sampling from. there are 7 billion people on earth. the tallest person on earth is, say, a 1 in 7 billion chance: p(x>=tallest human) = 0.000000000142857. that's a very low probability, but there's one of them in the population, right? that's OK.
 
his "simplification" is incorrect, and z-scores have no "highest" (or "lowest") value; that's not a meaningful concept. there is no reason to believe 0 has a z-score of -3. it seems your teacher does not understand where z-scores come from or what the density function of a normal distribution is, or how we derive the p-values that go into the z-score table you use.

if you want i can walk you through the reasoning (this is not a high school topic--this is something that would be taught in a 2nd or 3rd year university stats course at the earliest and would require that you have knowledge of calculus), but an easier way to think about this might be considering the real world:

imagine adult male human height is normally distributed with a mean of 70 inches (5 foot 10) and sd of 2.75 inches. this is approximately correct, by the way. this means that 99.5% of adult males are between 5'2" and 6'6". but there are people in your city taller than 6'6". there's an NBA, right? now look up the tallest and shortest adult men in history. what would be their z-scores?

values outside three standard deviations are improbable, but the likelihood of at least one such improbable case varies depending on the number of units in the population you are sampling from. there are 7 billion people on earth. the tallest person on earth is, say, a 1 in 7 billion chance: p(x>=tallest human) = 0.000000000142857. that's a very low probability, but there's one of them in the population, right? that's OK.

While he may have taught us wrong or mislead us on this, I don't think he doesn't know what he is talking about. His degree is in statistics after all, and he is the head of the Math department at my school. There has to be some reason why he wanted my class to believe the standard normal went from -3 to 3. It may be because he didn't think a lot of people would understand, or it may be for another reason.

I am good on the explanation. It is likely that I will learn it eventually through my major, and I doubt my current knowledge of calculus is sufficient to understand. My calculus teacher doesn't even teach us using Leibniz Notation, so I will probably have some catching up to do in university. But that is another topic.

Your simplified explanation makes sense and is very helpful. I see now that there are rare outliers that still count, and the z values I looked up online are rounded. Thanks for all the help :)
 
I'm doing a trig course online. Are there any good free PC programs for working out problems and doing calculations? I have a TI-83 and a scientific calculator but I'd like to simplify my workflow.
 
Can anyone help me make sense of this?

48ChhqG.png


First 2 lines are fine, but then:
Line 3 - I don't understand how Z+Z* = 2Z and not 2Re(Z).
Line 4 - I tried opening it a bunch of times and I can't get back to Line 3 from there. Why does this work?
 
Can anyone help me make sense of this?

48ChhqG.png


First 2 lines are fine, but then:
Line 3 - I don't understand how Z+Z* = 2Z and not 2Re(Z).
Line 4 - I tried opening it a bunch of times and I can't get back to Line 3 from there. Why does this work?

upandaway, you are correct. Are there additional hypotheses that you didn't write?

To go from line 2 to line 3, we need to assume/know that the inner product between gamma*x and alpha*y is a real number.

To go from line 3 to line 4, we need to assume/know that gamma*x and alpha*y are linearly dependent and their inner product is a non-negative number (due to Cauchy-Schwarz inequality).
 
Oh sorry, the exercise is to prove that the schwarz inequality is an equality => x and y are dependent.

So we know:
|<x, y>| = ||x|| * ||y||

And they tell us to choose this gamma in the instructions
ACZjlaC.png


I see what you mean about line 4, we can use the equality there to go from 3 to 4, I totally missed that. I still don't understand lines 2->3 though.
 
Oh sorry, the exercise is to prove that the schwarz inequality is an equality => x and y are dependent.

So we know:
|<x, y>| = ||x|| * ||y||

And they tell us to choose this gamma in the instructions
ACZjlaC.png


I see what you mean about line 4, we can use the equality there to go from 3 to 4, I totally missed that. I still don't understand lines 2->3 though.


Nah, we cannot go from line 2 to line 3.

----

I'd begin the proof by saying that the statement holds trivially if x or y is the zero vector. Hence, we can assume that neither is the zero vector from now on.


Show that, with the particular choice of gamma (is this well-defined?) and the Cauchy-Schwarz equality assumption, the expression on line 2 leads to,

||gamma x - alpha y||^2 = ||x||^2 - 2 Re(alpha) ||x|| ||y|| + |alpha|^2 ||y||^2.


We still have freedom of choice for the scalar alpha. Can you think of what to choose so that the RHS above becomes 0, i.e.

||gamma x - alpha y||^2 = 0?


Are we then safe to conclude that x and y are linearly dependent?
 
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