I haven't done any maths beyond addition and subtraction for the last two years, essentially. So I've forgotten a lot of basic stuff, and while the internet is quite helpful I don't think I'm coming at this particular problem correctly.


At first glance it seems to be a combination problem. So, ₈C₃. Which ends up being 56, assuming I did it right. My question is, where does the probability of 0.4 come into play? Do I multiply 56 by 0.4?

If someone from SEN calls someone outside of SEN, the probability of the other person answering approaches 0% quite rapidly.

What's the probability of someone from SEN calling someone else from SEN? Does it still stand with SMS, Email or Facebooking?

I'm not the best at probability, but something like for A:

You want three answereds and five not-answered, no more no less, in whatever order
It's a .4 chance of answering and .6 of not-answering.

.4 * .4 * .4 * .6 * .6 * .6 * .6 * .6

B is basically asking when will she NOT answer exactly 8 times? Find .4^8 and then subtract it from 1.
C is just .4 * 8?

General rule for probability questions: If you're getting the probability of something is greater than 1, you did something wrong. Anything greater than 1 or less than 0 is not a probability.

Before I give the solutions, here some hints if you want to work things out yourself:
Kevin was on the right track for qusetion A but something else is necessary because those calls can occur in many different orders, which increases the probability. A combination becomes necessary because you are essentially arranging patterns of AAANNNNN where A is answering and N is not answering. Kevin's answer is correct for part B because the particular combination used there will be one.
In this problem, your random variable is following a binomial distribution with parameters n = 8 and p = .4



Post has been edited 6 time(s), last time on Oct 26 2012, 3:52 am by Mini Moose 2707. Reason: I stank at placing the decimal point whe

E(X) = np = 8*.4 = 3.2

np(1-p) is the variance, moose

I don't understand this. Why is the expected number not 8 * 0.4, that is, 3.2? .32 doesn't make any sense, for one thing.


EDIT NVM

Post has been edited 1 time(s), last time on Oct 26 2012, 3:14 am by Jack.

[10:09 am] Azrael -- Moose, I think you misplaced a decimal somewhere

Also, part of what you copied is striked out.

Ah, that explained why I was getting a different result too. Didn't notice the .6 was crossed out. K nvm then, thanks for the help everyone.

Using the Newton-Raphson method, carry out 4 iterations to find an improved estimate of the solution of x³ + 5x - 3 = 0. Begin with an initial value of x = 0.6.

x₀ = 0.6

f(x) = x³ + 5x - 3 = 0

f'(x) = 3x² + 5

x₁ = 0.6 - f(0.6) / f'(0.6)

f(0.6) = 0.216

f'(0.6) = 6.08

x₁ = 0.564473684

x₂ = 0.564473684 - ((0.564473684)³ + 5(0.564473684) - 3)/(3(0.564473684)² + 5)

x₂ = 0.564099772

x₃ = 0.564099772 - ((0.564099772)³ + 5(0.564099772) - 3)/(3(0.564099772)² + 5)

x₃ = 0.564099733

x₄ = 0.564099733 - ((0.564099733)³ + 5(0.564099733) - 3)/(3(0.564099733)² + 5)

x₄ = 0.564099733 (same answer, so I'd say it's accurate accurate enough for 4 iterations )
Am I doing this correctly?

Yes, this is Newton's Method. You can verify your approximation by substituting it into the original problem and seeing how accurate it is.

A note on calculator usage: if your calculator can store variables or previous answers and commands, you can use them to do the iterations quite easily. I did this problem now on my TI-83+ by just entering .6 by itself, then Ans-(Ans^3+5Ans-3)/(3Ans²+5) and pressing enter a bunch of times. I then verified it in the original equation with Ans^3+5Ans-3, which returned 0. Alternatively, if the equation were large and complicated, I might want to store .6 as the variable X on there and then reuse X in my entries, storing the result of each iteration as X.

My results:
x₀ = .6
x₁ = .5644736842
x₂ = .5640997728
x₃ = .564099733
x₄ = .564099733
The last result differs in more decimal places than my calculator shows by default, however it stores more than it displays. When I tried it in the original equation, my calculator returned just 0 because it was "close enough". (As far as I know, I don't know everything about how this thing works.)

Yep, my calculator does that too. I wrote it out fully because to the best of my knowledge maths tests generally require all your working is shown.

Yes, but once you write out a variable, you should be able to reuse it instead of writing out the actual values. EG:
x₂ = x₁ - ((x₁³ + 5x₁ - 3)/(3x₁² + 5))
x₂ = .5640997728
x₃ = x₂ - ((x₂³ + 5x₂ - 3)/(3x₂² + 5))

Otherwise whoever grades your exams is a jerk. Or if they're nice, you should be able to get away with
x₀ = .6
x_n+1 = x_n - ((x_n³ + 5x_n - 3)/(3x_n² + 5))
x₁ = .5644736842
x₂ = .5640997728
x₃ = .564099733
x₄ = .564099733

Post has been edited 1 time(s), last time on Nov 3 2012, 3:10 am by Mini Moose 2707.

A photographer is taking portraits and he knows that there is a relationship between the distance (x) in metres from the light sources, and the intensity (y), on a surface x metres from the light source.

X 00.75 01.00 01.25 1.5 1.75 2.00
Y 23.56 15.30 10.95 8.33 6.61 5.41

a) Draw three different graphs to investigate the relationship between x and y.

I can understand how to draw one graph (plotting the points given above) but how could I get three graphs?

b) Decide which model is appropriate for this relationship. Choose between:
1. y = mx + c
2. y = ae^kx
3. y = axⁿ

This one I can think of one way to work it out (put in some random variables and see what the shape of the graph is and whether or not it matches the graph from a) but I doubt that's how they want me to do it.

c) Calculate the values of the two constants for your choice, and give the equation of the relationship between x and y. Comment on the accuracy of your model.

I think I could do this by myself, but again I'm not entirely sure on the "proper" method.

d) If the photographer required a light intensity of 4, what is the maximum distance that the light source can be from the subject?

This would presumably be simple to do once you had the correct model and constants.

I'm not familiar with this kind of problem at all as I never did it back in HS, so I would appreciate either something I can read on this kind of problem or a general explanation of how it works. It looks simple but I don't have any understanding of the principles involved (I think)

Plot the data in Excel, and have it draw a line of best fit. Compare that line to the graphs of the three following equations.
1. y = mx + c
2. y = ae^kx
3. y = axⁿ
They then want you to calculate the above italicized and bolded constants, show the equation you arrived at, and discuss its accuracy.

They then want you to extrapolate beyond your data, plugging in 4 for Y in your function you found, which breaks the commandments of data analysis but I'm guessing they don't care.

EDIT: If you need the graphs, here you go. The blue line is your data, the equations, with their R² along the right are in the same order as given in the problem (and coincidentally also in order of increasing accuracy). The solid black line is the trend line for the linear (first), the dashed is your exponential (second), and the dotted, sitting right on the blue, is your power (third) method.



To find your x-value, just reverse-engineer for whichever equation you like.

Post has been edited 1 time(s), last time on Nov 5 2012, 11:19 pm by poison_us.

I have a random variable X and its probability distribution, mean, and variance.

Calculate the expected value of the random variable Y = 1000X + 2

Calculate the variance of the random variable Y = 1000X + 2

How do I do the above?

Expected value is rather intuitive in comparison to the variance. If you multipy a random variable by 1000, logically the expected value will also be multiplied by 1000. If you add 2 to a random variable, the expected value will also increase by 2. We can also say that expected value is a linear function.

As such, E[Y] = E[1000X + 2]
E[Y] = E[1000X] + E[2] (since expected value is linear)
E[Y] = 1000 * E[X] + 2 (expected value of 2 is 2 because 2 is a constant)

Now for the variance. Thinking logically and remembering that variance is a measure of spread, we can see that multiplying a random variable by 1000 will definitely increase the variance because the random values will get more spread out. If we add 2 to a random variable, the spread of the random values won't change, so we can deduce that adding two will not cause a change in variance. So, what remains is to figure out exactly how much the variance changes.

Variance, unlike expected value, is not linear. Fortunately, it so happens that we have a lovely theorem (page 35/68 in this .PDF so you can see it) that tells us that Var(cX) = c² * Var(X) and Var(X + c) = Var(X) for any constants c. Combining them, we can get Var(cX + d) = c² * Var(X). The proof of those follows from putting such values into the definition of variance and some algebra and is in the .PDF from Dartmouth that I linked.

Var(Y) = Var(1000X + 2)
Var(Y) = 1000² * Var(X)
Var(Y) = 1000000 * Var(X)

An airline's investigations into the weights of cases discovers that type A travel cases have a mean of 18.3 kg and standard deviaion sqrt13 kg and type B travel cases have a mean of 20.1 KG and standard deviation of sqrt15 kg. Five type A travel cases and four type B travel cases are randomly and independently chosen and put on a weighing machine.

a) Calculate the mean of the total weight of the luggage on the weighing machine.

I did ((18.3 x 5) + (20.1 x 4)) / 9
= 19.1 kg. I'm not 100% sure this is correct.

b) Calculate the variance of the total weight of the luggage on the weighing machine.
Here I'm not sure I understand the effect of different sample sizes. Do I calculate the variance of the combined sample size and divide by n, which in this case would be 9? That is,
sqrt13² + sqrt13² + sqrt13² + sqrt13² + sqrt13² + sqrt15² + sqrt15² + sqrt15² + sqrt15² / 9

Here, your function is essentially: Y = 5A + 4B

They want the total weight, so you don't have to divide by 9.
Where µ denotes mean: μ_Y = 5μ_A + 4μ_B

Nah, we don't have to get that into it. We aren't doing a Central Limit Theorem problem here, we are just working with a function of random variables.

The rule we need is that variances add. (They were nice enough to give you the standard deviations as square roots, so in this case it's a hint that you'll be using variances somewhere.)

Var(Y) = 5 * Var(A) + 4 * Var(B)

How can I justify my choice of model? I doubt "I graphed them all in excel and the power equation was the best fit" will be a sufficient answer. We can rule out the linear equation because, well, it's a line and the plotted points aren't in a straight line, but how do I prove that the equation cannot be the second (exponential) one?