Statistics question (random variable)?
For the following exercise, determine the range (possible values) of the random variable X. Wood paneling can be ordered in thicknesses of 1/14, 1/7, or 3/14 inch. The random variable is the total thickness of paneling in two orders. The range of X is?
Please can you help me in this exercise on the probability? Thanks in advance.?
I take it they are both, for some reason (train times? class schedules?) certain to arrive on the hour--it's just not certain which hour. So we could make a table of possible arrival-time combinations, where the rows represent the times for one person and the columns the times for the other. I'll just list the pair of times (hour only) for each entry: 3-3 3-4 3-5 4-3 4-4 4-5 5-3 5-4 5-5 We can make a corresponding table for the value of X in each case: 0 1 2 1 0 1 2 1 0 So, out of 9 equally likely cases, we have 3 cases where X=0, 4 cases where X=1, and 2 cases where X=2, and thus E(X) = (3 * 0 + 4 * 1 + 2 * 2)/9 = 8/9
Let Y e a discrete random variable with mean μ and variance σ^2. If a and b are constants, use theorems of?
Ok here are the theorems to be used!! Theorem 1: Let Y be a discrete random variable with probability function p(y) and c be a constant. Then E(c) = c. Theorem 2: Let Y be a discrete random variable with probability function p(y), g(Y) be a function of Y, and c be a constant. Then E[cg(Y)] = cE[g(Y)]. THeorem 3. Let Y be a discrete random variable with proability function p(y) and g1(Y), g2(Y)...gk(Y) be k functiosn of Y. Then E[g1(Y)+...gk(Y)] = E[g1(Y)] +....+ E[gk(Y). theorem 4. Let Y be a discrete random variable with probability function p(y) and mean E(y) = μ; then V(Y) = σ^2= E[(Y - μ)^2] = E(Y^2) -μ^2. Please state which theorem you used, if possible. thank you so muchhhh
Help with STATISTICS, how do I do these problems?
I just started Statistics and I think that I wasn't paying enough attention. Here are the problems I can't figure out: Many random number generators allow users to specify the range of the random numbers to be produced. Suppose that you specify that the random number Y can take any value between 0 and 2. Then teh density curve of the outcomes has constant height between 0 and 2, and height 0 elsewhere. (a) is the random variable Y discrete or continuous? Why? (b) what is the height of the density curbe between 0 and 2? Draw a graph of the density curve, (c) use your graphy from (b) and the fact that probability is area under the curve to find P(Y<1). then the next problem says find tehse probabilies as areas under the density curve you sketched in problem 48. (a) P (0.5
This is Variance Question please?
First off, you wrote f(x,y) = 2(1-x), but this is clearly not a function of y ... Regardless, you compute the mean of this distribution with the integral: int xf(x)dx = int (2x-2x^2)dx = x^2-2/3x^3 The bounds of the integral are [0,1] since that is the support of the distribution: Mean = int_0^1 xf(x)dx = x^2-2/3x^3 |_0^1 = 1-2/3 = 1/3 Now that you know the mean M=1/3, you can compute the variance using the second moment: E[X^2] = int_0^1 x^2f(x)dx = int_0^1 (2x^2-2x^3) dx = 2/3x^3-1/2x^4 |_0^1 = 2/3-1/2 = 1/6 a) The variance is: s = Var[X] = E[X^2] - E[X]^2 = 1/6 - (1/3)^2 = 2/9. Chebyshev's inequality is usually written as Pr( |X-M| > ks) < 1/k^2. Using k=2, M=1/3, s=2/9 as we found earlier, just plug in to get Pr( |X-1/3| > 4/9 ) < 1/4 Pr( X-1/3 < -4/9 ) + Pr( X-1/3 > 4/9 ) < 1/4 Pr( X < -1/9 ) + Pr( X > 7/9 ) < 1/4 Note that the first part is 0, since the probability density function f(x) is only nonzero for x in [0,1]. Reducing Pr( X < -1/9 ) to 0, we're left with the statement Pr( X > 7/9 ) < 1/4 Now we just compute that probability: int_7/9^1 (2-2x)dx = 2x-x^2 |_7/9^1 = (2-1) - (14/9 - 49/81) = 4/81 b) Obviously 4/81 is less than 1/4 so the inequality is satisfied. The third part just asks for P( X > 1/10 ) since X is measured in units of 5000 and 500 is 1/10 of 5000. Once again, use an integral: int_.1^1 (2-2x)dx = 2x-x^2|_.1^1 = (2-1) - (2/10 - 1/100) = 81/100 c) So the probability of making at least 500 is 81/100. Good luck with math.
HELP With Statistics... Binomial Distribution?
The key requirements for a binomial distribution are 1. A two outcome situation (which is what gives it the name Binomial) 2. A fixed number of trials. 3. A fixed probability of success at each trial. 4. Independence of successive results. A) It is not a two outcome situation. Look no further. B) "Today's sample is 20 pairs" suggests that it is not always 20 so fixed number of trials would not be the case. This could be a red herring. Also wear in the machinery might gradually increase the probability of defects but probably so slowly as to be negligible This one is not easy to decide. On balance I would say yes it is binomial. C) I would think that all requirements are met with this one.
Please help with statistics question?
A crook who never heard of Benford's law might choose the first digits of his faked invoices so that all of 1, 2, 3, 4, 5, 6, 7, 8, and 9 are equally likely. Call the first digit of a randomly chosen fake invoice W for short. 1. What is the probability distribution for the random variable W? 2. Find P(W ≥ 5). Round your result to 3 decimal places. Compare your result with the Benford's law probability.