Ok thank you. So I have the PMF function. The problem asks to compute probabilities using...
1) the CLT (approximation)
2) exact distribution
What is the difference. I thought the CLT allowed the use of binomial dist because of the large n. How do I "approximate PMF" using CLT.
Oops, should be Combination I am guessing?
Using that I get P(X = 1,2,3,...) = .20, .26, .21, .13, ...
These just don't seem correct to me. Seem way to high??
Homework Statement
Of the people passing through an airport metal detector, 0.5% activate it; Let X denote the number among a randomly selected group of 500 who activate it.
1) What is the PMF of X
i) Using th CLT (approximate PMF)
ii) Using the exact distribution of X
2) P(X = 5)...
Ohhh okay, I don't think I really understood what the question was asking. So the question is asking for the TOTAL or COMBINED score distribution given that correlation coefficient is .75, which is why I solved for V(X+Y).
Soo then what about the E(X+Y), is that simply E(X) + E(Y) = 1000...
Ya sorry about the notation, I'm not up to speed with this syntax
So now my question is regarding part two... How would I find E(W) and V(W) when W = the sum of the weighted averages of Y (as apposed to part 1 where Y^bar was just the sum of the Y's divided by n = 4). Basically I'm confused...
Sure does, but how to I express mean and variance IN TERMS of mu and sigma^2
Heres what I got outa the book...
so Y^bar = Sum(Y_i/n)
E(Y^bar) = E(Sum(Y_i/n)) = (1/4)Sum(E(Y_i)) = (1/4)Sum(mu)
V(Y^bar) = V(Sum(Y_i/n)) = (1/16)V(Sum(Y_i)) = (1/16)(Sum(Y_i) + 2 SumSumcov(Y_i, Y_j)) =...
Homework Statement
Let Y_1, Y_2, Y_3, Y_4 be IID RV from a population with mean mu and variance sigma^2. Let Y^bar = .25(Y_1+Y_2+Y_3+Y_4) denote the average of these four RV's.
1)What are the expected value and variance of Y^bar in terms of mu and sigma^2
2)Consider a different...
I must be missing something... My thought was I solve V(X+Y) for cov(X,Y) and then plug it into my .75 = ... equation.
cov(X,Y) = [V(X+Y) - V(X) - V(Y)] = V(X+Y) - 20000
.75 = [V(X+Y) - 20000]/(sigma_x*sigma_y) = V(X+Y) - 20000/(100*100)
therefore V(X+Y) = .75(10000)+20000 = 27500 I...
So cov(X,Y) = E(XY) - E(X)E(Y)
I looked through my book and searched the web but I do not see how to solve E(XY). I know E(XY) = x*y*p(x,y) but how do I do it when all I know is that X & Y ~ N(500,10000)?
Ok, thanks...
The question continues...
2) Next assume that the correlation coefficient between the math and verbal scores is .75, Find the mean and variance of the resulting distribution
so... I got .75 = cov(X,Y)/(sigma_x * sigma_y) = [E(XY) - E(X)E(Y)]/(sigma_x * sigma_y)
where...
"How is the overall SAT score determined from the math and verbal scores? If it is their sum or their arithmetic average, then NO, you are incorrect. "
Im not sure I follow... I do not know how the OVERALL scores are determined. I gave the problem word for word as it was given.
Are you...
Homework Statement
Math and verbal SAT scores are each N(500, 10000)
1)If the math and verbal SAT scores were independently distributed, which is not the case, then what would be the distribution of the overall SAT scores? Find its mean and variance.
Homework Equations
The...
Homework Statement
X refers to score distribution in Math and Y refers to score distribution in Stat in a certain degree course exam. It is known that X~N(mean = 62, sigma=7) while Y~N(mean = 68, sigma=10). If X and Y are independent, find (i) P[X+Y>120]; (ii) P[X<Y]; (iii) P[X+Y>140]...
Thanks for the quick reply...
From my limited knowledge of Poisson distribution, lambda represents the average rate of success, in the case of the problem at hand, an average of 4 successful sales per week.
So my thought was that given the average rate of sales = 4 per week, find the amount...
Homework Statement
On the average, a grocer sells 4 of a certain article per week. How many of these should he have in stock so that the chance of his running of stock within a week will be less than 0.01? Assume Poisson distribution.
Homework Equations
The Attempt at a...