Chapter 8.4, Problem 6E

-x² The normal distribution has p(x) = e 2 determine the CDF in terms Erf, mean and standard deviation.

Find the probability in tossing a fair coin four times, there will appear a) 3H and 1T b) 2T and 2H using binomial distribution and assume coin has p(H)=1/3.

The joint pdf of random variables X=1, 2 and Y=1, 2, 3 is P(X,Y)= X 10.05 Find (a) The value of k. (c) P(X>1, Y <2). Y 0.2 0.18 0.15] (b) the marginal probability function of X and Y. (d) Ex, Hy

The conditional probability function for the random variables X and Y is 0 P(Y/X) = x0 [0.9 10.1 y 1 2 0.1 0 0.8 0.1 2 0 0.1 0.9. With P(x=0)=0.2, P(x-1)=0.4. Find P(X,Y), Hx, My, E(XY), OXY.

If X is a continuous random variable having pdf as shown. Find 1. The constant k. 2. P(X>0). 3. X, X2,0%. k p(x) 4 k/2 X -3 -1 0 1 2

Given a normally distributed variable X with mean 10 and standard deviation 4, find: 1. P(X5).

I need some assistance solving Part B of this question. Refer to the excel data in the image provided to answer Part B. SoftBus Company sells PC equipment and customized software to small companies to help them manage their day-to-day business activities. Although SoftBus spends time with all customers to understand their needs, the customers are eventually on their own to use the equipment and software intelligently. To understand its customers better, SoftBus recently sent questionnaires to a large number of prospective customers. Key personnel—those who would be using the software—were asked to fill out the questionnaire. SoftBus received 82 usable responses, as shown in the file. You can assume that these employees represent a random sample of all of SoftBus's prospective customers. SoftBus believes it can afford to spend much less time with customers who own PCs and score at least 4 on PC Knowledge. Let's call these the "PC-savvy" customers. On the other hand, SoftBus believes it…

Suppose you are gambling on a roulette wheel. Each time the wheel is spun, the result is one of the outcomes 0, 1, and so on through 36. Of these outcomes, 18 are red, 18 are black, and 1 is green. On each spin you bet $5 that a red outcome will occur and $1 that the green outcome will occur. If red occurs, you win a net $4. (You win $10 from red and nothing from green.) If green occurs, you win a net $24. (You win $30 from green and nothing from red.) If black occurs, you lose everything you bet for a loss of $6. a.  Use simulation to generate 1,000 plays from this strategy. Each play should indicate the net amount won or lost. Then, based on these outcomes, calculate a 95% confidence interval for the total net amount won or lost from 1,000 plays of the game. (Round your answers to two decimal places and if your answer is negative value, enter "minus" sign.) Lower Limit                             Upper Limit

B1 The x distribution is a special case of Gamma distribution (not to be confused with gamma function; see below). The density function of the Gamma distribution with parameters and k is given by where -1 e -x/0 (x) = = г(k) Øk if x > 0, and otherwise, г(k) = √ ₁ k-1-x dx x' e is the gamma function. (a) For every k ≥ 1, 0 > 0, find the mode of the density. Hint: The algebra can be simplified by appropriate use of logarithms. ~ Now suppose that X1,..., Xn id Exp(\) and that we have a prior belief in A which is consistent with a prior distribution X. Gamma(a, b), for some a, ß, i.e. the prior density of is Baxa-1-BA T(a) e = so 01/ẞ and k = a. (b) Write down the likelihood, and show that the posterior distribution for \ is also a Gamma distribution, but with parameters a +n and B + Σ Xi. (c) Find the mode of the posterior distribution and examine the behaviour as n → ∞.