Question

if you can answer any of these i’d would appreciate it greatly

1. List the reasons why the expectation and the variance are used as measures of the center and the spread of a probability distribution. 2. The random variable Z in Table 8.4 takes the values 3000 and 4000 with probability of 0.5. Compute MAD(2) and StDev(z). State in words why MAD(Z) and StDev(2) take the values that they do. 3. As an alternative to the expectation, one could employ the "average value" of a random variable, namely, the sum of the values, divided by their number. By this measure, the average value of the random variable Y in Figure 8.2 would be -$1000+$40000003000 = $ Intuitively, why is the expectation more descriptive than the average? 4. Suppose that X is equally likely to take the values 1, 2, 3, 4, 5, and 6. Specify the probability distribution of X. Compute E(X), Var(X), and StDev(x).

Answer 1

4)

"The mean (expectation) of the given distribution is: mu = 7/2 In order to find the mean (expectation) of the given distribution, we will use the following formula mu = Sigma x middot p(x) So, first we need to multiply each value of X by each probability P(X), then add these results together. In this example we have mu = 1 middot 1/6 + 2 middot 1/6 + 3 middot 1/6 + 4 middot 1/6 + 5 middot 1/6 + 6 middot 1/6 um = 7/2 In order to find the variance (sigma^2) of the given distribution, we will use the following formula sigma^2 = Sigma x^2 middot p(x) - mu^2 In this example mu = 7/2 (use: ""Find the Mean"" option in this calculator for the step-by-step explanation on how to find mean) Now we will find the sum Sigma x^2 middot p(x) - 1^2 middot 1/6 + 2^2 middot 1/6"

Please post rest question again .