Partially Correct Expression 3: Your answer is incorrect. Suppose that X, Y, and Z are jointly distributed random variables, that is, they are defined on the same sample space. Suppose that we also have the following. E (X)=9 Var (X)=37 E(Y)=-5 E (Z)=-6 Var(Y)=23 Var (Z)=24 Compute the values of the expressions below. E (Y+2)= -3 – 5X+Z E 51 4- Var (-5Y)= 119 E(42?)= 240
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- Suppose that the genders of the three children of a certain family are soon to be revealed. Outcomes are thus triples of "girls" (g) and "boys" (b), which we write gbg, bbb, etc. For each outcome, let R be the random variable counting the number of boys in each outcome. For example, if the outcome is ggb, then R(ggb)=1. Suppose that the random variable X is defined in terms of R as follows: X=R-R-4. The values of X are thus: Outcome bggggbbgbbbgbbbggggbggbb Value of -4-4-6-6 -10 -4-4-6 Calculate the probability distribution function of X, i.e. the function pv(x). First, fill in the first row with the values of X. Then fill in the appropriate probabilities in the second row. Value x of X MIO Submit Assignment Contie O 2021 McGraw-Hill Education. All Rights Reserved. Terms of Use Privacy 96A biased die is thrown thirty times and the number sixes seen is eight. If the die is thrown a further twelve times find the expected numbers of sixes and the variance of the number of sixes. E(X)=3.2 E(X)=3.1 E(X)=3.3 None of the choices! please solve all parts
- Solve it correctly please. I will rate and review accordingly.In large corporations, an "intimidator" is an employee who tries to stop communication, sometimes sabotages others, and, above all, likes to listen to him or herself talk. Let x1 be a random variable representing productive hours per week lost by peer employees of an intimidator. x1: 7 3 6 2 2 5 2 A "stressor" is an employee with a hot temper that leads to unproductive tantrums in corporate society. Let x2 be a random variable representing productive hours per week lost by peer employees of a stressor. x2: 3 3 10 8 6 2 5 8 x1: 3.8571 s1: 2.1157 | x2: 5.625 s2: 2.8754 | Level of Significance: 0.05 | H0: ?1 = ?2; H1: ?1 < ?2 got the value of the sample test statistic wrong but its unchangeable. I need help finding the 90% confidence interval for ?1 − ?2 (rounded answer to 2 decimals)just this multichoice question please
- An ordinary (fair) coin is tossed 3 times. Outcomes are thus triples of "heads" (h) and "tails" (t) which we write hth, ttt, etc. For each outcome, let N be the random variable counting the number of heads in each outcome. For example, if the outcome is ttt, then N (ttt) = 0. Suppose that the random variable X is defined in terms of N as follows: X = 2N−2N2 - 1. The values of X are given in the table below. Outcome ttt hht tth htt hhh tht hth thh Value of X −1 −5 −1 −1 −13 −1 −5 −5 Calculate the probabilities P (X = x) of the probability distribution of X. First, fill in the first row with the values of X. Then fill in the appropriate probabilities in the second row. Value x of X P (X = x)Excel Inferences of population variance(s): Buyer’s Digest CaseBuyer’s Digest rates thermostats manufactured for home temperature control. In arecent test, 10 thermostats manufactured by ThermoRite were randomly selected andplaced in a test room that was maintained at a temperature of 68oF. The temperaturereadings of the ten thermostats are shown below:Thermostat 1 2 3 4 5 6 7 8 9 10Temperature 67.4 67.8 68.2 69.3 69.5 67.0 68.1 68.6 67.9 67.2 Buyer’s Digest gives an “acceptable” rating to a thermostat with a temperature variance of 0.5 or less. Given significance level of α = 0.05, use a formal hypothesis test to determine whether the ThermoRite products are “acceptable” in the quality measured in the variance. r the ThermoRite’s claim is valid.Refer to the photo
- Cola Weights For the four samples described in Exercise 1, the sample of regular Coke has a mean weight of 0.81682 lb, the sample of Diet Coke has a mean weight of 0.78479 lb, the sample of regular Pepsi has a mean weight of 0.82410 lb, and the sample of Diet Pepsi has a mean weight of 0.78386 lb. If we use analysis of variance and reach a conclusion to reject equality of the four sample means, can we then conclude that any of the specific samples have means that are significantly different from the others?An ordinary (fair) coin is tossed 3 times. Outcomes are thus triples of "heads" (h) and "tails" (1) which we write hth, ttt, etc. For each outcome, let N be the random variable counting the number of heads in each outcome. For example, if the outcome is htt, then N (htt) = 1. Suppose that the random variable X is defined in terms of N as follows: X=2N-N2 -4, The values of X are given in the table below. Outcome hht tth hth thh tht ttt hhh htt Value of x -4 -3 -4 -4 -3 -4 -7 -3 Calculate the probabilities P(X=*) of the probability distribution of X. First, fill in the first row with the values of X. Then fill in the appropriate probabilities in the second row. Value of x _ _ _ p(X=x) _ _ _An ordinary (fair) coin is tossed 3 times. Outcomes are thus triples of "heads" (h) and "tails" (t) which we write hth, ttt, etc. For each outcome, let N be the random variable counting the number of tails in each outcome. For example, if the outcome is hth, then N (hth) = 1. Suppose that the random variable X is defined in terms of N as follows: X=2N² -6N-1. The values of Xare given in the table below. Outcome thh tth hhh hth ttt htt hht tht Value of X -5 -5 − 1 -5 −1 -5 -5 -5 Calculate the probabilities P(X=x) of the probability distribution of X. First, fill in the first row with the values of X. Then fill in the appropriate probabilities in the second row. Value X of X P(X=x) 0 8 X