x < -1 x/4+1/2 -1
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Let X be a random variable with CDF:
![x < -1
x/4+1/2 -1< x < 1
1 < x
Fx(x) = {
1
Sketch the CDF and then find the following:
a) P[X < 1] and P[X < 1].
b) The PDF fx (x).
c) The mean E[X].
d) The variance o.
X](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F584ffa08-94ad-4eb4-a87d-fd258db9f802%2F46296444-c19c-4873-84e5-e7cb13656c9d%2Fzpw2m5m_processed.png&w=3840&q=75)
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- A class contains 40 students. The probability of failing in one of the courses for each student is 0.03, but there is no chance to fail in more than one course. You know that a student failed in a course, the probability that the fail reason in non-attendance is 0.04. All other fail reasons are low average. The examinations are private, so the results are mutually independent. Let X and Y be the numbers of failed students due to non-attendance and due to low average, respectively, in that class this semester. Find the joint MGFSuppose that in a year of 365 days, it rains on 62 days. There is lightning on 13 days. There is both lightning and rain on 12 days. Compare the chance that a rainy day has lightning to the chance that a day with lightning has rain. Show your work here P(lightning | rain) = P(rain | lightning)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 girls in each outcome. For example, if the outcome is bbb, then R(bbb)=0. Suppose that the random variable X is defined in terms of R as follows: X=R^2-2R-1. The values of X are given in the table below.
- 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 girls in each outcome. For example, if the outcome is bbg, then R(bbg) = 1. Suppose that the random variable X is defined in terms of R as follows: X=R- - 2R-4. The values of X are given in the table below. Outcome bbb ggb bbg gbg gbb bgg bgb gg Value of X-4 -4 -5 -4 -5 -4 -5 -1 Calculate the values of the probability distribution function of X, i.e. the function py. 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 Px (x) E 1:48 PM 3/21/2022 hp Compag LAI956X 立An ordinary (fair) coin is tossed 3 times. Outcomes are thus triples of "heads" (h) and "tails" (*) 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 -2. The values of X are given in the table below. Outcome ttt hth tht htt thh hhh hht tth Value of X -2 2 0 0 2 4 2 0 Calculate the probabilities P(X=*) of the probability distribution of X. First, fill in the first row with the valuesof X. Then fill in the appropriate probabilities in the second row. Value x 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 tth, then N (tth)=2. Suppose that the random variable X is defined in terms of N as follows: X=N²-2N-2. The values of X are given in the table below. Outcome ttt htt hhh tht tth hth hht thh Value of X 1 -2 -2 -2 -2 -3 -3 -3 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 0 0 0 0 00 X Ś
- The genotype of an organism can be either normal (wild type, W) or mutant (M). Each generation, a wild type individual has probability 0.03 of having a mutant offspring, and a mutant has probability 0.005 of having a wild type offspring.We throw a symmetrical dice twice. Let X denote the number of throws in which an odd number of dice fell, and let Y denote the remainder of the product of the dice's rolls divided by 3.Check whether the variables X and Y are (a) independent (b) uncorrelatedAn 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 X are 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 0 00 X
