Mathematics, 11.02.2020 04:02 nataliamontirl8152

Let a random experiment be the casting of a pair of fair dice, each having six faces, and let the random variable X denote the sum of the dice. (a) With reasonable assumptions, determine the pmf f (x) of X. Hint: Picture the sample space consisting of the 36 points (result on first die, result on second die), and assume that each has probabilit Find the probability of each possible outcome of X, namely, x = 2, 3, 4, . . . , 12. b) Draw a probability histogram for f(x). y 1/36.

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Answer from: ciaraberger

Pr(2) = 1/36

Pr(3) = 2/36 = 1/18

Pr(4) = 3/36 = 1/12

Pr(5) = 4/36 = 1/9

Pr(6) = 5/36

Pr(7) = 6/36 = 1/6

Pr(8) = 5/36

Pr(9) = 4/36 = 1/9

Pr(10) = 3/36 = 1/12

Pr(11) = 2/36 = 1/18

Pr(12) = 1/36.

b) See attached

Step-by-step explanation:

Since the dice are fair: We expect the following possible outcomes:

(1,1), (1,2), (1,3), (1,4), (1,5), (1,6)

(2,1), (2,2), (2,3), (2,4), (2,5), (2,6)

(3,1), (3,2), (3,3), (3,4), (3,5), (3,6)

(4,1), (4,2), (4,3), (4,4), (4,5), (4,6)

(5,1), (5,2), (5,3), (5,4), (5,5), (5,6)

(6,1), (6,2), (6,3), (6,4), (6,5), (6,6)

And sum gives us the following:

2,3,4,5,6,7

3,4,5,6,7,8

4,5,6,7,8,9

5,6,7,8,9,10

6,7,8,9,10,11

7,8,9,10,11,12

Therefore, the probability of each sum outcome is given as follow:

Pr(x = sum outcome of event) = number of event/total number of possible events. Thus:

Pr(2) = 1/36

Pr(3) = 2/36 = 1/18

Pr(4) = 3/36 = 1/12

Pr(5) = 4/36 = 1/9

Pr(6) = 5/36

Pr(7) = 6/36 = 1/6

Pr(8) = 5/36

Pr(9) = 4/36 = 1/9

Pr(10) = 3/36 = 1/12

Pr(11) = 2/36 = 1/18

Pr(12) = 1/36.

For the attached plot, see the R code below:

probability = c(1/36, 1/18, 1/12, 1/9, 5/36, 1/6, 5/36, 1/9, 1/12, 1/18, 1/36)

hist(probability, prob=T)

lines(density(probability))

Let a random experiment be the casting of a pair of fair dice, each having six faces, and let the ra

Answer from: Quest

Ifound this on google “a skewed (non-symmetric) distribution is a distribution in which there is no such mirror-imaging. a "skewed right" distribution is one in which the tail is on the right side. a "skewed left" distribution is one in which the tail is on the left side. the above histogram is for a distribution that is skewed right.” i hope this you for what your looking for

Answer from: Quest

b. 6

step-by-step explanation:

we are given the table representing an exponential function.

we can see from the table that, as x is increasing, the value of y is increasing in the power of 6.

i.e. the following relation is obtained,

x = 1, y = 6

x = 2, y = 36 = $6^{2}$

x = 3, y = 216 = $6^{3}$

x = 4, y = 1296 = $6^{4}$

x = 5, y = 7776 = $6^{5}$

so, the geometric series obtained is $6, 6^{2}, 6^{3}, 6^{4}, 6^{5}$ .

thus, the common ratio is given by $\frac{6^{2}}{6}$ = $\frac{36}{6}$ = 6.

hence, the common ratio is 6.

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Let a random experiment be the casting of a pair of fair dice, each having six faces, and let the ra...