Four points are chosen uniformly and independently from the interval [0, 10].

note: these are continuous random variables.

(a) what is the probability that the maximum of these values is below 5?
(b) what is the probability that the maximum of these values is between 5 and 7?
(c) find a formula for the pdf of the 3rd highest of these values.
(d) what is the probability that the 3rd highest of these values is between 5 and 7?
(e) what is the expectation of the 3rd highest of these values?

- four points means 4 values

- chosen uniformly means chosen consecutively or one after the other, without skipping any value

- chosen independently means that the presence of a particular value in one group does not influence its presence in another group

- continuous means they are numeric and flow in the same interval of 1 unit

- all possible 4-point groups are:

(1) 0,1,2,3

(2) 1,2,3,4

(3) 2,3,4,5

(4) 3,4,5,6

(5) 4,5,6,7

(6) 5,6,7,8

(7) 6,7,8,9

(8) 7,8,9,10

Step-by-step explanation:

(A) the probability that the maximum of these values is below 5 is 2/8 or 1/4

(B) the probability that the maximum of these values is between 5 and 7 is the same as the probability that the maximum of these values is 6

This probability is equal to 1/8

(C) find a formula for the probability distribution function of the 3rd highest of these values

From the 8 groups, the 3rd highest values are: 2,3,4,5,6,7,8,9

The probability distribution function of these values would enlist the probabilities of obtaining each of these values in third place, across the 8 groups

They all have the same probability which is 1/8

A formula for this PDF would be

Sigma(summation notation) from 1 to 8 of P (X1, X2, X3,...,X8)

Where X is the 3rd highest value in each group

X1 is the probability that a third highest value across the 8 groups is 2. That is, P(X=2)

Same goes for all the others

X8 is the probability that X is 9

(D) the probability that the third highest of these values is between 5 and 7 is the same as the probability that the third highest is 6

This is found in group 5 and the probability is 1/8

(E) the expectation or expected value of the third highest in each group is the mean of all third highest values.

E.V. = (2+3+4+5+6+7+8+9) ÷ 8

E.V. = 5.5

You are welcome.

answer 1 : option "c"

answer 2 : option "d"

answer 3 : option "b"

answer 4 : option "d"

answer 5 : option "b"

answer 6 : option "a"

answer 7 : option "c"

answer 8 : option "d"

answer 9 : option "c"

answer 10 : option "d"

answer 11 : option "c"

answer 12 option "b"

answer 13 option "c"

answer 14 option "c"

answer 15 option "d"

answer 16 option "a"

answer 17 option "b"

answer 18 option "c"

answer 19 option "a"

answer 20 option "a"

answer 21 option "c"

d