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.

step-by-step explanation:

if you want to solve percent error follow these steps.

1. subtract the accepted value from the experimental value.

2. take the absolute value of step 1.

3. divide that answer by the accepted value.

4. multiply that answer by 100 and add the % symbol to express the answer as a percentage.

if you want to solve percent of change just do this.

1. work out the difference (increase) between the two numbers you are comparing.

2. divide the increase by the original number and multiply the answer by 100. if your answer is a negative number then this is a percentage decrease.

hope this : )

step-by-step explanation:

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