Question

Let X1, X2, X3 and X4 be four random variables with mean 1, 2, 3 and 4 and standard deviations 1, 2, 3, 4. Which of the following is TRUE?
The random variables X1, X2, X3 and X4 are independent and identically distributed (IID)
The variance of the sum X1+X2+X3+X4 is 100
The sum X1+X4 has the same distribution as the sum X2+X3
The expected value of the weighted sum 4X1+3X2+2X3+X4 is 20

Answer 1

Answer:

the random variables X1,X2,X3 and X4 are independent and identically distributed (IID)

Answer 2

The random variables X1, X2, X3, and X4 are independent and identically distributed (IID)

The Favorable cases of X4 being the least is 0.25.

GIVEN: X1, X2, X3, and X4 be four random variables with means 1, 2, 3, and 4 and standard deviations 1, 2, 3, 4.
TO FIND True Statement.
SOLUTION:

As we are given,

X1, X2, X3, and X4 be four random variables with means 1, 2, 3, and 4 and standard deviations 1, 2, 3, 4.

As we know,

The variables can be arranged as,

o < o < o < o

X4 < o < o < o

X1, X2, and X3 can be arranged in 3 possible ways

i.e. 6 ways

Hence, there are 6 cases where X4 is the lowest out of all.

The total no. of cases in which X1, X2, X3, and X4 can be arranged is 4! = 24

Favorable cases for X4 being the least Is 24 and IID

Therefore,

The Favorable cases of X4 being the least is 6/24 or  0.25

Also, The random variables X1, X2, X3, and X4 are independent and identically distributed (IID)

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