Q5 of 77 Page 32

If X is a random-variable with probability distribution as given below:

The value of k and its variance are


The probability distribution of X is:



We Know that



Therefore


⇒ k+3k+3k+k=1


⇒ 8k=1





⇒ (1×k)+(2×3k)+(3×3k)+(4×k)


⇒ k+6k+9k+4k=20k




⇒ (12 ×k)+(22 ×3k)+(32 ×3k)+(42 ×k)


⇒ k+12k+27k+16k=56k



Var(X)=E(X2) - (E(X))2





More from this chapter

All 77 →
3

A random variable X takes the values 0, 1, 2, 3 and its mean is 1.3. If P (X =3) = 2P (X = 1) and P(X = 2) = 0.3, then P(X = 0) is

 4

A random variable has the following probability distribution:

The value of p is


 6

The probability distribution of a discrete random variable X is given below:

The value of E(X) is


 7

For the following probability distribution:

The value of E(X) is