Practicing Success

Target Exam

CUET

Subject

-- Mathematics - Section B2

Chapter

Probability Distributions

Question:

A random variable X has the following probability distribution:

X 0 1 2 3 4
P(X) 2k k 4k 6k 3k

The mean of the distribution is :

Options:

$\frac{43}{16}$

$\frac{39}{8}$

$\frac{31}{24}$

$\frac{39}{16}$

Correct Answer:

$\frac{39}{16}$