Demand of products per day for three days are 21, 19, 22 units and their respective probabilities are 0.29, 0.40, 0.35. Profit per unit is 0.50 paisa then expected profits for three days are ________.

(a)

21, 19, 22

(b)

21.5, 19.5, 22.5

(c)

0.29, 0.40, 0.35

(d)

3.045, 3.8, 3.85

Given E(X)=5 and E(Y)=−2, then E(X−Y) is ________.

(a)

3

(b)

5

(c)

7

(d)

-2

A formula or equation used to represent the probability distribution of a continuous random variable is called ________.

(a)

probability distribution

(b)

distribution function

(c)

probability density function

(d)

mathematical expectation

Which of the following is not possible in probability distribution?

(a)

\(\sum { p(x)\ge 0 } \)

(b)

\(\sum { p(x)=1 } \)

(c)

\(\sum { xp(x)=2 } \)

(d)

\(p(x)=-0.5\)

A discrete probability distribution may be represented by ________.

(a)

table

(b)

graph

(c)

mathematical equation

(d)

all of these

If c is a constant in a continuous probability distribution, then p(x = c) is always equal to ________.

(a)

zero

(b)

one

(c)

negative

(d)

does not exist

The distribution function F(x) is equal to ________.

(a)

\(P(X=x)\)

(b)

P(X\(\le\)x)

(c)

P(X\(\ge\)x)

(d)

all of these

A random variable X has the following probability distribution

X	0	1	2	3	4	5
P(X=x)	\(\frac{1}{4}\)	2a	3a	4a	5a	\(\frac{1}{4}\)

Then P(1≤X≤4) is

(a)

\(\frac{10}{21}\)

(b)

\(\frac{2}{7}\)

(c)

\(\frac{1}{14}\)

(d)

\(\frac{1}{2}\)

X is a discrete random variable. Which take values 0,1,2 and P(X = 0) = \(\frac{144}{169}\), P(X = 1) = \(\frac{1}{169}\), then the value of P(X = 2) is ________

(a)

\(\frac{145}{169}\)

(b)

\(\frac{24}{169}\)

(c)

\(\frac{2}{169}\)

(d)

\(\frac{143}{169}\)

X is a random variable. Taking the values 3, 4 and 12 with probabilities \(\frac{1}{3},\frac{1}{4}\)and \(\frac{5}{12}\).Then E(X) is

(a)

5

(b)

7

(c)

6

(d)

3