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(d) p(x) = -0.5 

p(x) = -0.5 is not possible since the probability cannot be negative.

• The mean np of the binomial distribution shows the expected number (average) of successes in n Bernoulli trials.
• The variance of the distribution is = npq. So its standard deviation is = \(\sqrt{npq}\)
• In the distribution np > npq.
• Probability of failure q = \(\frac{npq}{np}\)
• If p < 1/2, skewness of the distribution is positive.
• If p > 1/2, skewness of the distribution is
  negative.
• If p = 1/2, skewness of the distribution is zero and it becomes symmetrical distribution.

P(X = x) = p(x) = ⁿC_x px q^n-x

Where, n = Number of Bernoulli trials;

x = 0, 1, 2 …… n

p – Probability of success (0 <p < 1)

q = 1 – p

If a random variable X is an element of R or its subset, whose interval is (a, b), where a < b and is capable of assuming any value in the interval, then that random variable X is called a continuous random variable. Examples of continuous random variable are: age of a person, temperature of a place, production of a company.

 Variance of Random Variable X:

σ2 = V(X) = E(X – μ)²

= E (X²) – [E(X)]²

= Σx²p(x)- [Σx – p(x)]²

Mean of Random Variable X:

μ = E(X) = Σx . p(x)

If in the series of successes (S) and failures (F) obtained in n numbers of Bernoulli trials, the number of successes (S) is denoted by X, then X is called binomial random variable. X is capable of taking any value of a finite set {0, 1, 2, …, n}.

Let U be a sample space of a random experiment. A function which associates a real number with each outcome of U, is called a random variable and is denoted by symbol X. Symbolically, it is also represented by X: U → R.

An experiment, which has only two outcomes, namely Success or Failure, is called a dichotomous experiment.

(a) positive probabilities

The correct answer is : (a) zero

(b) probability density function

(d) all of these

(d) both (b) and (c)

(c) Expected value

(a) non-negative

(c) E(XY) = E(X) E(Y)

Given X is a random variable and Y = 2X + 1 and Var(X ) = 5 

Var (Y) = Var (2X + 1) = (2)² = 4 

Var X = 4(5) = 20

(b) cumulative probability

(b) 

The sample space is HHH, HHT, HTH, HTT, THH, THT, TTH and TTT 

Number of heads is 0, 1, 2, 3 with probability 1/8, 3/8, 3/8 and 1/8

1. P(X ≤ x) 

2. discrete and continuous 

3. discrete probability function 

4. distribution function 

5. continuous probability function, integrating the density function

6. probability differential 

7. µ_x

8. the spread, dispersion of the density 

9. √Var[X] 

10. center of gravity, the moment of inertia.

The correct answer is : (c) mean

(c) probability density function

(c) 7 

E(X – Y) = E(X) – E (Y) = 5 – (-2) = 7 .

Discrete Random Variable: 

•  A variable which can take only certain values.

•  The value of the variables can increase incomplete numbers

•  Binomial, Poisson, Hypergeometric probability distributions come under this category 

•  Example: Number of students who opt for commerce in class 11, say 30, 35, 40, 45, and 50.

Continuous random variable: 

•  A variable which can take any value in a particular limit. 

•  Its value increases infractions but not in jumps. 

•  Normal, student’s t and chi-square distribution come under this category. 

•  Example: Height, Weight and age of family members: 50.5 kg, 30 kg, 42.8 kg and 18.6 kg.

•  Property 1: The distribution function F is increasing, (i.e) if x ≤ y, then F(x) ≤ F(y) 

•  Property 2: F(x) is continuous from right, (i.e) for each x ∈ R, F (x⁺ ) = F (x) 

•  Property 3: F (∞) = 1 

•  Property 4: F (-∞) = 0 

•  Property 5: F'(x) = f(x) 

•  Property 6: P(a ≤ X ≤ b) = F(b) – F(a)

Let X denote the number on the top side of the unbiased die. 

The probability mass function is given by the following table.

The expected value for the random variable X is

E(X) = ∑_xxP_x(x) 

= (1 x 1/6) + (2 x 1/6) + (3 x 1/6) + (4 x 1/6) + (5 x 1/6) + (6 x 1/6)

= 1/6(1 + 2 + 3 + 4 + 5 + 6)

= 21/6 = 7/2 = 3.5

Let X be the random variable which denotes the gain in the investment. It is given that X takes the value 5000 with probability 0.62 and -8000 with a probability 0.38. 

(Note that we take -8000 since it is a loss) 

The probability distribution is given by

E(X) = (0.38) (-8000) + (0.62) (5000)

= -3040 + 3100

= 60

Hence the expected gain is Rs. 60

The properties of Mathematical expectation are as follows: 

(i) E(a) = a, where ‘a’ is a constant

(ii) Addition theorem: For two r.v’s X and Y, E(X + Y) = E(X) + E(Y)

(iii) Multiplication theorem: E(XY) = E(X) E(Y) 

(iv) E(aX) = aE(X), where ‘a’ is a constant 

(v) For constants a and b, E(aX + b) = a E(X) + b

(b) continuous random variable

(b) continuous random variable

E(X) = (-1) (0.2) + 0(0.3) + 1(0.5) = -0.2 + 0.5 = 0.3 

So E(3X + 1) = 3 E(X) + 1 = 3(0.3) + 1 = 1.9 

E(X² ) = (-1)² (0.2) + 0² (0.3) + 1² (0.5) = 0.2 + 0.5 = 0.7 

Var(X) = E(X² ) – [E(X)]² = 0.7 – (0.3)² = 0.61

(a) Given X is a discrete random variable. The probability distribution can be written as

We know that Σp(x) = 1 

⇒ 2k + 3k + 4k = 1 

⇒ 9k = 1 ⇒ k = 1/9 

(b) P(X > 2) = P(X = 3) + P(X = 5) 

= 3k + 4k 

= 7k

= 7/9