Question

Provide an example of a probability distribution of discrete random variable, Y, that takes any 4 different integer values between 1 and 20 inclusive; and present the values of Y and their corresponding (non-zero) probabilities in a probability distribution table.

Calculate: a) E(Y)

b) E(Y2 ) and

c) var(Y).

d) Give examples of values of ? and ? , both non-zero, for a binomial random variable X. Use either the binomial probability formula or the binomial probability cumulative distribution tables provided in class calculate:

a) ?(? = ?0) where ?0 is an integer of your own choice satisfying 0 < ?0 < ?. b) ?(? > ?0)

e) Suggest any value, ?0, of the standard normal probability distribution (correct to two decimal places), satisfying 1.10 < ?0 < 2.5 and then calculate:

a) P(Z> −?0) and b) P (Z< 0.8?0)

Answer 1

Solution :

The discrete random variable with it's probability distribution table is given below :

Y = y	P(Y = y)
0	0.2
1	0.4
2	0.1
3	0.3

a) The E(Y) is given by,

E(Y) = Σ y. P(Y = y)

E(Y) = (0 × 0.2) + (1 × 0.4) + (2 × 0.1) + (3 × 0.3)

E(Y) = 0 + 0.4 + 0.2 + 0.9

E(Y) = 1.5

b) E(Y²) is given by,

E(Y²) = Σ y². P(Y = y)

E(Y²) = (0² × 0.2) + (1² × 0.4) + (2² × 0.1) + (3² × 0.3)

E(Y²) = 0 + 0.4 + 0.4 + 2.7

E(Y²) = 3.5

c) Var(Y) is given by,

Var(Y) = E(Y²) - (E(Y))²

Var(Y) = 3.5 - (1.5)²

Var(Y) = 1.25

d)(a) Let n = 4 and p = 0.5 and xo = 1

Now we have to find P(X = 1).

Using binomial probability cumulative distribution table we get,

for n = 4, p = 0.5

P(X = 1) = 0.25

d)(b) We have to find P(X > 1).

P(X > 1) = 1 - P(X ≤ 1)

P(X > 1) = 1 - [P(X = 0) + P(X = 1)]

Using binomial probability cumulative distribution table we get,

for n = 4, p = 0.5

P(X = 1) = 0.25 and P(X = 0) = 0.0625

Hence, P(X > 1) = 1 - [0.25 + 0.0625]

P(X > 1) = 1 - 0.3125

P(X > 1) = 0.6875

e) Let z0 = 2.0

a) We have to find P( Z > -2).

Using Standard Normal Table we get,

P(Z > -2) = 0.9772

b) We have to find P(Z < 0.8×2).

P(Z < 0.8 × 2) = P(Z < 1.6)

Using Standard Normal Table we get,

P(Z < 1.6) = 0.9452