Discrete Probability Distributions

Question 1

Random variables

Answer 1

When al the possible outcomes of a random situation are numbers

E.g number thrown on a die

No. Calls to a call centre in an hour

Question 2

Two types of random variable

Answer 2

Continuous random variable
Any value within a range (time for particular train journey )

Discrete random variable
Some values in a range - number of defective items in a batch (integer )

Question 3

Expectation,expected value or mean

Answer 3

The long run average value of a random variable of interest

Called the mean, expected value or expectation

E(X)

E(X) = ExP(X)

(E= sum of) with

Question 4

Chebyshevs theorem

Answer 4

The probability that a random variable lies within Kō is at least 1-1/K2

E.g. The probability that a random variable lied within 2 standard deviations of its mean is atleast

1-1/2^2 = 0.75

Question 5

Variance of a random variable

Answer 5

E(X^2)-E(X)^2

Where E(X^2) = Ex^2p(X)

Question 6

Binomial distribution number of ways x objects can be chosen from n

Answer 6

NCX= N!/X!(N-X)!

E.g. Company produces 6 products and wishes to promote only three

How many ways can it do this

6C3= 6!/3!(6-3)!

= 6x5x4x3x2x1/ (3x2x1)(3x2x1)=20

Question 7

Binomial distribution probability of x success out of n trials

Answer 7

P(x)= NCX(P^x(1-p)^n-x)

PPPP(X times )(success)

(1-p) n-x times (failure)

E.g. 8 students sit exam
Pass -o.6
Fail-0.4

3 pass?
8C3= 56

8C3P^3(1-p)^5

=56(0.6)^3(0.4)^5=0.123863

Question 8

Binomial distribution SPSS

Answer 8

Transform>compute variable

In excel use function BINOMIAL.DIST(x,n,p true)

True indicates cumulative probability

Use false for the probability of exactly x successes

Question 9

Cumulative binomial probabilities

Answer 9

Probability of 3 or less(e.g.)

P(x(less than or equal to) 3) = p(0)+p(1)+(p2)+(p3)

E.g. 45% voters are Tory
Probability that in a sample of 8 voters, 2 or less are Tory

P(x(less than or equal to) 2)

P(0) 8C0 0.45^0 0.55^8=….
P(1) 8C1 0.45^1 0.55^7
P(2). 8C2. 0.45^2 0.55^6

Add up answers together

Question 10

Expected value and variance of binomial variable

Answer 10

Expected value
U= E(X)=np

Variance= o^2= np(1-p)

E.g. 40% patients need treatment
100 patients

N=100
P=0.4

Mean u=100x40
Variance= 100 x 0.4 x 0.6 = 24