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6.9 – Discrete Random Variables IBHLY2 - Santowski

6.9 – Discrete Random Variables IBHLY2 - Santowski

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Page 1: 6.9 – Discrete Random Variables IBHLY2 - Santowski

6.9 – Discrete Random Variables

IBHLY2 - Santowski

Page 2: 6.9 – Discrete Random Variables IBHLY2 - Santowski

(A) Random Variables

 Now we wish to combine some basic statistics with some basic probability we are interested in the numbers that are associated with situations resulting from elements of chance i.e. in the values of random variables

We also wish to know the probabilities with which these random variables take in the range of their possible values i.e. their probability distributions

Page 3: 6.9 – Discrete Random Variables IBHLY2 - Santowski

(A) Random Variables

So 2 definitions need to be clarified:

(i) a discrete random variable is a variable quantity which occurs randomly in a given experiment and which can assume certain, well defined values, usually integral examples: number of bicycles sold in a week, number of defective light bulbs in a shipment

discrete random variables involve a count

 

(ii) a continuous random variable is a variable quantity which occurs randomly in a given experiment and which can assume all possible values within a specified range examples: the heights of men in a basketball league, the volume of rainwater in a water tank in a month

continuous random variables involve a measure

Page 4: 6.9 – Discrete Random Variables IBHLY2 - Santowski

(B) CLASSWORK

CLASSWORK: (to review the distinction between the 2 types of random variables)

Math SL text, pg 710, Chap29A, Q1,2,3

Math HL Text, p 728, Chap 30A, Q1,2,3

Page 5: 6.9 – Discrete Random Variables IBHLY2 - Santowski

(C) The Distributions of Random Variables

For any random variable, there is an associated probability distribution:

 the discrete probability distribution is associated with discrete random variables so this probability distribution describes a discrete random variable in terms of the probabilities associated with each individual value that the variable may takethe normal distribution is associated with continuous random variables

we will initially consider only discrete data and their associated probability distributions

Page 6: 6.9 – Discrete Random Variables IBHLY2 - Santowski

(C) The Distributions of Random Variables

We will toss three coins. The random variable, X, will represent the number of heads obtained. Construct a table and a graph to represent the discrete probability distribution

the probability of exactly 1 head in three tosses will be written as P(X=1) (which I can read as the probability that my random variable (# of heads) has the value 1 i.e. 1 head is tossed)It can be determined in many different ways I will use binomial probability distribution ((p + q)3) from our last section C(3,1) x (0.5)1 x (0.5)3-1 = 3 x 0.5 x 0.25 = 0.375

Or I could use a GDC and determine binompdf(3,0.5,1) = 0.375(Or I could use the Fundamental Counting Principle ==> p(H) x p(T) x p(T) x C(3,1) ) = 0.375

Likewise, I could do similar calculations to find the associated probabilities for 0,1,2,3 Heads ==> I will write this as P(X = x) and equate it to C(3,x) x (0.5)x x (0.5)3-x, x = 0,1,2,3

Page 7: 6.9 – Discrete Random Variables IBHLY2 - Santowski

(C) The Distributions of Random Variables

I get the following table: I get the following graph:

x P(X=x)

0 0.125

1 0.375

2 0.375

3 0.125

0

0.1

0.2

0.3

0.4

Pro

bability

0 1 2 3Number of Heads

Binomial Probability DistributionP(X = x)

Page 8: 6.9 – Discrete Random Variables IBHLY2 - Santowski

(C) The Distributions of Random Variables

ex 2. Of the 15 light bulbs in a box, 5 are defective. Four bulbs are chosen at random from the box. Let the random variable, X, represent the number of defective bulbs selected. Construct a table and graph to represent this distribution.

(NOTE: the events are NOT independent (as selecting a defective bulb first, now influences the probabilities of the selection of a second bulb therefore, binompdf(4,1/3,x) will give us different answers than the following approach:

the number of ways of selecting x defective bulbs from the 5 is C(5,x)

the number of ways of selecting (4 - x) non-defective bulbs is C(10, 4-x)

the number of ways of selecting 4 bulbs from 15 is C(15,4)

so our basic probability formula would be (# of specific events) (total # of events) = [C(5,x) x C(10,4-x)] [C(15,4)]

Page 9: 6.9 – Discrete Random Variables IBHLY2 - Santowski

(C) The Distributions of Random Variables

I get the following table: I get the following graph:

x P(X = x)

0 0.154

1 0.440

2 0.330

3 0.073

4 0.004

0

0.1

0.2

0.3

0.4

0.5

Pro

bability

0 1 2 3 4# of Defective Bulbs

Selection of Defective Light Bulbs

Page 10: 6.9 – Discrete Random Variables IBHLY2 - Santowski

(D) Homework

SL Math text, Chap 29B, p712, Q1-7

HL Math text, Chap 30B, p730, Q1-11