Week 5 Discrete Random Variables and Probability Distributions Statistics for Social Sciences

Chapter Goals After completing this chapter, you should be able to: Defined discrete random variables and probability distributions Calculate the probability, mean and standard deviation of discrete random variable Interpret the mean and standard deviation for a discrete random variable

Revisit random experiment What would probably happen when we: toss a coin? toss two coins? We have more than one possible outcomes from the random experiment. The outcome is not known in advance. Tossing a coin, possible event s could be {H} or {T} For two coins tossing, possible events are {HH}, {HT}, {TH} or {TT} Difficult to predict which outcome will occur when an experiment is performed.

Random variable  The outcome of such events can only be presented in terms of probabilities.  If the outcomes and their probabilities for an experiment are known, can find out what will happen, on average, if that experiment is performed. The outcomes of such a chance experiment can be related to the concept of a random variable. We shall consider probability in the context of a random variable and understand the notion of a probability distribution.

The outcomes of a random experiment is often not a numerical. Toss a coin: S={H,T} Sit for a test: S={pass, fail} Interest rate: S={up, down, unchanged} to determine what will happen, on average, to these situation, translate outcomes from the sample space into number Like the qualitative variable: gender (male, female), a numerical number will be assigned to all outcomes. ‘0’ to head and ‘1’ to tail. Discrete random variable

Introduction to Probability Distributions Random Variable Represents a possible numerical value from a random experiment Random Variables Discrete Random Variable Continuous Random Variable

Probability Distribution and Discrete random variable Tossing 2 coins: sample space {HH, HT, TH, TT} Let X denote the number of head observes from an experiment. The numerical value for observing head can take 0 (TT), 1 (HT, TH) and 2 (HH). The value takes is random.  Since X depends on chance - the outcome of an experiment, X is a random variable. X= (0,1,2) If each event in an experiment is assigned a numerical value, the result of the experiment we observe is a random variable. Look at what we expect to happen by constructing a probability distribution. A listing of all the outcomes of an experiment and the probability associated with each outcome.

Discrete Random Variables Set of possible values from the experiment’s outcomes can be listed Can only take on a countable number of values Examples: Roll a dice twice Let X be the number of times 4 comes up (then X could be 0, 1, or 2 times) Toss a coin 3 times. Let X be the number of heads (then X = 0, 1, 2, 3)

Discrete Probability Distribution x Value Probability of x 0 1/4 = /4 = /4 =.25 Experiment: Toss 2 Coins. Let X = # heads. T T 4 possible outcomes T T H H HH Probability Distribution x Probability Show P(x), i.e., P(X = x), for all values of x:

Discrete Random Variable example Experiment: toss 3 coins at the same time. Let x=number of ‘Head’ observed  S={hhh, hht, hth, htt, thh, tht, tth, ttt} The probability distribution for observing number of head. P(at least 1 H) = P(X=1) +P(X=2) +P(X=3) = 3/8+3/8+1/8 = 7/8 XFrequency of observing XProbability of observing X i.e. P(X) 011/8 133/ /8

No of siblings (X)FrequencyRelative Frequency 0303/ / / / /200 Total2000 Suppose one student is randomly selected from UNIMAS. Let x indicate the no of siblings by the selected student. The value depends on the outcome of the experiment - which student is selected. So x is called a discrete random variable. By pairing the no of siblings (x) and the relative frequency for each x, a probability distribution is constructed.

Probability Distribution Required Properties Probability distribution of x [P(x)] describes how the probabilities are distributed over all the possible values of x. P(x)  0 for any value of x  P(x) = 1 Both conditions must be fulfilled to represent the probability distribution of x.

Calculating Discrete Random probability Breakdown per week (X) Probability Find the probability that the number of breakdowns for this machine during a given week is: P(X=2) = 0.35 between 0 to 2 where 0 and 2 are included P(0  X  2) = P(X=0) + P(X=1) + P(X=2) = = 0.70 P(X>1) = P(x=2) + P(X=3) = = 0.65 at most 1 = P(X≤1) = P(X=0) + P(X=1) = 0.35

Probability Distribution of the Number of Children per Household a. P(X<3) = P(X=0)+P(X=1)+P(X=2) b. P(X  3) = P(X=3)+P(X=4)+P(X=5) c. P(X=2)= d. P(1<X<4)= No of children per household (x)P(x)

Nearly 34% being single-vehicle and 31% being two- vehicle households. However, nearly 35% of all households have three or more vehicles. The national average is 2.28 (how to get this figure?) vehicles per household, By pairing the number of vehicles per household as the variable x with the probability for each value of x, a probability distribution is created. This is much like the relative frequency distribution.

Mean or Expected Value for a random variable X – E(X) Expected Value (or mean) of a discrete random variable  It is a long run average, expected value or mean value. Example: Toss 2 coins, x = # of heads, compute expected value of x: E(x) = (0 x.25) + (1 x.50) + (2 x.25) = 1.0 x P(x)

Find the expected value of the following distribution. The formula is E(X)=  XP(X) = 0(0.1) + 1(0.15) +2(0.3) +3(0.25) +4(0.2) = 2.30 In the long run, the mean or the expected value of this discrete random variable (such as number of accident in a given day) is 2.3 cases. Of course the accident will never have 2.3 cases in any one accident. X01234 P(X)

X= number of fish caught by fishermen in a 6-hour period. Find the probability that a fisherman catches one or more fish in a 6-hour period. Find the probability that a fisherman catches at most two fish in a 6- hour period. Find the expected value for the number of fish caught per fisherman in a 6-hour period. X01234 or more %

Application of expected value The following table shows the probability distribution for X (amount a player ‘gains’ on a single play). The mean amount a player will ‘gain’ per bet is: Expected value= (RM2*0.3 )+ (-RM1*0.7) = -RM1 Over a large number of plays (that means in the long run), a player can expect to lose an average of RM1 in the bet. X= amount gainRM2RM-1 Probability0.40.6

Variance and Standard Deviation Variance of a discrete random variable X Standard Deviation of a discrete random variable X

Standard Deviation Example Example: Toss 2 coins, X = # heads, compute standard deviation (recall E(x) = 1) Possible number of heads = 0, 1, or 2

Practical use of standard deviation for a discrete random variable Suppose you decide to invest RM100 in a scheme that you hope will make some money. You have two choices of investment plans, and you must decide which one to invest. The possible net gains one year later and their probabilities are as follows: PLAN 1PLAN 2 X=net gainsProbX= net gainsProb RM RM200.3 RM RM100.2 RM00.994RM40.5

Plan 1 X=net gainsProb P(x) xP(x) (X-  ) 2 *P(x) RM ( ) 2 *0.001=24900 RM ( ) 2 *0.005=4900 RM (0-10) 2 *0.994 =  29900=172.9 Plan 2 X= net gainP(x)xP(x) (X-  ) 2 *P(x) (20-10) 2 *0.3= (10-10) 2 *0.2= (4-10) 2 *0.5 =  48=6.9

Suppose that the probability function for the number of errors, X, on pages from Social Sciences textbooks is P(0) =0.81 P(1) =0.17 P(2)=0.02 Find (i) mean number of errors per page [µ = E(X) =  xP(X)]. (ii) variance of the number of errors [  2 = E(X- µ) 2 P(X)]