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SSF1063: Statistics for Social Sciences LU5: Probability Distribution of Discrete Random Variables 18 th February 2008.

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Presentation on theme: "SSF1063: Statistics for Social Sciences LU5: Probability Distribution of Discrete Random Variables 18 th February 2008."— Presentation transcript:

1 SSF1063: Statistics for Social Sciences LU5: Probability Distribution of Discrete Random Variables 18 th February 2008

2 Definitions Random Variables  A variable whose value is determined by the outcome of a random experiment Discrete Random Variables  Variables with countable values

3 Table 5.1 (p. 189) Frequency and Relative Frequency Distributions of the Number of Vehicles Owned by Families

4 Discrete Random Variable Other examples includes:  The number of cars sold at a dealership during a given month  The number of babies being born during a given week in a local hospital  The number of houses in a certain block  The number of fish caught on a fishing trip  The number of complaints received at the office of an airline on a given day

5 Continuous Random Variable A variable that assumes any value contained in one or more intervals Examples:  The height of a person  The time taken to complete an examination  The weight of a fish  The price of a house 0250

6 Probability Distribution of a Discrete Random Variable Lists all the possible values that the random variable can assume and their corresponding probabilities

7 Probability Distribution of a Discrete Random Variable Write the probability distribution of the discrete random variable for Table 5.1

8 The Two Characteristics The probability distribution of a discrete random variable has two characteristics:  The probability assigned to each value is 0≤P(x)≤1 for each x  The sum of all probabilities assigned to all possible values of x is equal to 1.0; ΣP(x) = 1.0  These are the two conditions that a probability distribution must satisfy

9 The Two Characteristics Find the probability of the following: P(2) = P(>2) = P(≤1) = P(>4) =

10 Graphical Presentation of Probability Distribution

11 Validity of Probability Distribution xP(x) 01230123 0.08 0.11 0.39 0.27 xP(x) 23452345 0.25 0.34 0.28 0.13 xP(x) 789789 0.70 0.50 -0.20

12 Example 1 The following table lists the probability distribution of the number of breakdowns per week for a machine based on past data. a. Present this probability distribution graphically b. Find the probability that the number of breakdowns for this machine during a given week is: i. exactly 2ii 0 to 2 iii. more than 1iv. at most 1 Breakdowns per week0123 Probability0.150.200.350.30

13 Example 1 i. exactly 2 ii. 0 to 2 iii. more than 1 iv. at most 1

14 Example 2 According to a survey, 60% of students suffer from Math anxiety. Two students are randomly selected from this university. Let x denote the number of students in this sample who suffer from math anxiety. Develop the probability distribution of x. N = student selected does not suffer from Math anxiety M = student selected suffer from Math anxiety

15 Example 2

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17 Example 3 Five percent of all cars manufactured at a large auto company are lemons. Let x denote the number of lemons in this sample. Write the probability distribution of x. Draw a tree diagram for this problem.

18 Mean of a Discrete Random Variable Mean of the probability distribution (also known as ‘expected value’) The value that we expect to observe per repetition, on average If a man sells 1.6 cars per week, on average, does is mean that he sells 1.6 cars every week?

19 Mean of a Discrete Random Variable The mean of discrete random variable x is denoted by E(x) or μ E(x) = ΣxP(x)

20 Standard Deviation of a Discrete Random Variable Measures the spread of its probability distribution, denoted by σ Higher value of σ indicates that x can assume values over a larger range about the mean Smaller value for σ indicates that values of x are clustered closely about the mean Formula: σ = √ Σx 2 P(x) – μ 2

21 Example 4 In Panda Electronics, a few defective parts will usually go into a shipment undetected. Let x denote the number of defective parts in a shipment of 400. The following table gives the probability distribution of x. x012345 P(x)0.020.200.30 0.100.08

22 Example 4 σ = √ Σx 2 P(x) – μ 2

23 Factorials 8! is read 8 factorials Value is obtained by multiplying all the integers from that number to 1. 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1 0! always equals to 1. Evaluate 4!, 6!, 9!, 11! Evaluate (12 – 4)! Evaluate (5 – 5)!

24 Combinations Gives the number of ways x elements can be selected from n elements. Denoted n C x The formula is given: n C x = n! x!(n – x)! x = The number of elements selected per selection n = The total number of elements

25 Example 5 An ice cream parlor has six flavors of ice cream. Kristen wants to buy two flavors of ice cream. If she randomly selects two flavors out of six, how many possible combinations are there?

26 Example 6 Three members of jury will be randomly selected from five people. How many different combinations are possible?

27 Example 7 What is the probability of a player (who plays this lottery once) wins if he needs to select 6 number from 1 – 49 ?

28 Permutations The concept is similar to combinations but with one major difference – order of selection of important. n P x The formula is given: n P x = n! (n – x)! x = The number of elements selected per selection n = The total number of elements

29 Example 8 A club has 20 members. They are to select three office holders – president, secretary and treasurer – for the next year. Three names will be drawn randomly where the first person will be selected becomes the president, the second is the secretary and the third one takes over as treasurer. Therefore, the order is important.

30 Example 9 An ice cream shop offers 25 flavors of ice cream. How many ways are there to select 2 different flavors from these 25 flavors? How many permutations are possible? A ski patrol unit has nine members available for duty and two of them are to be sent to rescue and injured skier. In how many ways two of these nine members be selected? Suppose the order of selection is important. How many arrangements are possible in this case?

31 Binomial Probability Distribution The binomial probability distribution is most widely used discrete probability distribution Applied to find the probability that an outcome will occur x times in n performances of experiment

32 Rules of the Binomial Experiment A binomial experiment must satisfy ALL FOUR conditions 1.n identical trials 2.Each trial has only TWO outcomes; success and failure 3.Success = p, Failure = q, p + q =1. Constant probability 4.Trials are independent. One trial does not affect the outcome of the other

33 Rules of the Binomial Experiment Is this a binomial experiment? Probability for a defective DVD player is 0.05. What is the probability of getting exactly ONE defective DVD player in the sample of 3 DVD player? Check if the situation satisfy the FOUR rules.

34 0.05 x 0.95 x 0.95 0.95 x 0.05 x 0.95 0.95 x 0.95 x 0.05 0.1354

35 The Binomial Formula P(x) = n C x ·p x ·q n-x n = total number of trials p = probability of success q = probability of failure (1 – p) x = number of successes in n trials n – x = number of failures in n trials

36 The Binomial Formula The probability of exactly ONE DVD player is defective: P(x) = n C x ·p x ·q n-x  p = success = defective DVD player = 0.05  q = failure = non-defective DVD player = 0.95  n = total number of trials = 3  x = number of successes in n trial P(ONE defect) = 3 C 1 ·0.05 1 ·0.95 3-1 = 3 x 0.05 x 0.95 2 = 0.1354

37 Example 10 2% of packages delivered by Donkey Express do not reach their destinations on time. Suppose a corporation mails 10 packages through Donkey Express. Find: The probability exactly ONE package will not reach its destination on time The probability at most ONE of these packages will not reach the receiver on time

38 Example 10 Success, p = package fail to be delivered on time Failure, q = package delivered on time Thus, p = 0.02 and q = 1 – 0.02 = 0.98 n = 10 Find p(x=1) Find p(x≤1) = p(x=0) + p(x=1) P(x) = n C x ·p x ·q n-x

39 Graphical Presentation of Probability Distribution 64% of adults in Singapore complained that the working hours is too long. Choose 3 adults randomly and see if they have the same opinion. Write the probability distribution and draw a bar graph to illustrate it.

40 Graphical Presentation of Probability Distribution

41 Table of Binomial Probabilities 30% adults in Malaysia say they cannot live without cell phones. If 6 adults are selected randomly, find the probability that exactly 3 adults hold the said opinion.

42 Table of Binomial Probabilities P(at most 2 hold the said opinion) P(at least 3 hold the said opinion) P(exactly 5 hold the said opinion) P(more than 4 hold the said opinion)

43 Shape of the Binomial Distribution Distribution is skewed to the right if p < 0.50

44 Shape of the Binomial Distribution Distribution is symmetric if p = 0.50

45 Shape of the Binomial Distribution Distribution is skewed to the left if p > 0.50

46 Mean and Standard Deviation of the Binomial Distribution Mean, μ = np Standard deviation, σ = √npq If an experiment has 60 trials, with p=0.85 and q=0.15, what is the mean and the standard deviation of the distribution?

47 The Poisson Probability Distribution Useful for occurrences that are random and with unpredictable patterns THREE conditions to apply the Poisson Probability Distribution: 1. Discrete random variable 2. The occurrences are random 3. The occurrences are independent

48 The Poisson Probability Distribution P(x) = λ x ·e -λ x! λ = mean number of occurrences in the interval (weeks, month, 5m, two-hours) e = 2.71828

49 Example 11 On average, a household receives 9.5 telemarketing calls a week. Using the Poisson distribution formula, find the probability that a randomly selected household receives: Exactly 6 telemarketing calls a week At most ONE call a week P(x) = λ x ·e -λ x!

50 Table of Poisson Probabilities On average, two new accounts are opened per day at Donkey Bank. Find the probabilities that on a given day, the number of new accounts opened at this bank will be: Exactly 6 At most 3 At least 7

51 Mean and Standard Deviation of the Poisson Distribution Mean, μ = λ Variance, σ 2 = λ Standard deviation, σ = √ λ Given λ is 0.9. Find the mean, variance and standard deviation for the probability distribution.


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