Discrete Probability Distributions - Lab Report Example

Discrete probability distributions Introduction Probability distributions are classified as either discrete or continuous probability distributions. Where a variable can take on any value between two specified values, it is a continuous variable. Otherwise, it is a discrete variable. Discrete probability distributions The various types of discrete probability distributions are explained below; a) Random variables This is a numerical description of an experiment’s outcome. Random variables are either discrete or continuous. A discrete random variable assumes either a finite number of values or an infinite sequence of values. On the other hand, a continuous random variable assumes any numerical value in an interval or collection of intervals. Discrete random variables i) A discrete random variable with a finite number of values can take on any values within a given value e.g. let x = number of TVs sold at the store in one day where x can take on 5 values (0, 1, 2, 3, 4). ii) Discrete random variables with an infinite sequence of values on the other hand has no upper limit on the value the variable takes e.g. let x= number of customers arriving in one day, where x can take on the values of 0,1,2,3, …. Though we can count the customers arriving, there is no finite upper limit on the number that might arrive. The table below aims at differentiating the various types of random variables and where each might be applied, Random variables Question Random variable x Type Family size x= Number of dependents in family reported on tax return Discrete Distance from home to store X= Distance in miles from home to the store site Continuous Own dog or cat X= 1 if own no pet; =2 if own dog(s) only; = if own cat(s) only; = 4 if own dog(s) and cat(s) Discrete b) Discrete probability Distributions This is probability distribution for a random variable that describes how probabilities are distributed over the values of the random variable. A discrete probability distribution can therefore be described using a graph, table or equation. Discrete probability distribution is defined by a probability function that is denoted by f(x), which provides the probability for each value of the random variable. The conditions required for a discrete probability function are F(x) ≥ 0 ∑f(x) =1 Example: Using past data on TV sales of JSL appliances, we can develop a tabular representation of the probability distribution for TV sales as shown below; Units sold Number of days x F(x) 0 80 0 0.40 1 50 1 0.25 2 40 2 0.20 3 10 3 0.50 4 20 4 0.10 200 1.00 NB//f(x) = is arrived at after dividing the total number of days by the number of days for specific units sold. The probability distribution in the above table can be representedgraphically as shown below; Values of random variable x (TV sales) Discrete Uniform probability distribution This is the simplest form of a discrete probability distribution and is given by the function below; F(x)= 1/n Note that the values of the random variable are equally likely n=the number of values the random variable may assume c) Expected value and variance The expected value is the mean of a random variable and is a measure of its central location. It is denoted by E(x) = m = Sxf(x). The variance on the other hand summarizes the variability in the values of a random sample and is defined by; Var(x) = s 2 = S(x - m) 2 f(x) In addition, the standard deviation represented by s is the positive square root of variance. Given the JS appliances data, the expected value of a discrete variable could be calculated as depicted in the table below; x f(x) xf(x) 0 0.40 00 1 0.25 0.25 2 0.20 0.40 3 0.05 0.15 4 0.10 0.40 E(x)= 1.20 Based on the calculations above, the expected number of TVs sold in a day is 1.20 On the other hand, the variance and the standard deviation of a discrete random variable based on the JSL appliances data above will be calculated as shown in the table below; x x-m (x-m)2 f(x) (x-m)2f(x) 0 -1.2 1.44 0.40 0.576 1 -0.2 0.04 0.25 0.010 2 0.8 0.64 0.20 0.128 3 1.8 3.24 0.05 0.162 4 2.8 7.84 0.10 0.784 Variance of daily sales=2 = 1.660 TVs squared Standard deviation of daily sales = 1.2884 TVs We can compute the expected value, variance and standard deviation by using excel as can be shown in the formula worksheet below; Using the excel, we can compute the expected value, variance and standard deviation for the JSL data above as shown in the value worksheet below; A B C 1 Sales Probability Sq.Dev.from mean 2 0 0.40 1.44 3 1 0.25 0.04 4 2 0.20 0.64 5 3 0.05 3.24 6 4 0.10 7.84 7 8 Mean 1.2 9 Variance 1.66 10 Std.Dev. 1.2884 d) Binomial probability distribution Binomial experiments have the following properties; i) Binomial experiments consist of a sequence of n identical trials ii) There two possible outcomes in a binomial experiment in each trial.ie, Success and failure iii) Binomial experiments are characterized by stationarity assumption. In other words, the probability of a success which is denoted by P does not change from trial to trial. iv) The trials in a binomial experiment are independent Binomial probability distribution is mainly concerned with the number of successes that occur in the n trials. In this regard, x denotes the number of successes that occur in n trials. Binomial probability distribution; f(x)= n! Px(1-P)(n-x) Where: x! (n-x)! f(x) = the probability of x success in n trials n = the number of trials p = the probability of success on any one trial In the function, n! Represents the number of experimental outcomes providing exactly x x! (n-x)!successes in n trials On the other hand,Px(1-P)(n-x) is the probability of a particular sequence of trial outcomes with x successes in n trials. Example of binomial probability distribution Evans electronics is concerned about a low retention rate for its employees. In recent years, the management has seen a turnover of 10% of the hourly employees annually. Thus for any hourly employee chosen at random, management estimates a probability of 0.10 that the person will not be with the company next year. Choosing 3 hourly employees at random, what is the probability that 1 of them will leave the company this year? Solution: f(x)= n! Px(1-P)(n-x) Where: x! (n-x)! Let p= 0.10, n= 3, x= 1 f (1) = 3! (0.1)1(0.9)2=3(0.1) (0.8) =2.43 1! (3-1)! We can also use a tree diagram to illustrate this as shown below; 1st Worker 2ndworker 3rd Worker x prob Leaves 0.1 3 0.0010 Leaves0.1 stays 0.9 2 0.0900 Leaves 0.1 2 0.0810 Stays 0.9 Stays 0.9 1 0.0090 Leaves 0.1 Leaves 0.1 2 0.0090 stays 0.9 Leaves0.1 stays 0.9 1 0.0810 Stays0.9 Leaves 0.1 1 0.0810 Stays 0.9 2 0.0010 We can also use excel to compute binomial probabilities as shown in the formula worksheet below; Using the above formula sheet, we can compute the binomial probabilities that one employee will leave as shown in the value worksheet below; A B 1 3 =Number of trials (n) 2 0.1 =Probability of success (p) 3 4 x f(x) 5 0 0.729 6 1 0.243 7 2 0.027 8 3 0.001 We can also use excel to compute cumulative binomial probabilities as shown in the formula worksheet below; A B 1 3 =Number of trials (n) 2 0.1 =Probability of success (p) 3 4 x Cumulative probability 5 0 =BINOMDIST(A5,$A$1,$A$A2,TRUE) 6 1 =BINOMDIST(A6,$A$A1,$A$2, TRUE) 7 2 =BINOMDIST(A7,$A$A1,$A$2, TRUE) 8 3 = BINOMDIST(A8,$A$A1,$A$2, TRUE) In this regard, the cumulative binomial probabilities for the employees’ turnover have been calculated as shown in the table below; Binomial probability distribution’s expected value is given by E(x) =m=np. On the other hand, variance is given by Var(x) = s2 = np (1 - p) while standard deviation is given by S= sqrt (1-p). Based on the above example, the expected value will be given by E(x) =  = 3(.1) = .3 employees out of 3. The variance is given by Var(x) =  2 = 3(.1)(.9) = .27 while standard deviation is given by S= SQRT (3(.1)(.9))=.52 employees. e)Poisson probability distribution a Poisson distributed valuable is often useful when estimating the number of occurrences over a specified interval of time or space. A Poisson probability distribution is a discrete random variable which may assume an infinite sequence of values (x=0,1,2….). An example of a Poisson distributed random variable may be the number of knotholes in 24 linear feet of pine board or the number of vehicles arriving at a toll booth in one hour. A Poisson experiment has two properties including; i) The probability of an occurrence is the same for any two intervals of equal length ii) The occurrence or non-occurrence in any intervals is independent of the occurrence or non-occurrence in any other interval. The Poisson probability function is f(x) =( mxe-m)/x! Where; f(x)= Probability of x occurrences in an interval m = mean number of occurrences in an interval e =2.71828 Example of Poisson probability function At Mercy hospital, patients arrive at the emergency room at the average rate of 6 patients per hour on weekend evenings. What is the probability of 4 arrivals in 30 minutes on a weekend evening? Solution: m = 6/hour = 3/half-hour, x = 4 f(4) = (34(2.71828)-3)/4! =0.1680 We can use excel to compute Poisson probabilities as shown in the formula worksheet below; Similarly, we can solve the above problem using excel as shown in the value worksheet below; The Poisson distribution can be represented graphically as shown below; The table below shows how we can use excel to compute cumulative Poisson probabilities The cumulative Poisson probabilities have been computed using excel as shown in the value worksheet below; References: Read More