**Discrete Random Variable.**

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Name Shakeel Nouman Religion Christian Domicile Punjab (Lahore) Contact # 0332-4462527. 0321-9898767 E.Mail [email protected] [email protected]

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### Discrete Random Variables:

Shakeel Nouman M.Phil Statistics Discrete Random Variables Random Variables By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer### PowerPoint Presentation:

Using Statistics Expected Values of Discrete Random Variables The Binomial Distribution Other Discrete Probability Distributions Continuous Random Variables Using the Computer Summary and Review of Terms Random Variables 3 Random Variables By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer### 3-1 Using Statistics:

Consider the different possible orderings of boy (B) and girl (G) in four sequential births. There are 2*2*2*2=2 4 = 16 possibilities, so the sample space is: BBBB BGBB GBBB GGBB BBBG BGBG GBBG GGBG BBGB BGGB GBGB GGGB BBGG BGGG GBGG GGGG If girl and boy are each equally likely [P(G)=P(B) = 1/2], and the gender of each child is independent of that of the previous child, then the probability of each of these 16 possibilities is: (1/2)(1/2)(1/2)(1/2) = 1/16. 3-1 Using Statistics Random Variables By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer### Random Variables :

Now count the number of girls in each set of four sequential births: BBBB (0) BGBB (1) GBBB (1) GGBB (2) BBBG (1) BGBG (2) GBBG (2) GGBG (3) BBGB (1) BGGB (2) GBGB (2) GGGB (3) BBGG (2) BGGG (3) GBGG (3) GGGG (4) Notice that: each possible outcome is assigned a single numeric value, all outcomes are assigned a numeric value, and the value assigned varies over the outcomes . The count of the number of girls is a random variable : A random variable, X, is a function that assigns a single, but variable, value to each element of a sample space. Random Variables Random Variables By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer### Random Variables (Continued):

Random Variables (Continued) BBBB BGBB GBBB BBBG BBGB GGBB GBBG BGBG BGGB GBGB BBGG BGGG GBGG GGGB GGBG GGGG 0 1 2 3 4 X Sample Space Points on the Real Line Random Variables By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer### Random Variables (Continued):

Since the random variable X = 3 when any of the four outcomes BGGG, GBGG, GGBG, or GGGB occurs, P(X = 3) = P(BGGG) + P(GBGG) + P(GGBG) + P(GGGB) = 4/16 The probability distribution of a random variable is a table that lists the possible values of the random variables and their associated probabilities . x P(x) 0 1/16 1 4/16 2 6/16 3 4/16 4 1/16 16/16=1 4 3 2 1 0 0 . 4 0 . 3 0 . 2 0 . 1 N u m b e r o f g i r l s , x P ( x ) 0 . 0 6 2 5 0 . 2 5 0 0 0 . 3 7 5 0 0 . 2 5 0 0 0 . 0 6 2 5 P r o b a b i l i t y D i s t r i b u t i o n o f t h e N u m b e r o f G i r l s i n F o u r B i r t h s Random Variables (Continued) Random Variables By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer### Example 3-1:

Consider the experiment of tossing two six-sided dice . There are 36 possible outcomes . Let the random variable X represent the sum of the numbers on the two dice : 2 3 4 5 6 7 1,1 1,2 1,3 1,4 1,5 1,6 8 2,1 2,2 2,3 2,4 2,5 2,6 9 3,1 3,2 3,3 3,4 3,5 3,6 10 4,1 4,2 4,3 4,4 4,5 4,6 11 5,1 5,2 5,3 5,4 5,5 5,6 12 6,1 6,2 6,3 6,4 6,5 6,6 x P(x) * 2 1/36 3 2/36 4 3/36 5 4/36 6 5/36 7 6/36 8 5/36 9 4/36 10 3/36 11 2/36 12 1/36 1 1 2 1 1 1 0 9 8 7 6 5 4 3 2 0 . 1 7 0 . 1 2 0 . 0 7 0 . 0 2 x p ( x ) P r o b a b i l i t y D i s t r i b u t i o n o f S u m o f T w o D i c e Example 3-1 Random Variables By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer### Example 3-2:

Probability of at least 1 switch: P(X ³ 1) = 1 - P(0) = 1 - 0.1 = .9 Probability Distribution of the Number of Switches x P(x) 0 0.1 1 0.2 2 0.3 3 0.2 4 0.1 5 0.1 1 Probability of more than 2 switches: P(X > 2) = P(3) + P(4) + P(5) = 0.2 + 0.1 + 0.1 = 0.4 5 4 3 2 1 0 0 . 4 0 . 3 0 . 2 0 . 1 0 . 0 x P ( x ) T h e P r o b a b i l i t y D i s t r i b u t i o n o f t h e N u m b e r o f S w i t c h e s Example 3-2 Random Variables By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer### Discrete and Continuous Random Variables:

A discrete random variable: has a countable number of possible values has discrete jumps (or gaps) between successive values has measurable probability associated with individual values counts A continuous random variable: has an uncountably infinite number of possible values moves continuously from value to value has no measurable probability associated with each value measures (e.g.: height, weight, speed, value, duration, length) Discrete and Continuous Random Variables Random Variables By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer### Rules of Discrete Probability Distributions:

The probability distribution of a discrete random variable X must satisfy the following two conditions. Rules of Discrete Probability Distributions Random Variables By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer### Cumulative Distribution Function:

The cumulative distribution function , F(x) , of a discrete random variable X is: x P(x) F(x) 0 0.1 0.1 1 0.2 0.3 2 0.3 0.6 3 0.2 0.8 4 0.1 0.9 5 0.1 1.0 1 5 4 3 2 1 0 1 . 0 0 . 9 0 . 8 0 . 7 0 . 6 0 . 5 0 . 4 0 . 3 0 . 2 0 . 1 0 . 0 x F ( x ) C u m u l a t i v e P r o b a b i l i t y D i s t r i b u t i o n o f t h e N u m b e r o f S w i t c h e s Cumulative Distribution Function Random Variables By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer### Cumulative Distribution Function:

x P(x ) F(x) 0 0.1 0.1 1 0.2 0.3 2 0.3 0.6 3 0.2 0.8 4 0.1 0.9 5 0.1 1.0 1 The probability that at most three switches will occur: T h e P r o b a b i l i t y T h a t a t M o s t T h r e e S w i t c h e s W i l l O c c u r 5 4 3 2 1 0 0 . 4 0 . 3 0 . 2 0 . 1 0 . 0 x P ( x ) Cumulative Distribution Function Note: P(X < 3) = F(3) = 0.8 = P(0) + P(1) + P(2) + P(3) Random Variables By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer### Using Cumulative Probability Distributions (Figure 3-8):

x P(x) F(x) 0 0.1 0.1 1 0.2 0.3 2 0.3 0.6 3 0.2 0.8 4 0.1 0.9 5 0.1 1.0 1 The probability that more than one switch will occur: 5 4 3 2 1 0 0 . 4 0 . 3 0 . 2 0 . 1 0 . 0 x P ( x ) The Probability That More than One Switch Will Occur Using Cumulative Probability Distributions (Figure 3-8) Note: P(X > 1) = P(X > 2) = 1 – P(X < 1) = 1 – F(1) = 1 – 0.3 = 0.7 Random Variables By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer### Using Cumulative Probability Distributions (Figure 3-9):

x P(x) F(x) 0 0.1 0.1 1 0.2 0.3 2 0.3 0.6 3 0.2 0.8 4 0.1 0.9 5 0.1 1.0 1 The probability that anywhere from one to three switches will occur: 5 4 3 2 1 0 0 . 4 0 . 3 0 . 2 0 . 1 0 . 0 x P ( x ) The Probability That Anywhere from One to Three Switches Will Occur Using Cumulative Probability Distributions (Figure 3-9) Note: P(1 < X < 3) = P(X < 3) – P(X < 0) = F(3) – F(0) = 0.8 – 0.1 = 0.7 Random Variables By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer### 3-2 Expected Values of Discrete Random Variables:

5 4 3 2 1 0 The mean of a probability distribution is a measure of its centrality or location, as is the mean or average of a frequency distribution. It is a weighted average , with the values of the random variable weighted by their probabilities. The mean is also known as the expected value (or expectation ) of a random variable, because it is the value that is expected to occur, on average. The expected value of a discrete random variable X is equal to the sum of each value of the random variable multiplied by its probability. x P(x) xP (x) 0 0.1 0.0 1 0.2 0.2 2 0.3 0.6 3 0.2 0.6 4 0.1 0.4 5 0.1 0.5 1.0 2.3 = E(X) = m 2.3 3-2 Expected Values of Discrete Random Variables Random Variables By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer### A Fair Game:

Suppose you are playing a coin toss game in which you are paid $1 if the coin turns up heads and you lose $1 when the coin turns up tails. The expected value of this game is E(X) = 0. A game of chance with an expected payoff of 0 is called a fair game . x P(x) xP (x) -1 0.5 -0.50 1 0.5 0.50 1.0 0.00 = E(X)=m -1 1 0 A Fair Game Random Variables By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer### Expected Value of a Function of a Discrete Random Variables:

Number of items, x P(x) xP (x) h(x) h(x)P(x) 5000 0.2 1000 2000 400 6000 0.3 1800 4000 1200 7000 0.2 1400 6000 1200 8000 0.2 1600 8000 1600 9000 0.1 900 10000 1000 1.0 6700 5400 Example 3-3 : Monthly sales of a certain product are believed to follow the given probability distribution. Suppose the company has a fixed monthly production cost of $8000 and that each item brings $2. Find the expected monthly profit h(X), from product sales. The expected value of a function of a discrete random variable X is: The expected value of a linear function of a random variable is: E(aX+b)=aE(X)+b In this case: E(2X-8000)=2E(X)-8000=(2)(6700)-8000=5400 Expected Value of a Function of a Discrete Random Variables Note: h(X) = 2X – 8000 where X = # of items sold Random Variables By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer### Variance and Standard Deviation of a Random Variable:

The variance of a random variable is the expected squared deviation from the mean: The standard deviation of a random variable is the square root of its variance: Variance and Standard Deviation of a Random Variable Random Variables By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer### Variance and Standard Deviation of a Random Variable – using Example 3-2:

Number of Switches, x P(x) xP(x) (x-m) (x-m) 2 P(x-m) 2 x 2 P(x) 0 0.1 0.0 -2.3 5.29 0.529 0.0 1 0.2 0.2 -1.3 1.69 0.338 0.2 2 0.3 0.6 -0.3 0.09 0.027 1.2 3 0.2 0.6 0.7 0.49 0.098 1.8 4 0.1 0.4 1.7 2.89 0.289 1.6 5 0.1 0.5 2.7 7.29 0.729 2.5 2.3 2.010 7.3 s m m 2 2 2 2 01 2 2 2 2 7 3 2 3 2 2 01 = = - = - å = = - = å é ë ê ù û ú - å é ë ê ù û ú = - = V X E X x all x P x E X E X x all x P x xP x all x ( ) [( ) ] ( ) ( ) . ( ) [ ( )] ( ) ( ) . . . Table 3-8 Variance and Standard Deviation of a Random Variable – using Example 3-2 Recall: = 2.3. Random Variables By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer### Variance of a Linear Function of a Random Variable:

The variance of a linear function of a random variable is: Number of items, x P(x) xP(x) x 2 P(x) 5000 0.2 1000 5000000 6000 0.3 1800 10800000 7000 0.2 1400 9800000 8000 0.2 1600 12800000 9000 0.1 900 8100000 1.0 6700 46500000 Example 3-3: Variance of a Linear Function of a Random Variable Random Variables By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer### Some Properties of Means and Variances of Random Variables:

The mean or expected value of the sum of random variables is the sum of their means or expected values: For example: E(X) = $350 and E(Y) = $200 E(X+Y) = $350 + $200 = $550 The variance of the sum of independent random variables is the sum of their variances: For example: V(X) = 84 and V(Y) = 60 V(X+Y) = 144 Some Properties of Means and Variances of Random Variables Random Variables By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer### Chebyshev’s Theorem Applied to Probability Distributions:

Chebyshev’s Theorem applies to probability distributions just as it applies to frequency distributions. For a random variable X with mean m, standard deviation s, and for any number k > 1: At least Lie within Standard deviations of the mean 2 3 4 Chebyshev’s Theorem Applied to Probability Distributions Random Variables By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer### Using the Template to Calculate statistics of h(x):

Using the Template to Calculate statistics of h(x) Random Variables By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer### 3-3 Bernoulli Random Variable:

If an experiment consists of a single trial and the outcome of the trial can only be either a success * or a failure, then the trial is called a Bernoulli trial. The number of success X in one Bernoulli trial, which can be 1 or 0, is a Bernoulli random variable. Note: If p is the probability of success in a Bernoulli experiment, the E(X) = p and V(X) = p(1 – p). * The terms success and failure are simply statistical terms, and do not have positive or negative implications. In a production setting, finding a defective product may be termed a “success,” although it is not a positive result. 3-3 Bernoulli Random Variable Random Variables By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer### The Binomial Random Variable :

Consider a Bernoulli Process in which we have a sequence of n identical trials satisfying the following conditions: 1. Each trial has two possible outcomes, called success *and failure . The two outcomes are mutually exclusive and exhaustive . 2. The probability of success , denoted by p , remains constant from trial to trial. The probability of failure is denoted by q , where q = 1-p . 3. The n trials are independent . That is, the outcome of any trial does not affect the outcomes of the other trials. A random variable, X, that counts the number of successes in n Bernoulli trials, where p is the probability of success* in any given trial, is said to follow the binomial probability distribution with parameters n (number of trials) and p (probability of success). We call X the binomial random variable . * The terms success and failure are simply statistical terms, and do not have positive or negative implications. In a production setting, finding a defective product may be termed a “success,” although it is not a positive result. The Binomial Random Variable Random Variables By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer### Binomial Probabilities (Introduction):

Suppose we toss a single fair and balanced coin five times in succession, and let X represent the number of heads. There are 2 5 = 32 possible sequences of H and T (S and F) in the sample space for this experiment. Of these, there are 10 in which there are exactly 2 heads (X=2): HHTTT HTHTH HTTHT HTTTH THHTT THTHT THTTH TTHHT TTHTH TTTHH The probability of each of these 10 outcomes is p 3 q 3 = (1/2) 3 (1/2) 2 =(1/32), so the probability of 2 heads in 5 tosses of a fair and balanced coin is: P(X = 2) = 10 * (1/32) = (10/32) = .3125 Binomial Probabilities (Introduction) Random Variables By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer### Binomial Probabilities (continued):

P(X=2) = 10 * (1/32) = (10/32) = .3125 Notice that this probability has two parts: In general: 1. The probability of a given sequence of x successes out of n trials with probability of success p and probability of failure q is equal to: p x q (n-x) 2. The number of different sequences of n trials that result in exactly x successes is equal to the number of choices of x elements out of a total of n elements. This number is denoted: Binomial Probabilities (continued) Random Variables By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer### The Binomial Probability Distribution:

The binomial probability distribution : where : p is the probability of success in a single trial, q = 1-p , n is the number of trials, and x is the number of successes. The Binomial Probability Distribution Random Variables By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer### The Cumulative Binomial Probability Table (Table 1, Appendix C):

Cumulative Binomial Probability Distribution and Binomial Probability Distribution of H,the Number of Heads Appearing in Five Tosses of a Fair Coin Deriving Individual Probabilities from Cumulative Probabilities The Cumulative Binomial Probability Table (Table 1, Appendix C) Random Variables By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer### Calculating Binomial Probabilities - Example:

60% of Brooke shares are owned by LeBow . A random sample of 15 shares is chosen. What is the probability that at most three of them will be found to be owned by LeBow ? Calculating Binomial Probabilities - Example Random Variables By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer### Mean, Variance, and Standard Deviation of the Binomial Distribution:

Mean, Variance, and Standard Deviation of the Binomial Distribution Random Variables By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer### Calculating Binomial Probabilities using the Template:

Calculating Binomial Probabilities using the Template Random Variables By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer### Shape of the Binomial Distribution:

4 3 2 1 0 0 . 7 0 . 6 0 . 5 0 . 4 0 . 3 0 . 2 0 . 1 0 . 0 x P ( x ) B i n o m i a l P r o b a b i l i t y : n = 4 p = 0 . 5 4 3 2 1 0 0 . 7 0 . 6 0 . 5 0 . 4 0 . 3 0 . 2 0 . 1 0 . 0 x P ( x ) B i n o m i a l P r o b a b i l i t y : n = 4 p = 0 . 1 4 3 2 1 0 0 . 7 0 . 6 0 . 5 0 . 4 0 . 3 0 . 2 0 . 1 0 . 0 x P ( x ) B i n o m i a l P r o b a b i l i t y : n = 4 p = 0 . 3 1 0 9 8 7 6 5 4 3 2 1 0 0 . 5 0 . 4 0 . 3 0 . 2 0 . 1 0 . 0 x P ( x ) B i n o m i a l P r o b a b i l i t y : n = 1 0 p = 0 . 1 1 0 9 8 7 6 5 4 3 2 1 0 0 . 5 0 . 4 0 . 3 0 . 2 0 . 1 0 . 0 x P ( x ) B i n o m i a l P r o b a b i l i t y : n = 1 0 p = 0 . 3 1 0 9 8 7 6 5 4 3 2 1 0 0 . 5 0 . 4 0 . 3 0 . 2 0 . 1 0 . 0 x P ( x ) B i n o m i a l P r o b a b i l i t y : n = 1 0 p = 0 . 5 2 0 1 9 1 8 1 7 1 6 1 5 1 4 1 3 1 2 1 1 1 0 9 8 7 6 5 4 3 2 1 0 0 . 2 0 . 1 0 . 0 x P ( x ) B i n o m i a l P r o b a b i l i t y : n = 2 0 p = 0 . 1 2 0 1 9 1 8 1 7 1 6 1 5 1 4 1 3 1 2 1 1 1 0 9 8 7 6 5 4 3 2 1 0 0 . 2 0 . 1 0 . 0 x P ( x ) B i n o m i a l P r o b a b i l i t y : n = 2 0 p = 0 . 3 2 0 1 9 1 8 1 7 1 6 1 5 1 4 1 3 1 2 1 1 1 0 9 8 7 6 5 4 3 2 1 0 0 . 2 0 . 1 0 . 0 x P ( x ) B i n o m i a l P r o b a b i l i t y : n = 2 0 p = 0 . 5 Binomial distributions become more symmetric as n increases and as p .5 . p = 0.1 p = 0.3 p = 0.5 n = 4 n = 10 n = 20 Shape of the Binomial Distribution Random Variables By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer### 3-5 Negative Binomial Distribution:

The negative binomial distribution is useful for determining the probability of the number of trials made until the desired number of successes are achieved in a sequence of Bernoulli trials. It counts the number of trials X to achieve the number of successes s with p being the probability of success on each trial. 3-5 Negative Binomial Distribution Random Variables By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer### Negative Binomial Distribution - Example:

Negative Binomial Distribution - Example Example: Suppose that the probability of a manufacturing process producing a defective item is 0.05. Suppose further that the quality of any one item is independent of the quality of any other item produced. If a quality control officer selects items at random from the production line, what is the probability that the first defective item is the eight item selected. Here s = 1, x = 8, and p = 0.05. Thus, Random Variables By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer### Calculating Negative Binomial Probabilities using the Template:

Calculating Negative Binomial Probabilities using the Template Random Variables By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer### 3-6 The Geometric Distribution:

Within the context of a binomial experiment, in which the outcome of each of n independent trials can be classified as a success ( S ) or a failure ( F ), the geometric random variable counts the number of trials until the first success .. 3-6 The Geometric Distribution Random Variables By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer### The Geometric Distribution - Example:

Example: A recent study indicates that Pepsi-Cola has a market share of 33.2% (versus 40.9% for Coca-Cola). A marketing research firm wants to conduct a new taste test for which it needs Pepsi drinkers. Potential participants for the test are selected by random screening of soft drink users to find Pepsi drinkers. What is the probability that the first randomly selected drinker qualifies? What’s the probability that two soft drink users will have to be interviewed to find the first Pepsi drinker? Three? Four? The Geometric Distribution - Example Random Variables By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer### Calculating Geometric Distribution Probabilities using the Template:

Calculating Geometric Distribution Probabilities using the Template Random Variables By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer### 3-7 The Hypergeometric Distribution:

The hypergeometric probability distribution is useful for determining the probability of a number of occurrences when sampling without replacement . It counts the number of successes ( x ) in n selections, without replacement , from a population of N elements, S of which are successes and (N-S) of which are failures . 3-7 The Hypergeometric Distribution Random Variables By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer### The Hypergeometric Distribution - Example :

Example: Suppose that automobiles arrive at a dealership in lots of 10 and that for time and resource considerations, only 5 out of each 10 are inspected for safety. The 5 cars are randomly chosen from the 10 on the lot. If 2 out of the 10 cars on the lot are below standards for safety, what is the probability that at least 1 out of the 5 cars to be inspected will be found not meeting safety standards? Thus, P(1) + P(2) = 0.556 + 0.222 = 0.778. The Hypergeometric Distribution - Example Random Variables By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer### Calculating Hypergeometric Distribution Probabilities using the Template:

Calculating Hypergeometric Distribution Probabilities using the Template Random Variables By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer### 3-8 The Poisson Distribution:

The Poisson probability distribution is useful for determining the probability of a number of occurrences over a given period of time or within a given area or volume. That is, the Poisson random variable counts occurrences over a continuous interval of time or space. It can also be used to calculate approximate binomial probabilities when the probability of success is small ( p£0.05 ) and the number of trials is large ( n³20 ). where m is the mean of the distribution (which also happens to be the variance) and e is the base of natural logarithms ( e=2.71828... ). 3-8 The Poisson Distribution Random Variables By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer### The Poisson Distribution - Example:

Example 3-5: Telephone manufacturers now offer 1000 different choices for a telephone (as combinations of color, type, options, portability, etc.). A company is opening a large regional office, and each of its 200 managers is allowed to order his or her own choice of a telephone. Assuming independence of choices and that each of the 1000 choices is equally likely, what is the probability that a particular choice will be made by none, one, two, or three of the managers? n = 200 m = np = (200)(0.001) = 0.2 p = 1/1000 = 0.001 P e P e P e P e ( ) . ! ( ) . ! ( ) . ! ( ) . ! . . . . 0 2 0 1 2 1 2 2 2 3 2 3 0 2 1 2 2 2 3 2 = = = = - - - - = 0.8187 = 0.1637 = 0.0164 = 0.0011 The Poisson Distribution - Example Random Variables By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer### Calculating Poisson Distribution Probabilities using the Template:

Calculating Poisson Distribution Probabilities using the Template Random Variables By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer### The Poisson Distribution (continued):

Poisson assumptions: The probability that an event will occur in a short interval of time or space is proportional to the size of the interval. In a very small interval, the probability that two events will occur is close to zero. The probability that any number of events will occur in a given interval is independent of where the interval begins. The probability of any number of events occurring over a given interval is independent of the number of events that occurred prior to the interval. The Poisson Distribution (continued) Random Variables By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer### The Poisson Distribution (continued):

2 0 1 9 1 8 1 7 1 6 1 5 1 4 1 3 1 2 1 1 1 0 9 8 7 6 5 4 3 2 1 0 0 . 1 5 0 . 1 0 0 . 0 5 0 . 0 0 X P ( x ) m = 1 0 1 0 9 8 7 6 5 4 3 2 1 0 0 . 2 0 . 1 0 . 0 X P ( x ) m = 4 7 6 5 4 3 2 1 0 0 . 4 0 . 3 0 . 2 0 . 1 0 . 0 X P ( x ) m = 1 . 5 4 3 2 1 0 0 . 4 0 . 3 0 . 2 0 . 1 0 . 0 X P ( x ) m = 1 . 0 The Poisson Distribution (continued) Random Variables By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer### Discrete and Continuous Random Variables - Revisited:

A discrete random variable : counts occurrences has a countable number of possible values has discrete jumps between successive values has measurable probability associated with individual values probability is height A continuous random variable : measures (e.g.: height, weight, speed, value, duration, length) has an uncountably infinite number of possible values moves continuously from value to value has no measurable probability associated with individual values probability is area For example: Binomial n=3 p=.5 x P(x) 0 0.125 1 0.375 2 0.375 3 0.125 1.000 3 2 1 0 0 . 4 0 . 3 0 . 2 0 . 1 0 . 0 C 1 P ( x ) B i n o m i a l : n = 3 p = . 5 For example: In this case, the shaded area epresents the probability that the task takes between 2 and 3 minutes. 6 5 4 3 2 1 0 . 3 0 . 2 0 . 1 0 . 0 M i n u t e s P ( x ) M i n u t e s t o C o m p l e t e T a s k Discrete and Continuous Random Variables - Revisited Random Variables By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer### From a Discrete to a Continuous Distribution:

6 . 5 6 . 0 5 . 5 5 . 0 4 . 5 4 . 0 3 . 5 3 . 0 2 . 5 2 . 0 1 . 5 1 . 0 0 . 1 5 0 . 1 0 0 . 0 5 0 . 0 0 M i n u t e s P ( x ) M i n u t e s t o C o m p l e t e T a s k : B y H a l f - M i n u t e s 0.0 . 0 1 2 3 4 5 6 7 M i n u t e s P ( x ) M i n u t e s t o C o m p l e t e T a s k : F o u r t h s o f a M i n u t e M i n u t e s P ( x ) M i n u t e s t o C o m p l e t e T a s k : E i g h t h s o f a M i n u t e 0 1 2 3 4 5 6 7 The time it takes to complete a task can be subdivided into: Half-Minute Intervals Quarter-Minute Intervals Eighth-Minute Intervals Or even infinitesimally small intervals: When a continuous random variable has been subdivided into infinitesimally small intervals, a measurable probability can only be associated with an interval of values, and the probability is given by the area beneath the probability density function corresponding to that interval. In this example, the shaded area represents P(2 £ X £ 3). Minutes to Complete Task: Probability Density Function 7 6 5 4 3 2 1 0 Minutes f ( z ) From a Discrete to a Continuous Distribution Random Variables By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer### 3-9 Continuous Random Variables:

A continuous random variable is a random variable that can take on any value in an interval of numbers. The probabilities associated with a continuous random variable X are determined by the probability density function of the random variable. The function, denoted f(x) , has the following properties. 1. f(x) ³ 0 for all x . 2. The probability that X will be between two numbers a and b is equal to the area under f(x) between a and b . 3. The total area under the curve of f(x) is equal to 1.00. The cumulative distribution function of a continuous random variable: F(x) = P(X £ x) =Area under f(x) between the smallest possible value of X (often -¥) and the point x . 3-9 Continuous Random Variables Random Variables By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer### Probability Density Function and Cumulative Distribution Function:

F(x) f(x) x x 0 0 b a F(b) F(a) 1 b a } P(a £ X £ b) = Area under f(x) between a and b = F(b) - F(a) P(a £ X £ b)=F(b) - F(a) Probability Density Function and Cumulative Distribution Function Random Variables By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer### 3-10 Uniform Distribution:

The uniform [ a,b ] density : 1/(a – b) for a £ X £ b f(x)= 0 otherwise E(X) = (a + b)/2; V(X) = (b – a) 2 /12 { b b1 a x f ( x ) The entire area under f(x) = 1/(b – a) * (b – a) = 1.00 The area under f(x) from a1 to b1 = P(a1£X£ b1) = (b1 – a1)/(b – a) 3-10 Uniform Distribution a1 Uniform [a, b] Distribution Random Variables By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer### Uniform Distribution (continued):

The uniform [0,5] density : 1/5 for 0 £ X £ 5 f(x)= 0 otherwise E(X) = 2.5 { 6 5 4 3 2 1 0 - 1 0 . 5 0 . 4 0 . 3 0 . 2 0 . 1 0.0 . x f ( x ) U n i f o r m [ 0 , 5 ] D i s t r i b u t i o n The entire area under f(x) = 1/5 * 5 = 1.00 The area under f(x) from 1 to 3 = P(1£X£3) = (1/5)2 = 2/5 Uniform Distribution (continued) Random Variables By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer### Calculating Uniform Distribution Probabilities using the Template:

Calculating Uniform Distribution Probabilities using the Template Random Variables By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer### 3-11 Exponential Distribution:

The exponential random variable measures the time between two occurrences that have a Poisson distribution. 3 2 1 0 2 1 0 f ( x ) E x p o n e n t i a l D i s t r i b u t i o n : l = 2 Time 3-11 Exponential Distribution Random Variables By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer### Exponential Distribution - Example:

Example The time a particular machine operates before breaking down (time between breakdowns) is known to have an exponential distribution with parameter l = 2. Time is measured in hours. What is the probability that the machine will work continuously for at least one hour? What is the average time between breakdowns? F x e P X x e P X e x x ( ) ( ) ( ) . ( )( ) = - Þ ³ = ³ = = - - - 1 1 1353 2 1 l l E X ( ) . = = = 1 1 2 5 l Exponential Distribution - Example Random Variables By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer### Calculating Exponential Distribution Probabilities using the Template:

Calculating Exponential Distribution Probabilities using the Template Random Variables By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer### PowerPoint Presentation:

M.Phil (Statistics) GC University, . (Degree awarded by GC University) M.Sc (Statistics) GC University, . (Degree awarded by GC University) Statitical Officer (BS-17 ) ( Economics & Marketing Division ) Livestock Production Research Institute Bahadurnagar ( Okara ), Livestock & Dairy Development Department, Govt. of Punjab Name Shakeel Nouman Religion Christian Domicile Punjab (Lahore) Contact # 0332-4462527. 0321-9898767 E.Mail [email protected] [email protected] Random Variables By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer## View More Presentations

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