Probability Distributions Hana Alay Wedad Alsayegh Probability Distributions
Probability Distributions Hana Alay Wedad Alsayegh
Probability Distributions Outline Binominal Probability Distributions Poisson Probability Distributions
Probability Distributions Definition of random variable and probability distributions Determine when a potential probability distribution satisfy the requirements Compute mean and SD by a given probability distribution
Definitions Random Variable -Is a variable (represented by x) that has a single numerical value, determined by chance, for each outcome of a procedure. -Random variables may be discrete or continuous Probability Distribution Is a description that gives a probability for each value of the random variable. It is often expressed in the format of a table, formula, or graph.
Definitions Discrete Random Variable Has a collection of values that is finite or countable. (If there are infinitely many values, the number of values is countable if it is possible to count them individually, such as the number of tosses of a coin before getting heads) Continuous Random Variable Has infinitely many values, and the collection of values is not countable. (It is possible to count the individual items because at last some of them are in a continuous scale, such as body temperature)
Probability Distributions Requirements There is a numerical (not categorial) random variable x, and its number values are associated with corresponding probabilities. ΣP(x) = 1 where x assumes all possible events ( the sum of all probabilities must be 1, but sums as 0. 999 or 1. 001 are acceptable because they result from rounding errors) 0 ≤ P(x) ≤ 1 for every individual value of the random variable x (each probability value must be between 0 and 1, inclusive)
Example x = number of girls in two births The above x is a random variable because its numerical values depend on chance. With two births, the number of girls can be 0, 1 or 2. X p(x) B+B=0 1/4 0 0. 25 2/4 1 0. 50 G+B=1 B+G 2 0. 25 G+G=2 1/4
The table is probability distribution because it satisfies the 3 requirements: 1 - x is a numerical random variable and its values associated with probabilities 2 - ΣP(x) = 0. 25 + 0. 50 + 0. 25 = 1 3 - each value of P(x) is between 0 and 1 - The random variable x is a discrete random variable because it has 3 possible values and 3 is a finite number. X p(x) 0 0. 25 1 0. 50 2 0. 25
Probability Formula
Example: Hospital Job Interview Mistakes Hiring managers were asked to identify the biggest mistakes that job applicants make during an interview, and the provided table is based on their responses. Does the table describe a probability distributions? x p(x) Inappropriate attire 0. 50 Being late 0. 44 Lack of eye contact 0. 33 Checking phone 0. 30 Total 1. 57
ANSWER It does not describe a probability distribution because it violates the 3 requirements. x is categorial variable instead of numerical. Also, the sum of all probabilities is 1. 57
Parameters of Probability Distribution with probability distribution we have a description of a population instead of a sample, so the values of the mean, SD, and variance are parameters, not statistics. The mean, SD, and variance of a discrete probability distribution can be found with the following formulas:
Round-Off Rules Round results by carrying one more decimal place than the number of decimal places used for the random variable x. If the values of x are integers, round the mean, SD and variance to one decimal place. Exception to round off rule Sometimes result in misleading values. For example, with 4 engine jets the mean number of jet engines working successfully throughout a flight is 3. 99971486, which becomes 4. 0 when rounded, but it is misleading because it suggest that all jet engines always working successfully. Here we need more precision to correctly reflect the true mean, such as the precision in 3. 999714
Expected Values The mean of a discrete random variable x is theoretical mean outcome for infinitely many trials. We can think of that mean as the expected value in the sense that is it the average value that we would expect to get if the trials could continue indefinitely. The expected value of a discrete random variable x is denoted by E, and it is the mean value of the outcomes, so E = �� and E can also be found by evaluating Note: E need not to be a whole number. The expected number of girls in 5 births is 2. 5, even though five births can never result in 2. 5 girls
Example: Find the mean, variance and SD for the probability distribution described in the table �� 0 0. 25 0 x 0. 25 =0. 0 1 0. 50 1 x 0. 50=0. 50 2 0. 25 2 x 0. 25= 0. 50 Total 1. 00 0. 50
SIGNIFICANT VALUES
Example: Identifying significant results with the range rule of thumb In the previous example: �� = 1. 0 girl, SD �� = 0. 7 girl. Use those results and the range rule of thumb to determine whether 2 girls is a significantly high number of girls. ANSWER: �� + 2�� = 1 + 2(0. 7) = 2. 4 girls Significantly high number of girls are 2. 4 and above. Based on these results, we conclude that 2 girls is not a significantly high number of girls (because is not greater than or equal to 2. 4)
Identifying Significant Results With Probabilities
EXPECTED VALUE FORMULAS
5 -2 Binomial probability distribution
• Describe a binomial probability distribution. Describe OUTLINE • Find probability values for a binomial distribution. Find Compute
Frequency distribution where only 2 outcomes are possible in a given number of trials. Binomial probability distribution Common discrete distribution. Results from a procedure that meets four requirements.
The procedure has a fixed number of trials. Binomial Probability Distribution Requirements The trials must be independent. Each trial must have all outcomes classified into two categories. The probability of a success remains the same in all trials.
What makes (x) a binomial variable X= # of “heads” after 6 flips of a coin. Fixed number of trials (6). 2. Each trial can be classified as either success (heads) and failure (tails). 3. Probability of a success on each trial is constant (0. 6) 4. Made up of a finite number of independent trials. 1.
Dependent vs. independent If Y= # of kings after taking 2 cards from a standard deck without replacement. P(king on 1 st trial)= 4/52 P(king on 2 nd trial)= ? If 1 st trial outcome king P(2 nd trial)= 3/51 If 1 st trial outcome not king P(2 nd trial)= 4/51
n= # of trials. x= specific # of successes (between 0 – n). How to calculate binomial distribution p= probability of success in one n trial. q= probability of failure in one n trial (q= 1 -p). P(x)= probability of getting exactly x successes among the n trials.
How To Calculate Binomial Distribution Formula:
Can How To Calculate Binomial Distribution be divided : 1. n Cx = 2. px 3. qn-x n!/ (n-x)!x! Multiply (combination rule) the three factors together.
Practice Find the probability of getting 3 heads in 5 times of flipping the coin, given that n= 5 x= 3 p = 0. 6 q= 1 - p= 1 - 0. 6= 0. 4 n Cx = n!/ (n-x)!x! = 5!/ (5 -3)!3! 5*4*3! / 2! 3! = 20/2= px = (0. 6)3= 0. 216 10 qn-x = (0. 4)2= 0. 16 P(3)= 10*0. 216*0. 16= 0. 346
Binomial probability table at the end of the book (Appendix A).
WHAT IF THE PROCEDURE DOESN’T MEET THE REQUIREMENTS 01 02 03 If n is not fixed geometric distribution. If it is dependent hypergeometric distribution. If more than 2 outcomes multinomial distribution.
Binomial Distribution Mean µ= np Variance SD σ2 = npq σ = √npq
EXAMPLE A dice is rolled 21 times. Find the mean, variance and SD if a success is considered rolling a 5. n= 21 p= 1/6 q= 1 - (1/6)= 5/6
EXAMPLE CONT. Mean µ= np 21*(1/6)= 3. 5 Variance σ2 = npq 21*(1/6)*(5/6)= 2. 91 6 SD σ = √npq √ 2. 91 6 ≃ 1. 708 Out of rolling a dice 21 times, we would expect to see 3. 5 fives rolled on average with a SD of 1. 708
EXAMPLE CONT. Unusual events ( Range Rule of Thumb) µ± 2 σ 3. 5 -2(1. 708)= 0. 084 3. 5+2(1. 708)= 6. 916 Out of rolling a dice 21 times, it would be unusual to get 0 fives or 7 or more fives. It could happens, but less likely.
Test Your Knowledge Assuming the probability of a pea having a green pod is 0. 75, use the binomial probability of getting exactly 3 peas with green pods when 5 offspring peas are generated. Find P(3) given that: n= 5 x= 3 p= 0. 75 q= 1 -p= 0. 25
The probability of getting exactly 3 peas with green pods among 5 offspring peas is 0. 264. ANSWER:
Test your knowledge Find mean, and SD for the previous example given that: n= 5 x= 3 p= 0. 75 q= 1 -p= 0. 25
If ANSWER 75% of peas have green pods and 5 offspring peas are generated, we expect to get around (5*0. 75=3. 75≃) 3. 8 peas with green pods, with SD of (0. 968 ≃) 1
P(x) has a specific number P(at least 3) P(x > 3)= P(4)+P(5) P(at most 3) P(x ≥ 3)= P(3)+P(4)+P(5) (not more than that b/c n=5) P(more that 3) In the coin’s example: x=3 P(x ≤ 3)= P(0)+P(1)+P(2)+P(3) P(fewer than 3) P(x < 3)= P(0)+P(1)+P(2) NOTICE THAT
Poisson Probability Distribution
Is a discrete probability distribution that applies to occurrences of some event over a specified interval. The interval can be time, distance, area volume, or similar unite. A Poisson Probability Distribution Definition
Poisson probability distribution formula
The random variable x is the number of occurrences of the event in some interval. Requirements For Poisson Probability Distribution The occurrences must by random. the occurrences must be independent of each other. The occurrences must by uniformly distributed over the interval being used.
A particular poisson distribution is determined only by the mean. A poisson distribution has possible x value of 0, 1, 2, …. With no upper limit. Properties of The Poisson Probability Distribution
EXAMPLE Birth
In a recent year, there were 4229 births Riyadh medical center. Assume that the number of births each day is about the same, and assume that the Poisson distribution is a suitable model.
Find the mean number of births per day
Find the probability that on a randomly selected day, there are exactly 10 births. that is, find P(10), where P(x) is the probability of x births in a day
Thank you Questions ?
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