Probability.pptx Powerpoint presentaionc

Statistics and probability

1. Chapter 1 Random Variables and Probability Distributions Mr. Mark Christian S. Bien Subject Instructor

2. Exploring Random Variables Sample Space – the set of all possible outcomes of any experiment. (Ex. Tossing two coins – HH, HT, TH, TT) Variable – is a characteristic or attribute that can assume different values. We use capital letters to denote or represent variable.

3. Exploring Random Variables Random Variable – is a function that associates a real number to each element in the sample space. It is a variable whose values are determined by chance.

4. Exploring Random Variables Example : Suppose three coins are tossed. Let Y be the random variable representing the number of tails. Find the values of the random variable Y. Complete the table below. Possible Outcomes Value of a Random Variable Y HH HT TH TT

5. Exploring Random Variables Solution : So, the possible values of the random variable Y are 0, 1, and 2. Steps Solution Determine the sample space. Let H represent head and T represent tail. The sample space for this experiment is: S = {HH, HT, TH, TT} Count the number of tails in each outcome in the sample space and assign this number to this outcome. Possible Outcomes Value of a Random Variable Y HH HT 1 TH 1 TT 2

6. Exploring Random Variables Discrete Random Variable – it is a set of possible outcomes that is countable. Ex. Number of defective chairs produced in a factory. Continuous Random Variable – it is a set of possible outcomes on a continuous scale. Ex. Heights, weights and temperatures.

7. Constructing Probability Distributions Discrete Probability Distribution or Probability Mass Function – consists of values a random variable can assume and the corresponding probabilities of the values. Example : Suppose three coins are tossed. Let Y be the random variable representing the number of tails that occur. Find the probability of each of the values of the random variable Y.

8. Constructing Probability Distributions Solution :

9. Constructing Probability Distributions Solution : Number of Tails Y Probability P(Y) 1 1/4 1 1/4 2 2/4 or 1/2

10. Constructing Probability Distributions Solution : The Probability Distribution or the Probability Mass Function of Discrete Random Variable Y Number of Tails Y Probability P(Y) 1/4 1 1/4 2 2/4 or 1/2

11. Constructing Probability Distributions Properties of a Probability Distribution The probability of each value of the random variable must be between or equal to 0 and 1. In symbol, we write it as 0 < P(X) < 1. The sum of the probabilities of all values of the random variable must be equal to 1. In symbol, we write it as Σ P(x) = 1.

12. Computing the Mean of a Discrete Probability Distribution Steps in Computing the Mean of a Probability Distribution Construct the probability distribution for the random variable X. Multiply the value of the random variable X by the corresponding probability. Add the results obtained in Step 2.

13. Computing the Mean of a Discrete Probability Distribution The value obtained in Step 3 is called the mean of the random variable X or the mean of the probability distribution of X . In symbol, we have μ = Σ X • P(X).

14. Computing the Mean of a Discrete Probability Distribution Example: X P(X) X • P(X) 3 0.15 0.45 4 0.10 0.40 5 0.20 1.00 6 0.25 1.50 7 0.30 2.10 Σ X • P(X) = 5.45

15. Computing the Variance of a Discrete Probability Distribution Steps in Finding the Variance and the Standard Deviation of a Probability Distribution Find the mean of the probability distribution. Subtract the mean from each value of the random variable X. Square the results obtained in Step 2. Multiply the results obtained in Step 3 by the corresponding probability. Get the sum of the results obtained in Step 4.

16. Computing the Variance of a Discrete Probability Distribution Example: X P(X) X • P(X) X - μ (X – μ ) 2 (X – μ ) 2 • P(X) 1/10 -2.2 4.84 0.484 1 2/10 2/10 -1.2 1.44 0.288 2 3/10 6/10 -0.2 0.04 0.012 3 2/10 6/10 0.8 0.64 0.128 4 2/10 8/10 1.8 3.24 0.648 Σ X • P(X) = 22/10 = 2.2 Σ (X – μ ) 2 • P(X) = 1.56 μ = 2.2 Variance = 1.56 Standard Deviation = 1.25

17. Computing the Variance of a Discrete Probability Distribution Formula for the Variance and Standard Deviation of a Discrete Probability Distribution : Variance (σ 2 ) = Σ (X – μ ) 2 • P(X) Standard Deviation ( σ) = where: X = value of the random variable P(X) = probability of the random variable X μ = mean of the probability distribution  

18. Computing the Variance of a Discrete Probability Distribution Alternative Procedure in Finding the Variance and the Standard Deviation of a Probability Distribution Find the mean of the probability distribution. Multiply the square of the value of the random variable X by its corresponding probability. Get the sum of the results obtained in Step 2. Subtract the mean from the results obtained in Step 3.

19. Computing the Variance of a Discrete Probability Distribution Example: X P(X) X • P(X) X 2 • P(X) 1/10 1 2/10 2/10 0.2 2 3/10 6/10 1.2 3 2/10 6/10 1.8 4 2/10 8/10 3.2 Σ X • P(X) = 22/10 = 2.2 Σ X 2 • P(X) = 6.4 μ = 2.2 σ 2 = Σ X 2 • P(X) – μ 2 = 6.4 – (2.2) 2 Variance (σ 2 ) = 1.56 Standard Deviation ( σ) = 1.25

20. Computing the Variance of a Discrete Probability Distribution Alternative Formula for the Variance and Standard Deviation of a Discrete Probability Distribution : Variance (σ 2 ) = Σ X 2 • P(X) – μ 2 Standard Deviation ( σ) = where: X = value of the random variable P(X) = probability of the random variable X μ = mean of the probability distribution  

21. Computing the Mean,Variance and standard deviation of a Discrete Probability Distribution X P(X) X • P(X) X - μ (X – μ ) 2 (X – μ ) 2 • P(X) 12 0.27 13 0.25 14 0.18 15 0.15 16 0.15

22. Computing the Mean,Variance and standard deviation of a Discrete Probability Distribution X P(X) X • P(X) X - μ (X – μ ) 2 (X – μ ) 2 • P(X) 12 0.27 3.24 -1.66 2.76 0.74 13 0.25 3.25 -0.66 0.44 0.11 14 0.18 2.52 0.34 0.12 0.02 15 0.15 2.25 1.34 1.80 0.27 16 0.15 2.40 2.34 5.48 0.82 Σ X • P(X) = 13.66 Σ (X – μ ) 2 • P(X) = 1.96 μ = 13.66 Variance = 1.96 Standard Deviation = 1.40