ppt05. Computing the Variance of a Discrete Probability Distribution.pptx

Statistics and Probability

1. Computing the Variance of a Discrete Probability Distribution

2. Lesson Objectives At the end of this lesson, you should be able to: Illustrate and calculate the variance of a discrete random variable; Interpret the variance of a discrete random variable; and Solve problems involving variance of probability distributions.

3. Content Entry Card: Computing the variance Sample Problem 1: Number of Cars Sold Steps in Finding the Variance and Standard Deviation Formula for the Variance and Standard Deviation of a Discrete Probability Distribution Alternative Procedure in Finding the Variance and Standard Deviation of a Probability Distribution Sample Problem 2: Number of Heads Sample Problem 3: Number of Item Sold Seatwork Enrichment Assessment

4. Entry Card Compute the variance of this frequency distribution. Score Number of Students 5 3 8 5 10 4 12 5 15 3

5. The variance and standard deviation describe the amount of spread, dispersion, or variability of the items in distribution.

6. Problem 1: Number of Cars Sold The number of cars sold per day at a local car dealership, along with its corresponding probabilities, is shown in the succeeding table. Compute the variance and the standard deviation of the probability distribution. Number of Cars Sold X Probability P(X) 1 2 3 4 Number of Cars Sold X Probability P(X) 1 2 3 4

7. Steps in Finding the Variance and Standard Deviation Find the mean of the probability distribution. Subtract the mean from each value of the random variable X. Square the results obtain in Step 2. Multiply the results obtained in Step 3 by the corresponding probability. Get the sum of the result obtained in Step 4.

8. Step 1 Number of Cars Sold X Probability P(X) X ∙ P(X) 1 2 3 4 Number of Cars Sold X Probability P(X) X ∙ P(X) 1 2 3 4

9. Step 2 Number of Cars Sold X Probability P(X) X ∙ P(X) X - – 2.2 = -2.2 1 1 – 2.2 = -1.2 2 2 – 2.2 = -0.2 3 3 – 2.2 = 0.8 4 4 – 2.2 = 1.8 Number of Cars Sold X Probability P(X) X ∙ P(X) – 2.2 = -2.2 1 1 – 2.2 = -1.2 2 2 – 2.2 = -0.2 3 3 – 2.2 = 0.8 4 4 – 2.2 = 1.8

10. Step 3 Number of Cars Sold X Probability P(X) X ∙ P(X) X - -2.2 4.84 1 -1.2 1.44 2 -0.2 0.04 3 0.8 0.64 4 1.8 3.24 Number of Cars Sold X Probability P(X) X ∙ P(X) -2.2 4.84 1 -1.2 1.44 2 -0.2 0.04 3 0.8 0.64 4 1.8 3.24

11. Step 4 Number of Cars Sold X Probability P(X) X ∙ P(X) X - ∙ P(X) -2.2 4.84 0.484 1 0.2 -1.2 1.44 0.288 2 -0.2 0.04 0.012 3 0.8 0.64 0.128 4 1.8 3.24 0.648 Number of Cars Sold X Probability P(X) X ∙ P(X) -2.2 4.84 0.484 1 -1.2 1.44 0.288 2 -0.2 0.04 0.012 3 0.8 0.64 0.128 4 1.8 3.24 0.648

12. Step 5 Number of Cars Sold X Probability P(X) X ∙ P(X) X - ∙ P(X) -2.2 4.84 0.484 1 -1.2 1.44 0.288 2 -0.2 0.04 0.012 3 0.8 0.64 0.128 4 1.8 3.24 0.648 Number of Cars Sold X Probability P(X) X ∙ P(X) -2.2 4.84 0.484 1 -1.2 1.44 0.288 2 -0.2 0.04 0.012 3 0.8 0.64 0.128 4 1.8 3.24 0.648

13. Step 6 The variance of the probability distribution is 1.56. The standard deviation is  

14. Formula for the Variance and Standard Deviation of a Discrete Probability Distribution The variance of a discrete probability distribution is given by the formula: The standard deviation of a discrete probability distribution is given by the formula: where: X = value of the random variable P(X) = probability of the random variable X = mean of the probability distribution  

15. Alternative Procedure in Finding the Variance and 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

16. Step 1 Number of Cars Sold X Probability P(X) X ∙ P(X) 1 2 3 4 Number of Cars Sold X Probability P(X) X ∙ P(X) 1 2 3 4

17. Step 2 Number of Cars Sold X Probability P(X) X ∙ P(X) 1 2 3 4 Number of Cars Sold X Probability P(X) X ∙ P(X) 1 2 3 4

18. Step 3 X Probability P(X) X ∙ P(X) 1 2 3 4 X Probability P(X) X ∙ P(X) 1 2 3 4

19. Step 4 The variance is The standard deviation is  

20. Problem 2: Number of Heads When three coins are tossed, the probability distribution for the random variable X representing the number of heads that occur is given below. Compute the variance and standard deviation of the probability distribution. Number of Heads X Probability P(X) 1 2 3 Number of Heads X Probability P(X) 1 2 3

21. Step 1 Number of Heads X Probability P(X) X ∙ P(X) 1 2 3 Number of Heads X Probability P(X) X ∙ P(X) 1 2 3

22. Step 2 Number of Heads X Probability P(X) X ∙ P(X) 1 2 3 Number of Heads X Probability P(X) X ∙ P(X) 1 2 3

23. Step 3 Number of Heads X Probability P(X) X ∙ P(X) 1 2 3 Number of Heads X Probability P(X) X ∙ P(X) 1 2 3

24. Step 4 The variance is The standard deviation is  

25. Problem 3: Number of Items Sold The number of items sold per day at a retail store, with its corresponding probabilities, is shown in the table. Find the variance and standard deviation of the probability distribution. Number of Item Sold X Probability P(X) 19 0.20 20 0.20 21 0.30 22 0.20 23 0.10

26. Step 1 Number of Items Sold X Probability P(X) X ∙ P(X) 19 0.20 3.80 20 0.20 4.00 21 0.30 6.30 22 0.20 4.40 23 0.10 2.30 Number of Items Sold X Probability P(X) X ∙ P(X) 19 0.20 3.80 20 0.20 4.00 21 0.30 6.30 22 0.20 4.40 23 0.10 2.30

27. Step 1 Number of Items Sold X Probability P(X) X ∙ P(X) X- (X- )^2 (X- )^2∙ P(X) 19 0.20 3.80 19 – 20.80 = -1.8 (-1.8)^2 =3.24 3.24 (0.20)= 0.65 20 0.20 4.00 20-20.80= - 0.8 (-0.8)^2 = 0.64 0.64(0.20)= 0.13 21 0.30 6.30 21 – 20.80 = 0.2 (0.2)^2 = 0.04 0.04(0.30)= 0.01 22 0.20 4.40 22 – 20.80 = 1.2 (1.2)^2 = 1.44 1.44(0.20)= 0.29 23 0.10 2.30 23 – 20.80 = 2.2 (2.2)^2 = 4.84 4.84(0.10) = 0.48 (X- )^2∙ P(X) =1.56 Number of Items Sold X Probability P(X) X ∙ P(X) 19 0.20 3.80 19 – 20.80 = -1.8 (-1.8)^2 =3.24 3.24 (0.20)= 0.65 20 0.20 4.00 20-20.80= - 0.8 (-0.8)^2 = 0.64 0.64(0.20)= 0.13 21 0.30 6.30 21 – 20.80 = 0.2 (0.2)^2 = 0.04 0.04(0.30)= 0.01 22 0.20 4.40 22 – 20.80 = 1.2 (1.2)^2 = 1.44 1.44(0.20)= 0.29 23 0.10 2.30 23 – 20.80 = 2.2 (2.2)^2 = 4.84 4.84(0.10) = 0.48

28. Step 2 Number of Items Sold X Probability P(X) X ∙ P(X) 19 0.20 3.80 20 0.20 4.00 21 0.30 6.30 22 0.20 4.40 23 0.10 2.30 Number of Items Sold X Probability P(X) X ∙ P(X) 19 0.20 3.80 20 0.20 4.00 21 0.30 6.30 22 0.20 4.40 23 0.10 2.30

29. Step 3 Number of Items Sold X Probability P(X) X ∙ P(X) 19 0.20 3.80 72.20 20 0.20 4.00 80.00 21 0.30 6.30 132.30 22 0.20 4.40 96.80 23 0.10 2.30 52.90 Number of Items Sold X Probability P(X) X ∙ P(X) 19 0.20 3.80 72.20 20 0.20 4.00 80.00 21 0.30 6.30 132.30 22 0.20 4.40 96.80 23 0.10 2.30 52.90

30. Step 4 The variance is The standard deviation is  

31. QUIZ: Complete the table below and find the variance and standard deviation of the following probability distribution. X P(X) X ∙ P(X) ∙ P(X) 3 0.15 6 0.35 8 0.40 10 0.10 X P(X) X ∙ P(X) 3 0.15 6 0.35 8 0.40 10 0.10

32. X P(X) X ∙ P(X) ∙ P(X) 1 3 5 7 X P(X) X ∙ P(X) 1 3 5 7

33. Enrichment Problem Solving Find the variance and standard deviation of the probability distribution of the random variable X, which can take only the values 1, 2, and 3, given that P(1) = , P(2) = , and P(3) = .  

34. Violent Reactions? Clarifications?

35. Thank you for listening…