STATISTICS AND PROBABILITY GRADE 11 2024

pROBABILITY

1. PROBABILITY STATISTICS

2. FLASHBACK! What are the two types of random variable? DISCRETE VARIABLE CONTINUOUS VARIABLE

3. DISCRETE VARIABLE CONTINUOUS VARIABLE are random v ariables whose values are obtained by counting . These are variables that can take on a finite number of distinct values. are random variables that take an interminably uncountable number of potential values, regularly measurable amounts.

4. Source: https://www.facebook.com/CapizTourismAndCulturalAffairs The Suludnon , also known as the Panay-Bukidnon , Pan- ayanon , or Tumandok , are a culturally indigenous Visayan group of people who reside in the Capiz- Lambunao mountainous area and the Antique-Iloilo mountain area of Panay in the Visayan islands of the Philippines. Where their textiles truly stands out is through their embroidery, known as panubok . (https://narrastudio.com/blogs/journal/the-textile-and-epic-traditions-of-the-panay-Bukidnon)

5. LET'S PLAY Do you know me!

6. + Identify the mathematical notation Addition

7. Identify the mathematical notation ÷ Division

8. Identify the mathematical notation Square root

9. Identify the mathematical notation Vinculum

10. Identify the mathematical notation Summation (Sigma)

11. Identify the mathematical notation Population mean (Mu)

12. Identify the mathematical notation Sample mean (x-bar)

13. LESSON OBJECTIVE 1. Calculate the mean of a discrete random variable

14. Arithmetic Mean is the average of the numbers To calculate it: • add up all the numbers, • then divide by how many numbers there are.

15. Who’s the fastest solver? Ex. Given the values of the variables X and Y, evaluate the following summations.   X 1 = 5, X 2 = 4, X 3 = 3   Y 1 = 3, Y 2 = 1 Y 3 = 4   (1) Σ 𝑋 = (2) Σ( 𝑌 ) = (3) Σ ( 𝑋 + 𝑌) = (4) Σ( 𝑋𝑌) =

16. The symbol Σ (sigma) or summation symbol denotes a sum of a multiple terms or numbers. Ex. Given the values of the variables X and Y, evaluate the following summations.   X 1 = 5, X 2 = 4, X 3 = 3   Y 1 = 3, Y 2 = 1 Y 3 = 4   (1) Σ 𝑋 = 5 + 4 + 3 = 12 (2) Σ( 𝑌 ) = 3 + 1 + 4 = 8 (3) Σ ( 𝑋 + 𝑌) = (5+4) + (4+1) + (3+4) = 9+5+7 = 21 (4) Σ( 𝑋𝑌) = (5 *3)+(4*1) + (3*4) = 15 + 4 + 12 = 31

17. Who’s the faster solver? X 1 = 2, X 2 = 5, X 3 = 0   Y 1 = 3, Y 2 = 4 Y 3 = 1 1) Σ𝑋 2) ΣY 3) Σ (2+𝑋)

18. Mean of a discrete random VARIABLES The Mean (µ) of a discrete random variable is the central value or average of its corresponding probability mass function. It is also called as the Expected Value. It is computed using the formula: µ = Σ𝑋𝑃(𝑥) Where x is the outcome and p(x) is the probability of the outcome.

19. STEPS IN GETTING THE MEAN OF THE PROBABILITY DISTRIBUTION Construct the probability distribution for random variable. Multiply the value of random variable by the corresponding probability. Add all the results obtained in Step 2.

20. Reminder: Before solving for mean random variable make sure that the distribution given is a valid probability distribution.   What is a valid probability distribution? 1. Probability of each random variable should not be less than 0 and not more than 1 . 2. The sum of all the probability should always equal to 1.

21. x P(x) 10 0.12 11 0.14 12 0.10 13 0.20 14 0.15 15 0.16 16 0.08 17 0.05 Table 1 Let X be the random variable that represents the number of students who eat breakfast before going to school. Is table 1 a valid probability distribution? Why?

22. Table 2 Let X be the random variable that represents the number of tries before you win in 8-ball pool. Is table 2 a valid probability distribution? Why? X P(x) 1 -0.04 2 0.23 3 0.14 4 0.20 5 0.12 Σ𝑃(𝑥) = 0.65

23. Let’s solve for the Mean of the Probability Distribution Step 1: Construct the probability distribution for random variable. x P(x) 1 1/6 2 1/6 3 1/6 4 1/6 5 1/6 6 1/6 Consider rolling a die. What is the average number of pips (x) that would appear?

24. Step 2: Multiply the value of random variable by the corresponding probability. x P(x) x * P(x) 1 1/6 1 (1/6) = 1/6 2 1/6 2 (1/6) = 2/6 3 1/6 3 (1/6) = 3/6 4 1/6 4 (1/6) = 4/6 5 1/6 5 (1/6) = 6/6 6 1/6 6 (1/6) = 6/6

25. Step 3: Add all the results obtained in Step 2. x P(x) x * P(x) 1 1/6 1 (1/6) = 1/6 2 1/6 2 (1/6) = 2/6 3 1/6 3 (1/6) = 3/6 4 1/6 4 (1/6) = 4/6 5 1/6 5 (1/6) = 6/6 6 1/6 6 (1/6) = 6/6 21/6 or 3.5 µ = Σ 𝑋𝑃(𝑥) = 3.5

26. The teacher surveys TVL students if they eat breakfast before going to school. The number of student who eat breakfast are 10, 11, 12, 13, 14, 15, 16, or 17 with the probability of 0.12, 0.14, 0.10, 0.20, 0.15, 0.16, 0.08 and 0.05, respectively. Example 2

27. The teacher surveys TVL students if they eat breakfast before going to school. The number of student who eat breakfast are 10, 11, 12, 13, 14, 15, 16, or 17 with the probability of 0.12, 0.14, 0.10, 0.20, 0.15, 0.16, 0.08 and 0.05, respectively. Example 2 Step 1: Construct the probability distribution for random variable. X P(x) 10 0.12 11 0.14 12 0.10 13 0.20 14 0.15 15 0.16 16 0.08 17 0.05

28. Step 2 : Multiply the value of random variable by the corresponding probability. X P(x) X * P(x) 10 0.12 11 0.14 12 0.10 13 0.20 14 0.15 15 0.16 16 0.08 17 0.05 Table 1 Table 2 Table 3 Table 4

29. Step 2 : Multiply the value of random variable by the corresponding probability. X P(x) X * P(x) 10 0.12 1.20 11 0.14 1.54 12 0.10 1.20 13 0.20 2.60 14 0.15 2.10 15 0.16 2.40 16 0.08 1.28 17 0.05 0.85

30. Step 3 : Add all the results obtained in Step 2. X P(x) X * P(x) 10 0.12 1.20 11 0.14 1.54 12 0.10 1.20 13 0.20 2.60 14 0.15 2.10 15 0.16 2.40 16 0.08 1.28 17 0.05 0.85   µ = 13.17

31. What does the mean of a probability distribution tell us?   The significance of the mean of a probability distribution is that it gives the expected value of a discrete random variable.

32. Group Activity

33. Group 1: Manong Dodoy is a fisherman in living in Punta Cogon. He went to sea everyday to catch fish. The probability that he catches 10, 11, 12, 13, or 14 kilos of fish in a day are 0.15, 0.10, 0.20, 0.25 and 0.30, respectively. Find the average kilo of fish that Manong Dodoy catches everyday.

34. x P(x) X * P(x) 10 0.15   11 0.10 12 0.20 1.20 13 0.25   14 0.30 2.10     Complete the table and solve for mean random variable

35. Group 2: Manong Dodoy is a fisherman in living in Punta Cogon. He went to sea everyday to catch fish. The probability that he catches 10, 11, 12, 13, 14, 15, 16, or 17 kilos of fish in a day are 0.05, 0.17, 0.12, 0.15, 0.14, 0.10, 0.07 and 0.20, respectively. Find the average kilo of fish that Manong Dodoy catches everyday.

36. Instruction: Construct a probability distribution table and calculate the mean random variable. Answer the following question: 1. Is it a valid random distribution? Why? 2. What is the mean random variable of the given problem? 3. How do you interpret the mean of a probability distribution?

37. Group 3 Auntie Elena sells buko pie at Pueblo De Panay Terminal from Monday to Saturday. The probability that she sells 10, 12, 14, 16, 18 or 20 boxes of buko pie in a day, the sells are 0.15, 0.10, 0.20, 0.25 and 0.30, respectively. Find the average boxes of buko pie that Auntie Elena sells everyday.

38. X P(x) X * P(x) 10 0.04   12 0.1   14 0.2   16 0.2   18 0.3   20 0.16       Complete the table Answer the following question: 1. Is it a valid random distribution? Why? 2. What is the mean random variable of the given problem?

39. Group 4 Auntie Elena sells buko pie at Pueblo De Panay Terminal from Monday to Saturday. The probability that she sells 10, 12, 14, 16, 18 or 20 boxes of buko pie in a day, the sells are 0.15, 0.16, 0.08, 0.12, 0.22 and 0.27, respectively. Find the average boxes of buko pie that Auntie Elena sells everyday.

40. Group 4 Instruction: Construct a probability distribution table and calculate the mean random variable. Answer the following question: 1. Is it a valid random distribution? Why? 2. What is the mean of the discrete random variable of the given problem? 3. How do you interpret the mean of a probability distribution?

41. Finding Practical Applications The probability that you get a grade of 90 or higher in 5, 6, 7, or 8 subjects are 0.55, 0.20, 0.20 and 0.05, respectively. What is the mean of the probability distribution? x P(x) X * P(x) 5 0.55   6 0.20   7 0.20   8 0.05      

42. Generalization and Abstraction What is the mean of a discrete random variable? What are the steps in getting the mean of the probability distribution?

43. Assessment Direction: Choose the letter of the correct answer. Using the table 1 below, answer the following questions. X P(x) X * P(x) 3 0.12 0.36 5 0.11 (1) 7 0.23 (2) 10 0.24 2.4 12 0.3 (3)   (4) Table 1

44. Assessment Using the table 1 below, answer the following questions. X P(x) X * P(x) 3 0.12 0.36 5 0.11 (1) 7 0.23 (2) 10 0.24 2.4 12 0.3 (3)   (4) Tabl e 1 1 . What is the value of the missing data in (1)? A. 0.55 B. 0.11 C. 0.50 D. None of the choices

45. Assessment Using the table 1 below, answer the following questions. X P(x) X * P(x) 3 0.12 0.36 5 0.11 (1) 7 0.23 (2) 10 0.24 2.4 12 0.3 (3)   (4) Tabl e 1 2 . What is the value of the missing data in (2)? A. 1.50 B. 1.61 C. 1.65 D. None of the choices

46. Assessment Using the table 1 below, answer the following questions. X P(x) X * P(x) 3 0.12 0.36 5 0.11 (1) 7 0.23 (2) 10 0.24 2.4 12 0.3 (3)   (4) Tabl e 1 3 . What is the value of the missing data in (3)? A. 3.60 B. 2.60 C. 3.65 D. None of the choices

47. Assessment Using the table 1 below, answer the following questions. X P(x) X * P(x) 3 0.12 0.36 5 0.11 (1) 7 0.23 (2) 10 0.24 2.4 12 0.3 (3)   (4) Tabl e 1 4 . What is the value of the missing data in (4)? A. 9.50 B. 8.51 C. 8.52 D. None of the choices

48. Assessment 5. Which of the following is NOT a step in solving for the mean of a discrete random variable. A. Construct the probability distribution for random variable. B. Multiply the value of random variable by the corresponding probability. C. Divide the sum of the probability by the number of events. D. Add all the results obtained in Step 2.

49. Assignment 1. What is variance? 2. What the formula for computing the variance of probability distribution?

50. Good job!