![1.What is the mean?
𝝁 = [𝑿 ∗ 𝑷(𝑿)]
= 𝟎 + 𝟎. 𝟐 + 𝟎. 𝟔 + 𝟎. 𝟔 + 𝟎. 𝟒
𝝁 = 𝟏. 𝟖
SOLUTION:](https://image.slidesharecdn.com/statistics-lesson1-midterm-2-240513042704-3863d1c9/85/STATISTICS-LESSON-no-1-MIDTERM-2-pdf-14-320.jpg)
![1.What is the mean?
𝝁 = [𝑿 ∗ 𝑷(𝑿)]
= 𝟎 + 𝟎. 𝟎𝟓 + 𝟎. 𝟐𝟒 + 𝟎. 𝟗𝟎 + 𝟏. 𝟏𝟐 + 𝟏. 𝟏𝟎
𝝁 = 𝟑. 𝟒𝟏
SOLUTION:](https://image.slidesharecdn.com/statistics-lesson1-midterm-2-240513042704-3863d1c9/85/STATISTICS-LESSON-no-1-MIDTERM-2-pdf-20-320.jpg)
STATISTICS - LESSON no 1 - MIDTERM-2.pdf
- 1. LESSON 5: CALCULATING MEAN AND VARIANCE OF A DISCRETE RANDOM VARIABLE
- 2. THE MEAN a. The average of all possible outcomes. It is otherwise referred to as the “expected value” of a probability distribution. b. Means that if we repeat any given experiment infinite times, the theoretical mean would be the “expected value”. c. Any discrete probability distribution has a mean.
- 3. THE VARIANCE AND STANDARD DEVIATION The VARIANCE of a random variable displays the variability or the dispersions of the random variables. It shows the distance of a random variable from its mean.
- 4. THE VARIANCE AND STANDARD DEVIATION a. If the values of the variance and standard deviation are HIGH, that means that the individual outcomes of the experiment are far relative to each other. In other words, the values differ greatly. b. A large value of standard deviation (or variance) means that the distribution is spread out, with some possibility of observing values at some distance from the mean.
- 5. THE VARIANCE AND STANDARD DEVIATION c. If the variance and standard deviation are LOW, that means that the individual outcomes of the experiment are closely spaced with each other. In other words, the values are almost the same values or if they do differ, the difference is small. d. A small value of standard deviation (or variance) means that the dispersion of the random variable is narrowly concentrated around the mean.
- 6. Mean, variance, and standard deviation can be illustrated by looking pattern and analyzing given illustrations and diagrams.
- 7. During Town Fiesta, people used to go to Carnival that most folks call it “Perya”. Mang Ben used to play “Beto–beto” hoping that he would win. While he is thinking about what possible outcomes in every roll would be, he is always hoping that his bet is right. X 1 2 3 4 5 6 P(X) 𝟏 𝟔 𝟏 𝟔 𝟏 𝟔 𝟏 𝟔 𝟏 𝟔 𝟏 𝟔 EXAMPLE
- 8. 1.Is the probability of X lies between 0 and 1? 2.What is the sum of all probabilities of X? 3.Is there a negative probability? Is it possible to have a negative probability? 4.How will you illustrate the average or mean of the probabilities of discrete random variable? QUESTIONS:
- 9. 1.Is the probability of x lies between 0 and 1? Yes, the probability of X lies between 0 and 1. X 1 2 3 4 5 6 P(X) 𝟏 𝟔 𝟏 𝟔 𝟏 𝟔 𝟏 𝟔 𝟏 𝟔 𝟏 𝟔 SOLUTION:
- 10. 2. What is the sum of all probabilities of X? 𝑷 𝑿 = 𝑷 𝟏 + 𝑷 𝟐 + 𝑷 𝟑 + 𝑷 𝟒 + 𝑷 𝟓 + 𝑷 𝟔 = 𝟏 𝟔 + 𝟏 𝟔 + 𝟏 𝟔 + 𝟏 𝟔 + 𝟏 𝟔 + 𝟏 𝟔 = 𝟏+𝟏+𝟏+𝟏+𝟏+𝟏 𝟔 = 𝟔 𝟔 𝒐𝒓 𝟏 The sum of all probabilities of X is exactly 1. X 1 2 3 4 5 6 P(X) 𝟏 𝟔 𝟏 𝟔 𝟏 𝟔 𝟏 𝟔 𝟏 𝟔 𝟏 𝟔 SOLUTION:
- 11. 3. Is there a negative probability? Is it possible to have a negative probability? No negative probabilities because it is impossible to have it based on the characteristic of the probability of discrete random variables. SOLUTION:
- 12. In a seafood restaurant, the manager wants to know if their customers like their new raw large oysters. According to their sales representative, in the past 4 months, the number of oysters consumed by a customer, along with its corresponding probabilities, is shown in the succeeding table. Compute the mean, variance and standard deviation. Number of oysters consumed X Probabilit y P(X) 0 𝟐 𝟏𝟎 1 𝟐 𝟏𝟎 2 𝟑 𝟏𝟎 3 𝟐 𝟏𝟎 4 𝟏 EXAMPLE 1
- 13. (X) P(X) X ∗ P(X) 𝑋−𝜇. (𝑋 − 𝜇)² (𝑋 − 𝜇)² ∙ 𝑃(𝑋) 0 𝟎. 𝟐 𝟎 𝟎 − 𝟏. 𝟖 = −𝟏. 𝟖 −𝟏. 𝟖 𝟐 = 𝟑. 𝟐𝟒 𝟑. 𝟐𝟒 ∗ 𝟎. 𝟐 = 𝟎. 𝟔𝟖𝟒 1 𝟎. 𝟐 0.2 𝟏 − 𝟏. 𝟖 = −𝟎. 𝟖 −𝟎. 𝟖 𝟐 = 𝟎. 𝟔𝟒 𝟎. 𝟔𝟒 ∗ 𝟎. 𝟐 = 𝟎. 𝟏𝟐𝟖 2 𝟎. 𝟑 0.6 𝟐 − 𝟏. 𝟖 = 𝟎. 𝟐 𝟎. 𝟐 𝟐 = 𝟎. 𝟎𝟒 𝟎. 𝟎𝟒 ∗ 𝟎. 𝟑 = 𝟎. 𝟎𝟏𝟐 3 𝟎. 𝟐 0.6 𝟑 − 𝟏. 𝟖 = 𝟏. 𝟐 𝟏. 𝟐 𝟐 = 𝟏. 𝟒𝟒 𝟏. 𝟒𝟒 ∗ 𝟎. 𝟐 = 𝟎. 𝟐𝟖𝟖 4 𝟎. 𝟏 0.4 4−𝟏. 𝟖 = 𝟐. 𝟐 𝟐. 𝟐 𝟐 = 𝟒. 𝟖𝟒 𝟒. 𝟖𝟒 ∗ 𝟎. 𝟏 = 𝟎. 𝟒𝟖𝟒
- 14. 1.What is the mean? 𝝁 = [𝑿 ∗ 𝑷(𝑿)] = 𝟎 + 𝟎. 𝟐 + 𝟎. 𝟔 + 𝟎. 𝟔 + 𝟎. 𝟒 𝝁 = 𝟏. 𝟖 SOLUTION:
- 15. 2. What is the standard deviation? 𝜎2 = (𝑋 − 𝜇)²∙ 𝑃(𝑋) = 0.448 + 0.128 + 0.012 + 0.288 + 0.484 𝝈𝟐 𝟏. 𝟓𝟔 SOLUTION:
- 16. 3. What is the standard deviation? 𝜎 = σ (𝑋 − 𝜇)²∙ 𝑃(𝑋) 𝑂𝑅 𝜎 = 𝜎2 𝜎 = 1.56 𝝈 = 𝟏. 𝟐𝟓 SOLUTION:
- 17. Mr. Umali, a Mathematics teacher, regularly gives a formative assessment composed of 5 multiple-choice items. After the assessment, he used to check the probability distribution of the correct responses, and the data is presented below: TEST ITEM (X) Probability 𝑷(𝑿) 0 0.03 1 0.05 2 0.12 3 0.30 4 0.28 5 0.22 EXAMPLE 2
- 18. 1.What is the average or mean of the given probability distribution? 2.What are the values of the variance of the probability distribution? 3.What are the values of standard deviation of the probability distribution? QUESTIONS:
- 19. X P(X) X ∗ P(X) X - 𝛍 (X - 𝛍)² (X - 𝛍)² ∗ P(X) 0 0.03 0 ∗ 0.03 = 0 0 − 3.41 = −3.41 (−3.41)² = 11.63 11.63 ∗ 𝟎. 𝟎𝟑 = 𝟎. 𝟑𝟓 1 0.05 1 ∗ 0.05 = 0.05 1 − 3.41 = −2.41 (−2.41)² = 5.81 5. 𝟖𝟏 ∗ 𝟎. 𝟎𝟓 = 𝟎. 𝟐𝟗 2 0.12 2 ∗ 0.12 = 0.24 2 − 3.41 = −1.41 (−1.41)² = 1.99 1. 𝟗𝟗 ∗ 𝟎. 𝟏𝟐 = 𝟎. 𝟐𝟒 3 0.30 3 ∗ 0.30 = 0.90 3 − 3.41 = −0.41 (−0.41)² = 0.17 0. 𝟏𝟕 ∗ 𝟎. 𝟑𝟎 = 𝟎. 𝟎𝟓 4 0.28 4 ∗ 0.28 = 1.12 4 − 3.41 = 0.59 (0.59)² = 0.35 0.35 ∗ 𝟎. 𝟐𝟖 = 𝟎. 𝟏𝟎 5 0.22 5 ∗ 0.22 = 1.10 5 − 3.41 = 1.59 (1.59)² = 2.53 2.53 ∗ 𝟎. 𝟐𝟐 = 𝟎. 𝟓𝟔 𝛍 = 𝟑. 𝟒𝟏 𝝈𝟐 = 𝟏. 𝟓𝟗
- 20. 1.What is the mean? 𝝁 = [𝑿 ∗ 𝑷(𝑿)] = 𝟎 + 𝟎. 𝟎𝟓 + 𝟎. 𝟐𝟒 + 𝟎. 𝟗𝟎 + 𝟏. 𝟏𝟐 + 𝟏. 𝟏𝟎 𝝁 = 𝟑. 𝟒𝟏 SOLUTION:
- 21. 2. What is the standard deviation? 𝜎2 = (𝑋 − 𝜇)²∙ 𝑃(𝑋) = 𝟎. 𝟑𝟓 + 𝟎. 𝟐𝟗 + 𝟎. 𝟐𝟒 + 𝟎. 𝟎𝟓 + 𝟎. 𝟏𝟎 + 𝟓𝟔 𝝈𝟐 = 𝟏. 𝟓𝟗 SOLUTION:
- 22. 3. What is the standard deviation? 𝜎 = σ (𝑋 − 𝜇)²∙ 𝑃(𝑋) 𝑂𝑅 𝜎 = 𝜎2 𝜎 = 1.59 𝝈 = 𝟏. 𝟐𝟔 SOLUTION:
- 23. 1. The number of shoes sold per day at a retail store is shown in the table below. Illustrate the mean, variance, and standard deviation of this distribution. Write all the necessary formula and show the complete solution. Formula to be used: (a)Mean, (b)Variance,(c) Standard Deviation X 19 20 21 22 23 P(X) 0.4 0.2 0.2 0.1 0.1 ACTIVITY 1
- 24. 1. The Land Bank of the Philippines Manager claimed that each saving account customer has several credit cards. The following distribution showing the number of credits cards people own. a. Calculate the mean, variance, and standard deviation of a discrete random variable. X 0 1 2 3 4 P(X) 0.18 0.44 0.27 0.08 0.03 ASSIGNMENT
- 25. 2. Suppose an unfair die is rolled and let X be the random variable representing the number of dots that would appear with a probability distribution below. a. Calculate the mean, variance, and standard deviation of a discrete random variable. OUTCOME (X) 1 2 3 4 5 6 Probability 𝑃(X) 0.1 0.1 0.1 0.5 0.1 0.1
- 26. 1. The number of cellular phones sold per day at the E-Cell Retail Store with the corresponding probabilities is shown in the table below. Compute the mean, variance, and standard deviation and interpret the result. (𝑥) 15 18 19 20 22 Probability 𝑃(𝑥) 0.30 0.20 0.20 0.15 0.15