Mean and Variance of Discrete Random Variable.pptx

Mean and Variance of Discrete Random Variable

1. Mean and Variance of Discrete Random Variable

2. Illustrate the mean and variance of a discrete random variable; Calculate the mean and the variance of a discrete random variable; Interpret the mean and the variance of a discrete random variable; and Solve problems involving mean and variance of probability distributions. OBJECTIVES

3. Mean of the Discrete Random Variable Covid-19 is continuously spreading around the world, that is why reports regarding average infected people per country is being updated every day. For this kind of report, experts used Statistics and Probability to show reliable analysis in their data. In this lesson, you will learn how to compute the average or mean of a discrete probability distribution as well as the variance and standard deviation of a discrete random variable.

4. What’s New Consider the outcomes of a coin tossed as a random event. The probability of getting tail is ½ or 50%, and the probability of getting head is ½ or 50% also, but it is hard to predict the outcome that will occur. In this lesson, you will learn how to determine the likeli HOOD of the happening of an event. Mean of a Discrete Random Variable 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 = ∑ x p ( x ) Where x is the outcome and p(x) is the probability of the outcome.

5. What is It Examples: 1. Determine the mean or Expected Value of random variable below. Therefore, mean is 2 for the above random variable.

6. 2. Find the mean of the random variable Y representing the number of red color chocolates per 160-gram pack of colored chocolate packages that has the following probability distribution. Solution µ = ∑ ⟮ ( )⟯ = ∑ ⟮4(0.10) + 5(0.37) + 6(0.33) + 7(0.20)⟯ = ∑ ⟮0.40 + 1.85 + 1.98 + 1.40⟯ = 5.63 So, the mean of the probability distribution is 5.63. This implies that the average number of red chocolates per 160-gram is 5.63. Y 4 5 6 7 P(y) 0.10 0.37 0.33 0.20

7. 3. The probabilities that a customer will buy 1, 2, 3, 4, or 5 items in a grocery store are 3 10 , 1 10 , 1 10 , 2 10 , and 3 10 , respectively. What is the average number of items that a customer will buy? To solve the above problem, we will follow 3 steps below. STEPS IN FINDING THE MEAN Step 1: Construct the probability distribution for the random variable X representing the number of items that the customer will buy. Step 2: Multiply the value of the random variable X by the corresponding probability. Step 3: Add the results obtained in Step 2. Results obtained is the mean of the probability distribution.

8. Solution:

9. Solution Continuation So, the mean of the probability distribution is 3.1. This implies that the average number of items that the customer will buy is 3.1.

10. Variance and Standard Deviation of the Discrete Random Variable The variance and standard deviation describe the amount of spread, dispersion, or variability of the items in a distribution. How can we describe the spread or dispersion in a probability distribution? In this lesson, you will learn how to compute the variance and standard deviation of a discrete probability distribution. Now, let us find out how can we find the variance and standard deviation of a discrete probability distribution.

11. What’s New Variance and Standard Deviation of a Random Variable The variance and standard deviation are two values that describe how scattered or spread out the scores are from the mean value of the random variable. The variance, denoted as σ 2 , is determined using the formula: σ 2 = ∑( x − µ) 2 p(x) The standard deviation σ is the square root of the variance, thus, σ = √ ∑( x − µ)² p (x ) σ 2 - variance σ – standard deviation µ - mean p(x) – probability of the outcome

12. What is It Let’s try! Let’s have examples: 1. 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 by following the given steps. Write your answer in your answer sheets. Number of Cars Sold X Probability P(x) 10% 1 20% 2 30% 3 20% 4 20%

13. In solving the problem, let’s follow the steps below. STEPS IN FINDING THE VARIANCE AND STANDARD DEVIATION 1. Find the mean of the probability distribution. 2. Subtract the mean from each value of the random variable X. 3. Square the result obtained in Step 2. 4. Multiply the results obtained in Step 3 by the corresponding probability. 5. Get the sum of the results obtained in Step 4. Results obtained is the value of the variance of probability distribution.

14. Now let’s solve the problem.

15. Continuation

16. To Solve for Standard Deviation: Get the square root of the variance σ 2 = ∑( x − µ) 2 p(x) = 1.56 σ = √1.56 = 1.25 So, the variance of the number of cars sold per day is 1.56 and the standard deviation is 1.25.

17. 2. When three coins are tossed once, 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. Solution: Follow the steps in finding variance and standard deviation of the probability distribution.

18. To solve for Standard Deviation σ 2 = ∑( x − µ) 2 p(x) = 0.74 σ = √0.74 = 0.86 The mean in tossing 3 coins with probability of Head will show up is 0.86 and the variance is 0.74, then the standard deviation is 0.86.

19. What’s More Determine the mean or expected value of each Random Variable. Write your answer in your answer sheets. 1. 2. 3. s 3 4 12 20 P(s) 0.1 0.5 0.2 0.2 t 5 10 20 P(t) 50% 12% 38% w 1/12 1/6 1/3 1/2 P(w) 1/2 1/10 1/5 1/5

20. 4. Find the mean of the probability distribution of the random variable X, which can take only the values 1, 2, and 3, given that P(1) = 10 33 , P(2) = 1 3 , and P(3) = 12 33 . 5. The probabilities of a machine manufacturing 0, 1, 2, 3, 4, and 5 defective parts in one day are 0.75, 0.17, 0.04, 0.025, 0.01, and 0.005 respectively. Find the mean of the probability distribution. B. Determine the Variance and Standard Deviation of each random variable. Write your answer in your answer sheets. 1. x 1 2 3 4 5 P(x) 1 5 1 5 1 5 1 5 1 5

21. 2. 3. z 2 4 6 8 P(z) 0.6 0.1 0.2 0.1 m 1 3 5 7 P(m) 40% 25% 15% 20% 4. The random variable X, representing the number of nuts in a chocolate bar has the following probability distribution. Compute the variance and standard deviation.

22. 5. The number of items sold per day in a sari-sari store, with its corresponding probabilities, is shown in the table below. Find the variance and standard deviation of the probability distribution. . Number of Items Sold X Probability P(x) 19 0.20 20 0.20 21 0.30 22 0.20 23 0.10

23. Answer the following questions in your own understanding. 1. How to compute the mean of a discrete random variable? State the 3 steps. Write your answer in your answer sheets. 2. How to find the variance and standard deviation of a discrete random variable? Write your answer in your answer sheets. What I Have Learned

24. What I Can Do Make a study about how many sheets of paper you consumed weekly in answering your Self Learning Modules. Record the quantity (total number of sheets) per subject, then construct a probability distribution. Compute the mean, variance, and the standard deviation of the probability distribution you made. Interpret the result, then find out how many weeks you will consume 50 sheets of pad paper.

25. Assessment Find the mean, variance , and standard deviation of the following probability distribution then interpret the computed values. 1. Variable z representing the number of male teachers per Elementary school. 2. The number of mobile phones sold per day at a retail store varies as shown in the given probability distribution below. Find the expected number of mobile phones that will be sold in one day. z 2 3 4 5 6 P(z) 40% 32% 11% 9% 8% x 30 33 38 40 50 P(x) 0.2 0.2 0.35 0.23 0.02

26. 3. Number of monthly absences of Juan Dela Cruz based on his previous records of absences as presented in the probability distribution below. Number of Absences (X) Percent P(x) 3 25% 4 30% 5 30% 6 15% Number of Computer Sold X Probability P(x) 0.1 2 0.2 5 0.3 7 0.2 9 0.2 4. The number of computers sold per day at a local computer store, along with its corresponding probabilities, is shown in the table below.

27. 5. The number of inquiries received per day by the office of Admission in SHS X last enrolment is shown below. Number of Inquiries X Probability P(x) 22 0.08 25 0.19 26 0.36 28 0.25 29 0.07 30 0.05