Mean and Variance of Discrete Random Variables PDF

Mean and Variance of Discrete Random Variable OBJECTIVES ❖ 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 ❖ So...

Mean and Variance of Discrete Random Variable OBJECTIVES ❖ 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.  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.  What’s New 1 Consider the outcomes of a coin tossed as a random event. The probability of getting tail is 2 or 1 50%, and the probability of getting head is 2 or 50% also, but it is hard to predict the outcome that will occur. In this lesson, you will learn how to determine the likeliness 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.  What is It Examples: 1. Determine the mean or Expected Value of random variable below. Therefore, mean is 2 for the above random variable. 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. Y 4 5 6 7 P(y) 0.10 0.37 0.33 0.20 Solution µ = ∑ ⟮Y P(Y)⟯ = ∑ ⟮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. 3. The probabilities that a customer will buy 1, 2, 3, 4, or 5 items in a grocery 3 1 1 2 3 store are 10 10 10 10, and 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. Solution: 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.  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.  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 σ2, is determined using the formula: σ2 = ∑(X − µ)² P(X) The standard deviation σ is the square root of the variance, thus, σ = √ [∑(X − µ)² P(X)] σ2 – variance σ – standard deviation µ - mean P(X) – probability of the outcome  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) 0 10% 1 20% 2 30% 3 20% 4 20% 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. Now let’s solve the problem. Continuation To Solve for Standard Deviation: Get the square root of the variance σ2 = ∑(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. 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. To solve for Standard Deviation σ2 = ∑(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.