14:16 9 f 2.00 KB/S 93 Let Us Discover The mean of discrete random variable X is a weighted average of the possible values that the random variable can take. Unlike the sample mean of a group of observations, which gives each observation equal weight, the mean of random variable weights each outcome X, according to its possibility, P(X). It is also known as the expected value/outcome. The variance of discrete random variable X is a measure of how much the probability mass is spread out around the mean. The higher the variance, the farther away the result can be from its expected value. This implies that if the computed variance is high, the scores are very much scattered around the mean and if the computed variance is low, the data are closely concentrated around the mean. The square root of the variance is called the standard deviation. To solve for the mean and variance of a discrete random variable, the formulas to be used are: Mean Variance Standard Deviation H = E[X . P(X)] 62 = [[ X2 . P(X)] - 12 " = VE[ X2 . P(X)]- 12 where: mean of the discrete random variable X - possible outcome/random variable 02 variance of the discrete random variable P(X) - the probability of the outcome - standard deviation of the discrete random variable 1 GSC-CID-LRMS-ESSLM, v.r. 03.00, Effective June 14, 2021 Illustrative Example 1: Find the expected outcome or mean, variance, and standard deviation when a die is rolled many times. Let X be the possible outcome. Solution: Step 1. Construct a probability Step 2. Multiply the value of the random variable distribution (X) to its corresponding probability X P(X) X P(X) X . P(X) 1 = 0.167 = 0.167 2 3 1_ 6 = = 0.167 1 4 = 0.167 5 1_ 6 = 0.167 6 1 = 0.167 6 Step 3. Find the summation of the products of the values of the random variable (X) and the probabilities H = E[X . P(X)] [[X . P(X)] = =+-+-+-+-+. 6 6 21 U = 6 or 3. 5 Step 4. Squared each value of the random Step 5. Multiple the value of the random variable variable (X) to its corresponding probability P(X) X . P(X) X2 X P(X) X . P(X) X2 X2 . P(X) 1 2 4 2 3 9 3 9 16 16 16 E O14:179 f 0.37 KB/S So Ill 93 Illustrative Example 1: Find the expected outcome or mean, variance, and standard deviation when a die is rolled many times. Let X be the possible outcome. Solution: Step 1. Construct a probability Step 2. Multiply the value of the random variable distribution (X) to its corresponding probability P(X X P(X) X . P(X) = 0.167 2 = = 0.167 3 = 0.167 3 GIG alVale aIwaINaIN 1_ = 0.167 5 1 = 0.167 1 = 0.167 Step 3. Find the summation of the products of the values of the random variable (X) and the probabilities H = >[X . P(X)] E[X . P(X)] = 1 2 3 3 4 5 t- +-+- 6 6 6 6 21 H = - or 3. 5 Step 4. Squared each value of the random Step 5. Multiple the value of the random variable variable (X) to its corresponding probability X P(X X . P(X) X2 X P(X) X . P(X) X2 X2 . P(X) 2 9 9 16 16 5 25 5 25 6 36 36 Step 6. Apply the formula in finding the Step 7. Find the standard deviation by variance getting the square root of the 62 = [[ X2 . P(X)] - 12 variance 4 9 16 25 $6+6+6+6 -(3.50)2 = _[ X2 . P(X)] - 12 = 15.17 - 12.25 = V2.92 = 2. 917 or 2. 92 = 1. 709 or 1. 71 2 GSC-CID-LRMS-ESSLM, v.r. 03.00, Effective June 14, 2021 Step 8. Interpret the results When a die is rolled repeatedly the expected outcome is 3.5. This means that the distribution is balance at point 3.5 which is the mean or expected value. The variance of the distribution is 2.92 and the standard deviation is 1.71. The variance and standard deviation are both small numbers which indicates that the number of times in rolling a die are less disperse and that the average which is 3.5 do not differ or is very close from the number of times in rolling a die. E O14:18 N 19.0 KB/S 26 Ill 92 Let Us Do Activity 2: What's my Mean, Variance & Standard Deviation (MVSD)? Direction: Use the 8-step rule in solving the given problem. Step 1. Construct a probability distribution A Grade 11 class in Statistics and X P(X) Probability has 20 learners. The ages of these learners are as follows: One learner is 15 years old, four are 16 years old, nine are 17, three are 18, two are 19 and one is 20. Let X be the age of any learner measured in years only. Find the mean age of the learners and determine the variance and standard deviation. Interpret the results. 3 GSC-CID-LRMS-ESSLM, v.r. 03.00, Effective June 14, 2021 Step 2. Multiple the value of the random variable Step 3. Find the summation of the (X) to its corresponding probability products of the values of the X P(X) X . P(X) random variable (X) and the probabilities H = _[X . P(X)] Step 4. Squared each value of the random Step 5. Multiple the value of the random variable variable (X) to its corresponding probability P(X) X . P(X) X2 X P(X) X . P(X) X2 X2 . P(X) Step 6. Apply the formula in finding Step 7. Find the standard deviation by getting the variance the square root of the variance 62 = [[X2 . P(X)] - 12 E[ X2 . P(X)] - 12 Step 8. Interpret the results E O