022.03 IV. ASSESSMENT: I. The number of computers sold per day at a local computer store, along with itso corresponding probabilities, is shown in the table, Find the variance and standard deviation of the distribution. Number of Computers sold (x) Probability P(x) 0 20% 20% 2 35% 25% II.altiply the results obtained in Step 3 by the corresponding probability. Get the sum of the results obtained in Step 4. Results obtained is the value of the variance of probability distribution. STEPS SOLUTION 1. Find the mean of the probability Number of Probability X. P(x) distribution using the formula: Car Sold P(x) X H = EXP(x) 0.1 0 0.2 0.2 2 0.3 0.6 3 0.2 0.6 4 0.2 0.8 TOTAL 2.2 1 = EXP(x) = 2.2 2. Subtract the mean from each X P(x) X . P(x) X - H value of the random variable X. 0. 1 0 - 2.2 to - 2.2 I 0.2 0.2 1 - 2.2 - 1.2 2 0.3 0.6 2 - 2.2 2 - 0.2 3 0.2 0.6 3 - 2.2 = 0.8 4 0.2 0.8 4 - 2.2 21.8 3. Square the results obtained in X P(x) X . P(x) X - H (X - 1) 2 Step 2. O 0. 1 0 2.2 4.84 1 0.2 0.2 - 1.2 1.44 2 0.3 0.6 0.2 0.04 3 0.2 0.6 0.8 0.64 0.2 0.8 1 .8 3.24 4. Multiply the results obtained in X P(x) X.P(x) X-H (X - 1/2 (X - up . P(x) Step 3 by the corresponding 0 0.1 0 -2.2 4.84 0.484 probability. 1 0.2 0.2 -1.2 1.44 0.288 2 0.3 0.6 -0.2 0.04 0.012 3 0.2 0.6 0.8 0.64 0.128 4 0.2 0.8 1.8 3.24 0.648 5. Get the sum of the results X P(x) (X.P(x) X-u (X - 41) 2 (X - H)2 . P(x) obtained in Step 4. The result is 0.1 -2.2 4.84 0.484 the value of the variance. So, the 0.2 0.2 -1.2 1.44 0.288 formula is: 2 0.3 0.6 -0.2 0.04 0.012 3 0.2 0.6 0.8 0.64 0.128 4 0.2 0.8 1.8 3.24 0.648 02 : E(x - 1) p(x) TOTAL 1.56 oz= E(x - H) p(x) = 1.56 2022.03.22 20:51