Problem:

The probabilities that an online teacher communicates on 3, 4, 5, 6, 7, or 8 students in any day are 0.15, 0.10, 0.20, 0.30, 0.20, and 0.05, respectively. Find the Mean, Variance, and Standard Deviation of students that an online teacher communicates on a day. Fill in the incomplete information Steps Solution 1. Construct the probability Number of Students distribution for Probability (X) (X the random 3 0.15 variable X 0.10 representing the 5 0.20 number of 6 0.30 students that an 7 online teacher 0.20 0.05 communicates on a day 2. Multiply the Number of value of the Students (X) Probability (X) X . P(X) random variable 3 0. 15 X by the 4 0.10 corresponding 5 0.20 probability. 6 0.30 7 0.20 0.05 3. Add the results Number of obtained in Step Students (X) Probability (X) X . P(X) 1 . 3 0. 15 4 0.10 5 0.20 6 0.30 7 0.208 0.05 H = EX - P(X) The mean of the probability distribution is The result says that 4. Subtract the Number mean from each of Probability X . P(X) X - H value of the Students (X) random variable X. 0.15 0.10 0.20 0.30 0.20 0.05 5. Square the obtained result Number of Probability in Step 4. Students X . PPO) X - W (X - u]2 (X) (X 3 0.15 0.10 0.20 0.30 0.20 0.05 6. Multiply the results obtained Number Probability X . in Step 5 by the X - H Students PIX (X - 4)2 corresponding probability 0.15 0.10 0.20 0.30 0.20 0.05 7. Get the sum of the results Number Probability X - of X - H X - MY PXO) obtained in Step6. The result is Students the value of the 3 0.15 variance. 0.10 0.20 0.30 0.20 0.05 6'= [(X - M) . P(X) The variance of the probability distribution is The result says that 0 = VE(X -)' . P(X) 8. Get the square root of the variance in Step 7 to get the The standard deviation of the probability standard distribution is deviation. The result says that