Soemthing is not quite clear about frequency distribution
Hello, this explains a lot. But one more niggling thing. Where you calculated the expected E(A) value. How did the sum of probabilities in the sample space equal 7? 1:15 - 1:43. I tried assigning a unique probability for each value, and it totalled 1. If this was not what you did there, please help.
Hey Emmanuel,
Thank you for you question!
Let us examine the table that displays all 36 possible outcomes for the sum of two dice. The number 2 occurs only once in this table, so the probability of the sum being 2 is 1/36. The number 3 occurs 2 times in the table, so the probability of getting a sum of 3 is 2/36. If you sum the probabilities of obtaining all numbers from 2 to 12, you will definitely get a 1 as you exhaust all possibilities.
Now, we want to find the expected value E(A). This we do by taking the sum of all products between the outcome (say, 2) and its probability (P(2) = 1/36, as calculated above). This results in:
E(A) = P(2)*2 + P(3)*3 + ... + P(12)*12 = (1/36)*2 + (2/36)*3 ... + (1/36)*12 = 7.
Hope this clears things up!
Kind regards,
365 Hristina
Hello Hristina. This clears up a lot. Thanks.
