Business statistics problem | Business & Finance homework help

P(x=1) = 1/15; P(x=2) = 2/15; P(x=3) = 3/15; P(x=4) = 4/15; P(x=5) = 5/15

b. In order for a probability distribution to be considered valid, it must satisfy three conditions: it should be non-negative for all values, the sum of all its probabilities should equal 1 and each value must have a unique probability (no two values can have the same probability).

In this case, the probability distribution of x fulfills these conditions. All probabilities are greater than or equal to 0 as they are fractions with numerators no larger than 5 and denominators set at 15. Furthermore, when we add up all five of these fractions (1/15 + 2/15 + 3/15 + 4/15 + 5 / 15), we get an answer of 15 / 15 which simplifies to 1 – meaning that the sum of all probabilities in this distribution does indeed equal 1.

Finally, each value has a unique probability – none of them share any similarities or ratios between them which would indicate that they could be combined into one collective fraction. This means that our discrete probability distribution satisfies the properties necessary for it to be deemed valid and accepted by mathematicians as such.