Consider the trial of rolling a single die, with outcomes of 1, 2, 3, 4, 5, and 6. a. Construct a table representing the probability distribution. b. Find the mean for the probability distribution.

Respuesta :

a) see table below

b) mean probability of the distribution is 3.5

Explanation:

a) For a fair die, we have an equal chance of getting each of the numbers

Since the count of the numbers = 6

Probability of one of the numbers = 1/6

constructing a table representing the probability:

let x = 1, 2, 3, 4, 5, 6

P(x) = probability of each of the numbers occurring

b) To get the mean of the distribution, we will multiply x by P(x) for each of the numbers. Then we will sum together.

[tex]\begin{gathered} \operatorname{mean}\text{ = }\sum ^{}_{}x(Px) \\ \operatorname{mean}\text{ = }\frac{1}{6}(1)\text{ + }\frac{1}{6}(2)\text{ + }\frac{1}{6}(3)\text{+ }\frac{1}{6}(4)\text{+ }\frac{1}{6}(5)\text{+ }\frac{1}{6}(6) \\ \operatorname{mean}\text{ = }\frac{1}{6}(1\text{ + 2 + 3 + 4 + 5 + 6)} \\ \operatorname{mean}\text{ = }\frac{1}{6}(21) \\ \operatorname{mean}\text{ = 3.5} \end{gathered}[/tex]

mean probability of the distribution is 3.5

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