What is the expected value for the following probability distribution?

Mathematics

1 answer:

dolphi86 [110]1 year ago

5 0

To calculate the expected value of our dataset, first we need to find the missing probability. The sum of the probabilities of all outcomes must add up to 1

$\sum P(X)=1$

Therefore, 1 minus all the other probability values is going to be P(2).

$1-(0.04+0.05+0.2+0.1)=1-0.75=0.25$

The expected value is given by the sum of every value of the dataset multiplied by its respective probability

$E[X\rbrack=\sum P(x_i)x_i$

Using this formula in our problem, we have:

$\begin{gathered} E[X\rbrack=\sum_{i\mathop{=}1}^5P_i(x_i)x_i \\ =(0.4)\times(0)+(0.05)\times(1)+(0.25)\times(2)+(0.2)\times(3)+(0.1)\times(4) \\ =0+0.05+0.5+0.6+0.4 \\ =1.55 \end{gathered}$

The expected value of this dataset is 1.55.

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