The discrete random variable X is the number of passengers waiting at a bus stop. The table below shows the probability distribution for X. What is the expected value E(X) for this distribution? $$\begin{array} { l c c c c r }
X & 0 & 1 & 2 & 3 & \text { Total } \
P ( X ) & 20 & 40 & 30 & 10 & 1.00
\end{array}$$
A)1.1
B)1.3
C)1.7
D)1.9

Question

The discrete random variable X is the number of passengers waiting at a bus stop. The table below shows the probability distribution for X. What is the expected value E(X) for this distribution? $$\begin{array} { l c c c c r } 
X & 0 & 1 & 2 & 3 & \text { Total } \
P ( X ) & 20 & 40 & 30 & 10 & 1.00
\end{array}$$
A)1.1
B)1.3
C)1.7
D)1.9

Quizplus · Accepted Answer

The expected value or the mean of a discrete probability distribution can be found by multiplying each possible value of X by its probability and then adding up all the results. Mathematically,

E(X) = Σ[X * P(X)]

where Σ means sum, X takes each possible value of the random variable, and P(X) is the probability of X taking that value.

Using the given table:

E(X) = 0*0.2 + 1*0.4 + 2*0.3 + 3*0.1
     = 0 + 0.4 + 0.6 + 0.3
     = 1.3

Therefore, the expected value of X for this distribution is 1.3. The best choice is B.

The Discrete Random Variable X Is the Number of Passengers X0123 Total P(X)204030101.00\begin{array} { l c c c c r } X & 0 & 1 & 2 & 3 & \text { Total } \\P ( X ) & 20 & 40 & 30 & 10 & 1.00\end{array}XP(X)​020​140​230​310​ Total 1.00​

The Discrete Random Variable X Is the Number of Passengers $\begin{array} { l c c c c r } X & 0 & 1 & 2 & 3 & \text { Total } \\P ( X ) & 20 & 40 & 30 & 10 & 1.00\end{array}$