Fill in the missing value so that the following table represents a probability distribution.
$\begin{array} { c | c c c c }
x & - 4 & - 3 & - 2 & - 1 \
\hline P ( x ) & 0.19 & 0.28 & ? & 0.38
\end{array}$
A) 0.25
B) 0.15
C) 0.09
D) 0

Question

Fill in the missing value so that the following table represents a probability distribution. 
$\begin{array} { c | c c c c } 
x & - 4 & - 3 & - 2 & - 1 \
\hline P ( x ) & 0.19 & 0.28 & ? & 0.38
\end{array}$
A) 0.25
B) 0.15
C) 0.09
D) 0

Quizplus · Accepted Answer

The sum of the probabilities in a probability distribution must equal 1. Adding the given probabilities, 0.19 + 0.28 + 0.38 = 0.85. Therefore, the missing value must be 1 - 0.85 = 0.15 to make the total probability equal to 1.

Fill in the Missing Value So That the Following Table x−4−3−2−1P(x)0.190.28?0.38\begin{array} { c | c c c c } x & - 4 & - 3 & - 2 & - 1 \\\hline P ( x ) & 0.19 & 0.28 & ? & 0.38\end{array}xP(x)​−40.19​−30.28​−2?​−10.38​​

Fill in the Missing Value So That the Following Table $\begin{array} { c | c c c c } x & - 4 & - 3 & - 2 & - 1 \\\hline P ( x ) & 0.19 & 0.28 & ? & 0.38\end{array}$