Questions: Determine the required value of the missing probability to make the distribution a discrete probability distribution. x P(x) 3 0.15 4 ? 5 0.48 6 0.19 P(4)= (Type an integer or a decimal.)

Determine the required value of the missing probability to make the distribution a discrete probability distribution.

x P(x)
3 0.15
4 ?
5 0.48
6 0.19

P(4)=

(Type an integer or a decimal.)

Transcript text: Determine the required value of the missing probability to make the distribution a discrete probability distribution. \begin{tabular}{cc} \hline $\mathbf{x}$ & $\mathbf{P}(\mathbf{x})$ \\ \hline 3 & 0.15 \\ \hline 4 & $?$ \\ \hline 5 & 0.48 \\ \hline 6 & 0.19 \\ \hline \end{tabular} \[ P(4)= \] $\square$ (Type an integer or a decimal.)

Solution

Solution Steps

Step 1: Identify Known Probabilities

We are given the following probabilities for the discrete random variable $ x $:

\[ \begin{array}{cc} \hline \mathbf{x} & \mathbf{P}(\mathbf{x}) \\ \hline 3 & 0.15 \\ \hline 4 & ? \\ \hline 5 & 0.48 \\ \hline 6 & 0.19 \\ \hline \end{array} \]

Step 2: Calculate the Sum of Known Probabilities

To find the missing probability $ P(4) $, we first calculate the sum of the known probabilities:

\[ P(3) + P(5) + P(6) = 0.15 + 0.48 + 0.19 = 0.82 \]

Step 3: Determine the Missing Probability

Since the total probability of a discrete probability distribution must equal 1, we can find $ P(4) $ as follows:

\[ P(4) = 1 - (P(3) + P(5) + P(6)) = 1 - 0.82 = 0.18 \]

Step 4: Verify the Total Probability

Now, we verify that the total probability sums to 1:

\[ P(3) + P(4) + P(5) + P(6) = 0.15 + 0.18 + 0.48 + 0.19 = 1.00 \]