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The following table shows the probability distribution for a discrete random variable.

[tex]\[
\begin{tabular}{|c|c|c|c|c|c|c|}
\hline
$X$ & 24 & 26 & 27 & 32 & 35 & 39 \\
\hline
$P(X)$ & 0.16 & 0.09 & 0.18 & 0.12 & 0.24 & 0.21 \\
\hline
\end{tabular}
\][/tex]

What is the mean of this discrete random variable? That is, what is [tex]$E(X)$[/tex], the expected value of [tex]$X$[/tex]?

A. 32.63
B. 30.5
C. 31.47
D. 29.5



Answer :

GinnyAnswer
To find the mean (expected value) [tex]\(E(X)\)[/tex] of the discrete random variable [tex]\(X\)[/tex] given its probability distribution, we use the formula for the expected value of a discrete random variable. The formula is:

[tex]\[ E(X) = \sum{(X_i \cdot P(X_i))} \][/tex]

where [tex]\( X_i \)[/tex] are the values of the random variable and [tex]\( P(X_i) \)[/tex] are the corresponding probabilities.

Given the values:
[tex]\[ X = [24, 26, 27, 32, 35, 39] \][/tex]
and their respective probabilities:
[tex]\[ P(X) = [0.16, 0.09, 0.18, 0.12, 0.24, 0.21] \][/tex]

We will compute the weighted sum of the values:

[tex]\[ E(X) = (24 \cdot 0.16) + (26 \cdot 0.09) + (27 \cdot 0.18) + (32 \cdot 0.12) + (35 \cdot 0.24) + (39 \cdot 0.21) \][/tex]

Calculating each term individually:

[tex]\[ 24 \cdot 0.16 = 3.84 \][/tex]
[tex]\[ 26 \cdot 0.09 = 2.34 \][/tex]
[tex]\[ 27 \cdot 0.18 = 4.86 \][/tex]
[tex]\[ 32 \cdot 0.12 = 3.84 \][/tex]
[tex]\[ 35 \cdot 0.24 = 8.40 \][/tex]
[tex]\[ 39 \cdot 0.21 = 8.19 \][/tex]

Now, sum these values:

[tex]\[ E(X) = 3.84 + 2.34 + 4.86 + 3.84 + 8.40 + 8.19 \][/tex]

[tex]\[ E(X) = 31.47 \][/tex]

Therefore, the mean (expected value) [tex]\( E(X) \)[/tex] of this discrete random variable is:

[tex]\[ \boxed{31.47} \][/tex]

The correct answer is [tex]\( C. 31.47 \)[/tex].

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