Let's address each part of the question step-by-step.
### (a) Find the value of [tex]\( P(18) \)[/tex]
Given the probabilities [tex]\( P(14) = 0.12 \)[/tex], [tex]\( P(15) = 0.24 \)[/tex], [tex]\( P(16) = 0.15 \)[/tex], and [tex]\( P(17) = 0.24 \)[/tex]. The sum of all probabilities must equal 1 since [tex]\( X \)[/tex] is a random variable.
To find [tex]\( P(18) \)[/tex]:
[tex]\[
P(18) = 1 - \left( P(14) + P(15) + P(16) + P(17) \right)
\][/tex]
[tex]\[
P(18) = 1 - (0.12 + 0.24 + 0.15 + 0.24) = 1 - 0.75 = 0.25
\][/tex]
Thus, [tex]\( P(18) = 0.25 \)[/tex].
### (b) Complete the following table
Let's fill in the values for [tex]\( x \cdot P(x) \)[/tex] and [tex]\( x^2 \cdot P(x) \)[/tex] in the table:
[tex]\[
\begin{array}{|c|c|c|c|}
\hline
x & P(x) & x \cdot P(x) & x^2 \cdot P(x) \\
\hline
14 & 0.12 & 1.68 & 23.52 \\
\hline
15 & 0.24 & 3.6 & 54.0 \\
\hline
16 & 0.15 & 2.4 & 38.4 \\
\hline
17 & 0.24 & 4.08 & 69.36 \\
\hline
18 & 0.25 & 4.5 & 81.0 \\
\hline
\text{Total} & 1 & 16.26 & 266.28 \\
\hline
\end{array}
\][/tex]
### (c) Mean [tex]\( \mu \)[/tex]
The mean [tex]\( \mu \)[/tex] of a discrete random variable [tex]\( X \)[/tex] is calculated using the formula:
[tex]\[
\mu = E(X) = \sum (x \cdot P(x))
\][/tex]
From the table, sum of [tex]\( x \cdot P(x) \)[/tex] is:
[tex]\[
E(X) = 16.26
\][/tex]
Thus, the mean [tex]\( \mu \)[/tex] is 16.26.
### (d) [tex]\( E(X^2) \)[/tex]
The expectation [tex]\( E(X^2) \)[/tex] is calculated as:
[tex]\[
E(X^2) = \sum (x^2 \cdot P(x))
\][/tex]
From the table, sum of [tex]\( x^2 \cdot P(x) \)[/tex] is:
[tex]\[
E(X^2) = 266.28
\][/tex]
### (e) Variance [tex]\( \sigma^2 \)[/tex]
The variance [tex]\( \sigma^2 \)[/tex] of a random variable [tex]\( X \)[/tex] is given by:
[tex]\[
\sigma^2 = E(X^2) - (E(X))^2
\][/tex]
Plug in the values we have:
[tex]\[
\sigma^2 = 266.28 - (16.26)^2
\][/tex]
[tex]\[
\sigma^2 = 266.28 - 264.3876 = 1.8924
\][/tex]
Thus, the variance [tex]\( \sigma^2 \)[/tex] is [tex]\( 1.8924 \)[/tex].
