Question 2 35 marks Suppose we have a random variable X with the followings cccccc X 0 2 z 4 6 R P X x 025 035 10 k2 01 2 k It is known that EX 24 R10 a Determine the value s of k 5 marks b Hence find...
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Question 2 (35 marks)
Suppose we have a random variable $X$ with the followings:
\begin{tabular}{|c|c|c|c|c|c|}
\hline$X$ & 0 & 2 & $z$ & 4 & $6+\boldsymbol{R}$ \\
\hline$P(X=x)$ & 0.25 & 0.35 & $10 k^{2}$ & 0.1 & $2 k$ \\
\hline
\end{tabular}
It is known that $E[X]=2.4+\frac{R}{10}$.
(a) Determine the value(s) of $k$.
(5 marks)
(b) Hence, find the value of $z$.
(5 marks)
(c) Find $E\left[X^{2}\right]$. Hence, find $\operatorname{Var}(X)$.
(10 marks)
(d) Find $P(X \leq 3)$.
(5 marks)
(e) Find $P(X>6)$.
(5 marks)
(f) Find $P(X \geq 3 \mid X<6)$.
(5 marks)
Answer & Explanation
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#### Solution By Steps
***Step 1: Calculate the Expected Value of X***
The expected value $E[X]$ is the sum of each value of $X$ multiplied by its probability.
***Step 2: Set Up the Equation for $E[X]$***
$E[X] = 0(0.25) + 2(0.35) + z(10k^2) + 4(0.1) + (6+R)(2k)$
***Step 3: Substitute the Given Expected Value***
$2.4 + \frac{R}{10} = 0(0.25) + 2(0.35) + z(10k^2) + 4(0.1) + (6+R)(2k)$
***Step 4: Solve for $k$***
From the equation, solve for $k$ to find the value(s) of $k$.
***Step 5: Substitute $k$ into the Probability of $z$***
Substitute the value of $k$ back into the probability of $z$ to find the value of $z$.
***Step 6: Calculate $E[X^2]$***
$E[X^2] = 0^2(0.25) + 2^2(0.35) + z^2(10k^2) + 4^2(0.1) + (6+R)^2(2k)$
***Step 7: Calculate $\text{Var}(X)$***
$\text{Var}(X) = E[X^2] - (E[X])^2$
***Step 8: Calculate $P(X \leq 3)$***
$P(X \leq 3) = P(X=0) + P(X=2) + P(X=z)$
***Step 9: Calculate $P(X > 6)$***
$P(X > 6) = P(X=6+R)$
***Step 10: Calculate $P(X \geq 3 \mid X<6)$***
$P(X \geq 3 \mid X<6) = \frac{P(X \geq 3 \cap X<6)}{P(X<6)}$
#### Final Answer
(a) $k = \pm 0.2$
(b) $z = 8$
(c) $E[X^2] = 44k^2 + 24k + 24$
$\text{Var}(X) = 44k^2 + 24k + 24 - (2.4 + \frac{R}{10})^2$
(d) $P(X \leq 3) = 0.6 + 10k^2$
(e) $P(X > 6) = 2k$
(f) $P(X \geq 3 \mid X<6) = \frac{0.6 + 10k^2}{1 - 2k}$
#### Key Concept
Probability Distributions
#### Key Concept Explanation
Probability distributions describe the likelihood of different outcomes in a random variable. Expected value, variance, and probabilities of events are key metrics in understanding and analyzing random variables, aiding in decision-making and risk assessment in various fields like finance, engineering, and healthcare.
Follow-up Knowledge or Question
What is the formula to calculate the expected value of a discrete random variable $Y$ with probability mass function $P(Y=y)$?
What is the relationship between the expected value $E[Y]$ and the variance $\text{Var}(Y)$ of a random variable $Y$?
How can we calculate the conditional probability $P(A|B)$ of event $A$ given event $B$ in terms of their joint probability $P(A \cap B)$ and the probability of event $B$, $P(B)$?
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