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Question (a): Show that \(k = \frac{1}{2}\).YOUR ANSWER\(\int_{1}^{3} (kx - k) dx = 1\)
\(k \left[ \frac{x^2}{2} - x \right]_{1}^{3} = 1\)
\(k \left[ (\frac{9}{2} - 3) - (\frac{1}{2} - 1) \right] = 1\)
\(k (4 - 0.5) = 1\)
\(2k = 1\)
\(k = \frac{1}{2}\)YOUR ANSWERSOLUTION\(k = \frac{1}{2}\)25Sid, you followed the correct steps to integrate the function over its range and set the integrated value equal to 1, as the total area under a probability density function must equal 1. Your calculation to find \(k = \frac{1}{2}\) is correct, well done!

Question (b): Find \(E(X)\).YOUR ANSWER\(E(X) = \int_{1}^{3} x (kx - k) dx\)
\(= \int_{1}^{3} x (\frac{x}{2} - \frac{1}{2}) dx\)
\(= \frac{1}{2} \left[ \frac{x^3}{3} - \frac{x^2}{2} \right]_{1}^{3}\)
\(= \frac{1}{2} \left[ (\frac{27}{3} - \frac{9}{2}) - (\frac{1}{3} - \frac{1}{2}) \right]\)
\(= \frac{1}{2} (9 - 4.\overline{5})\)
\(= \frac{1}{2} \cdot 4.5\)
\(= 2.25\)SOLUTION\(E(X) = 2.25\)25Nicely done, Sid! Your calculation of the expected value \(E(X)\) correctly applied the definition, with integration performed properly and arithmetic done correctly.

Question (c): Work out the cumulative distribution function \(F(x)\).YOUR ANSWER\(F(x) = \int_{1}^{x} (\frac{t}{2} - \frac{1}{2}) dt\)
\(= \left[ \frac{t^2}{4} - \frac{t}{2} \right]_{1}^{x}\)
\(= \left[ \frac{x^2}{4} - \frac{x}{2} \right] - \left[ \frac{1}{4} - \frac{1}{2} \right]\)
\(F(x) = \frac{x^2}{4} - \frac{x}{2} + \frac{1}{4} \;\; \text{ for } 1 \leq x \leq 3\)SOLUTION\(F(x) = \frac{x^2}{4} - \frac{x}{2} + \frac{1}{4}, \; 1 \leq x \leq 3\)25Great job, Sid! You correctly determined the cumulative distribution function \(F(x)\) by integrating the probability density function within the appropriate limits and simplifying the expression.

Question (d): Show that the median value lies between 2.4 and 2.5.YOUR ANSWER\( F(2.4) = \frac{2.4^2}{4} - \frac{2.4}{2} + \frac{1}{4} \)
\( = 1.44 - 0.95 \)
\( = 0.49 \)

\( F(2.5) = \frac{2.5^2}{4} - \frac{2.5}{2} + \frac{1}{4} \)
\( = 1.5625 - 1 \)
\(= 0.5625\)SOLUTION\(F(2.4) < 0.5 < F(2.5)\), confirming the median lies between 2.4 and 2.5.25Excellent reasoning, Sid! Your calculations for \(F(2.4)\) and \(F(2.5)\) are correct, and you effortlessly showed that the median must lie between these two values since \(F(2.4)\) is less than 0.5 and \(F(2.5)\) is more than 0.5.

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