1. **Problem statement:** The random variable $X$ has a continuous uniform distribution over the interval $[-k, 5k]$ where $k > 0$. We need to find the probability density function (pdf), sketch it, find the mean, cumulative distribution function (cdf), and then work with $Y = X^2$ to find $E(Y)$ and the probability $P(Y < 2k^2)$. 2. **(a) Probability density function (pdf):** For a uniform distribution over $[a,b]$, the pdf is given by: $$f_X(x) = \frac{1}{b - a} \quad \text{for } a \leq x \leq b$$ Here, $a = -k$ and $b = 5k$, so: $$f_X(x) = \frac{1}{5k - (-k)} = \frac{1}{6k} \quad \text{for } -k \leq x \leq 5k$$ Outside this interval, $f_X(x) = 0$. 3. **(b) Sketch of pdf:** The pdf is a constant function $\frac{1}{6k}$ on the interval $[-k, 5k]$ and zero elsewhere. It looks like a rectangle starting at $x=-k$ and ending at $x=5k$ with height $\frac{1}{6k}$. 4. **(c) Mean of $X$:** The mean of a uniform distribution on $[a,b]$ is: $$\mu = \frac{a + b}{2}$$ So: $$\mu = \frac{-k + 5k}{2} = \frac{4k}{2} = 2k$$ 5. **(d) Cumulative distribution function (cdf):** The cdf $F_X(x)$ is defined as: $$F_X(x) = P(X \leq x) = \int_{-\infty}^x f_X(t) dt$$ For $x < -k$, $F_X(x) = 0$ because $X$ cannot be less than $-k$. For $-k \leq x \leq 5k$: $$F_X(x) = \int_{-k}^x \frac{1}{6k} dt = \frac{1}{6k} (x + k)$$ For $x > 5k$, $F_X(x) = 1$ because the entire probability is accumulated. So: $$F_X(x) = \begin{cases} 0 & x < -k \\ \frac{x + k}{6k} & -k \leq x \leq 5k \\ 1 & x > 5k \end{cases}$$ 6. **(e) Find $E(Y)$ where $Y = X^2$:** $$E(Y) = E(X^2) = \int_{-k}^{5k} x^2 f_X(x) dx = \int_{-k}^{5k} x^2 \frac{1}{6k} dx = \frac{1}{6k} \int_{-k}^{5k} x^2 dx$$ Calculate the integral: $$\int_{-k}^{5k} x^2 dx = \left[ \frac{x^3}{3} \right]_{-k}^{5k} = \frac{(5k)^3 - (-k)^3}{3} = \frac{125k^3 + k^3}{3} = \frac{126k^3}{3} = 42k^3$$ Therefore: $$E(Y) = \frac{1}{6k} \times 42k^3 = 7k^2$$ 7. **(f) Find $P(Y < 2k^2)$:** Since $Y = X^2$, the event $Y < 2k^2$ means: $$X^2 < 2k^2 \implies -\sqrt{2}k < X < \sqrt{2}k$$ We want: $$P(Y < 2k^2) = P(-\sqrt{2}k < X < \sqrt{2}k)$$ Since $X$ is uniform on $[-k, 5k]$, the interval $[-\sqrt{2}k, \sqrt{2}k]$ intersects $[-k, 5k]$ as follows: - Lower bound: $\max(-k, -\sqrt{2}k) = -k$ because $-k > -\sqrt{2}k$ (since $\sqrt{2} \approx 1.414 > 1$) - Upper bound: $\min(5k, \sqrt{2}k) = \sqrt{2}k$ So the probability is: $$P = \int_{-k}^{\sqrt{2}k} \frac{1}{6k} dx = \frac{1}{6k} (\sqrt{2}k + k) = \frac{1}{6k} k (\sqrt{2} + 1) = \frac{\sqrt{2} + 1}{6}$$ **Final answers:** - (a) $f_X(x) = \frac{1}{6k}$ for $-k \leq x \leq 5k$, else 0 - (b) pdf is a rectangle from $-k$ to $5k$ with height $\frac{1}{6k}$ - (c) Mean $= 2k$ - (d) $$F_X(x) = \begin{cases} 0 & x < -k \\ \frac{x + k}{6k} & -k \leq x \leq 5k \\ 1 & x > 5k \end{cases}$$ - (e) $E(Y) = 7k^2$ - (f) $P(Y < 2k^2) = \frac{\sqrt{2} + 1}{6}$