A4. Suppose that $$ X $$ is a continuous random variable whose probability density function is given by: $$ f_{X}(x)=\left\{\begin{array}{c} c\left(4 x-2 x^{2}\right) \text { for } 0 \leq x \leq 2 \ 0 \text { otherwise } \end{array}\right. $$ Find (a) the value of $$ c $$ [4] page 1 of 2 BSTD101 (b) $$ P(X \geq 1) $$ [4]

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**(a) Finding the value of $$ c $$:**

The density function must integrate to 1 over the entire real line. Since $$ f_{X}(x) = c(4x - 2x^2) $$ for $$ 0 \leq x \leq 2 $$ and 0 elsewhere, we have

$$
\int_{0}^{2} c(4x - 2x^2) \, dx = 1.
$$

Factor out $$ c $$:

$$
c \int_{0}^{2} (4x - 2x^2) \, dx = 1.
$$

Compute the integral:

1. Find the antiderivative:

$$
   \int (4x - 2x^2) \, dx = 2x^2 - \frac{2}{3}x^3 + C.
   $$

2. Evaluate from $$ x = 0 $$ to $$ x = 2 $$:

$$
   \left[2x^2 - \frac{2}{3}x^3 \right]_{0}^{2} = \left(2(2)^2 - \frac{2}{3}(2)^3\right) - \left(2(0)^2 - \frac{2}{3}(0)^3\right).
   $$

$$
   = \left(2 \cdot 4 - \frac{2}{3} \cdot 8\right) = 8 - \frac{16}{3}.
   $$

$$
   = \frac{24}{3} - \frac{16}{3} = \frac{8}{3}.
   $$

Thus, the normalization equation becomes:

$$
c \cdot \frac{8}{3} = 1.
$$

Solve for $$ c $$:

$$
c = \frac{3}{8}.
$$

---

**(b) Finding $$ P(X \geq 1) $$:**

We need to compute:

$$
P(X \geq 1) = \int_{1}^{2} f_{X}(x) \, dx = \int_{1}^{2} c(4x - 2x^2) \, dx.
$$

Substitute $$ c = \frac{3}{8} $$:

$$
P(X \geq 1) = \frac{3}{8} \int_{1}^{2} (4x - 2x^2) \, dx.
$$

Again, compute the integral using the antiderivative:

$$
\int (4x - 2x^2) \, dx = 2x^2 - \frac{2}{3}x^3.
$$

Evaluate from $$ x = 1 $$ to $$ x = 2 $$:

1. At $$ x = 2 $$:

$$
   2(2)^2 - \frac{2}{3}(2)^3 = 8 - \frac{16}{3} = \frac{24 - 16}{3} = \frac{8}{3}.
   $$

2. At $$ x = 1 $$:

$$
   2(1)^2 - \frac{2}{3}(1)^3 = 2 - \frac{2}{3} = \frac{6 - 2}{3} = \frac{4}{3}.
   $$

Subtract to find the definite integral:

$$
\frac{8}{3} - \frac{4}{3} = \frac{4}{3}.
$$

Multiply by $$ \frac{3}{8} $$:

$$
P(X \geq 1) = \frac{3}{8} \cdot \frac{4}{3} = \frac{4}{8} = \frac{1}{2}.
$$

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**Final Answers:**

- (a) $$ c = \frac{3}{8} $$
- (b) $$ P(X \geq 1) = \frac{1}{2} $$
**(a) $$ c = \frac{3}{8} $$ **(b) $$ P(X \geq 1) = \frac{1}{2} $$

A4. Suppose that is a continuous random variable whose probability density function is given by:

Find
(a) the value of
[4]
page 1 of 2
BSTD101
(b)
[4]

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A4. Suppose that is a continuous random variable whose probability density function is given by: Find (a) the value of [4] page 1 of 2 BSTD101 (b) [4]