A continuous random variable X has probability density function f(x) given by $$ f(x) = \begin{cases} 12x^{2}(1-x), & 0 \leq x \leq 1 \\ 0, & \text{for all other values of } x \end{cases}$$ (a) Find the mode of X. (b) Find the cumulative distribution function for the given probability density function. (c) Without calculating the median, show that the mode is greater than the median.

SSCE HSC Mathematics Advanced Absolute value functions: A continuous random variable X h

A continuous random variable X has probability density function f(x) given by $$ f(x) = \begin{cases} 12x^{2}(1-x), & 0 \leq x \leq 1 \\ 0, & \text{for all other values of } x \end{cases}$$ (a) Find the mode of X - HSC - SSCE Mathematics Advanced - Question 29 - 2023 - Paper 1

Question 29

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A continuous random variable X has probability density function f(x) given by $$ f(x) = \begin{cases} 12x^{2}(1-x), & 0 \leq x \leq 1 \\ 0, & \text{for all other va... show full transcript

Worked Solution & Example Answer:A continuous random variable X has probability density function f(x) given by $$ f(x) = \begin{cases} 12x^{2}(1-x), & 0 \leq x \leq 1 \\ 0, & \text{for all other values of } x \end{cases}$$ (a) Find the mode of X - HSC - SSCE Mathematics Advanced - Question 29 - 2023 - Paper 1

Step 1

Find the mode of X.

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Answer

To find the mode of the random variable X, we start by identifying where the function f(x) has maximum values. This occurs when the derivative of f(x) equals zero.

Calculate the derivative:

f'(x) = \frac{d}{dx}[12x^{2}(1-x)]\ = 12 (2x(1-x) + x^{2}(-1)) = 12(2x - 3x^{2}).$$ Set the derivative equal to zero to find critical points:

12(2x - 3x^{2}) = 0\ \Rightarrow 12x(2-3x) = 0.\

This gives us two solutions: $x = 0$ and $x = \frac{2}{3}$. Next, we need to evaluate the function at these points. We discard $x = 0$ since f(0) is not the maximum. Finally, we check the value of f at $x = \frac{2}{3}$:

f\left(\frac{2}{3}\right) = 12\left(\frac{2}{3}\right)^{2}\left(1 - \frac{2}{3}\right) = 12\left(\frac{4}{9}\right)\left(\frac{1}{3}\right) = \frac{16}{9}.$$

Thus, the mode of X is ( \frac{2}{3} ).

Step 2

Find the cumulative distribution function for the given probability density function.

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Answer

To find the cumulative distribution function F(x), we need to integrate the probability density function f(x):

F(x) = \int_{0}^{x} f(t) dt = \int_{0}^{x} 12t^{2}(1-t) dt.\

Now perform the integration:

Distribute the term:

= \int_{0}^{x} (12t^{2} - 12t^{3}) dt = \left[4t^{3} - 3t^{4}\right]_{0}^{x} = 4x^{3} - 3x^{4}.

Thus, the cumulative distribution function is:

F(x) = 4x^{3} - 3x^{4} \text{ for } 0 \leq x \leq 1.$$

Step 3

Without calculating the median, show that the mode is greater than the median.

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Answer

Given the mode ( \frac{2}{3} ), we need to show that the probability of the random variable being less than this value is greater than 0.5.

Using the cumulative distribution function we found:

F\left(\frac{2}{3}\right) = 4\left(\frac{2}{3}\right)^{3} - 3\left(\frac{2}{3}\right)^{4} = 4 \cdot \frac{8}{27} - 3 \cdot \frac{16}{81}.

Calculating this:

= \frac{32}{27} - \frac{48}{81} = \frac{32}{27} - \frac{16}{27} = \frac{16}{27} \approx 0.5926 > 0.5.

Therefore, the probability of the variable being less than ( \frac{2}{3} ) is greater than 0.5, confirming that the mode is indeed greater than the median.

A continuous random variable X has probability density function f(x) given by $$ f(x) = \begin{cases} 12x^{2}(1-x), & 0 \leq x \leq 1 \\ 0, & \text{for all other values of } x \end{cases}$$ (a) Find the mode of X - HSC - SSCE Mathematics Advanced - Question 29 - 2023 - Paper 1

Find the mode of X.

Find the cumulative distribution function for the given probability density function.

Without calculating the median, show that the mode is greater than the median.

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