Question

7. Let X be a continuous random variable with probability density function: 0, 5+2x if if 0 if x> 10 x<b f(x)=re, 76-0 5310 x 10 150 o Find the expected value and mode of random variable X. Part 3. Statistics The following sample is given 3.15 3.25 3.25 3.35 3.45 3.5 3.5

Answer 1

QUESTION 7.

Given a continuous random variable X with probability density function:

$f(x) = \frac{{5 &plus; 2x}}{{150}},0 \leqslant x \leqslant 10$

Therefore, the expected value is:

$E(X) = \int_0^{10} {xf(x)dx}$

  $\begin{gathered} = \int_0^{10} {x\left( {\frac{{5 &plus; 2x}}{{150}}} \right)} dx \hfill \\ = \frac{1}{{150}}\int_0^{10} {(5x &plus; 2{x^2})dx} \hfill \\ = \frac{1}{{150}}\left[ {\frac{5}{2}({x^2})_0^{10} &plus; \frac{2}{3}({x^3})_0^{10}} \right] \hfill \\ = \frac{1}{{150}}\left[ {\frac{{5 \times 100}}{2} &plus; \frac{{2 \times 1000}}{3}} \right] \hfill \\ = 6.11 \hfill \\ \end{gathered}$

Mode

The mode of a continuous random variable X with a probability density function f(x) is the value of x for which f(x) takes a maximum value.

We have:

$f(x) = \frac{{5 &plus; 2x}}{{150}}$

$\therefore f'(x) = \frac{2}{{150}}$

For the mode $f'(x) = 0$

$\therefore \frac{2}{{150}} = 0$

Since it's a straight line so it has no turning point. That is why differentiating and equating to zero gives an absurd result.

So on checking by putting the values of x in the function f(x), at x = 10 the function f(x) has the highest value of 25/150.

Therefore, the mode of the random variable X is 10.