In this problem we will be dealing with the probability density function (pdf) associated with a continuous random variable, X. Remember that the pdf is basically a function that assigns a probability density to the event of X taking a value x in the domain of X, with following two properties:
1 f(x) > 0, for all x
2 f(x) dx = 1.
f(x) does not give us the exact probability of X to take value x. Since the size of the domain of X is infinite, you cannot calculate the probability. Instead, you can calculate the probability of X to lie within a range:
Pr(a < X < b) = - / sa f(x) dx
Consider the following pdf function: 6x(1 – x) if 0 < x < 1,
f(x) = 0 otherwise. Calculate the probability of P(0.3 < X < 0.7)
a) 0.784
b) 0.568
c) 0.216
d) -0.568

Question

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Respuesta :

MrRoyal MrRoyal · Answer 1 · 2021-05-21T21:57:33+02:00

Answer:

[tex]b)\ 0.568[/tex]

Step-by-step explanation:

Given

[tex]f(x) = 6x(1- x);\ 0 \le x \le 1[/tex]

Required

[tex]P(0.3 < x < 0.7)[/tex]

From the question, we have:

[tex]P(a < x < b) = \int\limits^b_a {f(x)} \, dx[/tex]

So, we have:

[tex]P(0.3 < x < 0.7) = \int\limits^{0.7}_{0.3} {6x(1 - x)} \, dx[/tex]

Open bracket

[tex]P(0.3 < x < 0.7) = \int\limits^{0.7}_{0.3} {6x - 6x^2} \, dx[/tex]

Integrate

[tex]P(0.3 < x < 0.7) = \frac{6x^2}{2} - \frac{6x^3}{3}}|\limits^{0.7}_{0.3}[/tex]

[tex]P(0.3 < x < 0.7) = 3x^2 - 2x^3|\limits^{0.7}_{0.3}[/tex]

Substitute 0.7 and 0.3 for x

[tex]P(0.3 < x < 0.7) = (3*0.7^2 - 2*0.7^3) - (3*0.3^2 - 2*0.3^3)[/tex]

Using a calculator, we have:

[tex]P(0.3 < x < 0.7) = (0.784) - (0.216)[/tex]

Remove brackets

[tex]P(0.3 < x < 0.7) = 0.784 - 0.216[/tex]

[tex]P(0.3 < x < 0.7) = 0.568[/tex]