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Question: If probability density function $f(x)$ of a continuous random variable $x$ is defined by $f(x) = \be...

If probability density function f(x)f(x) of a continuous random variable xx is defined by f(x)={14,−2≤x≤20,otherwisef(x) = \begin{cases} \frac{1}{4}, -2 \leq x \leq 2 \\ 0, otherwise \end{cases} then P(x≤1)P(x \leq 1) is

A

14\frac{1}{4}

B

12\frac{1}{2}

C

13\frac{1}{3}

D

34\frac{3}{4}

Answer

34\frac{3}{4}

Explanation

Solution

The probability density function (PDF) is given by:

f(x)={14,−2≤x≤20,otherwisef(x) = \begin{cases} \frac{1}{4}, & -2 \leq x \leq 2 \\ 0, & otherwise \end{cases}

We need to find the probability P(x≤1)P(x \leq 1). For a continuous random variable, P(x≤1)P(x \leq 1) is the integral of the PDF from −∞-\infty to 11. Since f(x)f(x) is non-zero only for −2≤x≤2-2 \leq x \leq 2, we integrate from -2 to 1:

P(x≤1)=∫−21f(x) dx=∫−2114 dxP(x \leq 1) = \int_{-2}^{1} f(x) \, dx = \int_{-2}^{1} \frac{1}{4} \, dx

Evaluating the integral:

P(x≤1)=14∫−21dx=14[x]−21=14[1−(−2)]=14(3)=34P(x \leq 1) = \frac{1}{4} \int_{-2}^{1} dx = \frac{1}{4} [x]_{-2}^{1} = \frac{1}{4} [1 - (-2)] = \frac{1}{4} (3) = \frac{3}{4}

Therefore, P(x≤1)=34P(x \leq 1) = \frac{3}{4}.