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If the error involved in making a certain measurement is continuous random variable $\mathrm{X}$ with probability density function $f(x)=k\left(4-x^{2}\right)$ if $-2 \leq x \leq 2$
$=0 \quad$, otherwise
then, $\mathrm{P}[-1 < \mathrm{X} < 1]=$
Options:
$=0 \quad$, otherwise
then, $\mathrm{P}[-1 < \mathrm{X} < 1]=$
Solution:
1573 Upvotes
Verified Answer
The correct answer is:
$\frac{11}{16}$
Since the function represents a p.d.f.
$$
\begin{aligned}
& \int_{-2}^{2} k\left(4-x^{2}\right) d x=1 \\
\therefore & 2 \int_{0}^{2} k\left(4-x^{2}\right) d x=1 \Rightarrow 2 k\left[4 x-\frac{x^{3}}{3}\right]_{-0}^{2}=1 \\
\therefore & 2 k\left(8-\frac{8}{3}\right)=1 \Rightarrow 2 k\left(\frac{16}{3}\right)=1 \Rightarrow k=\frac{3}{32} \\
\therefore & P[-1 < x < 1] \\
&\left.=\int_{-1}^{1}\left(\frac{3}{32}\right)-4-x^{2}\right] d x=\frac{6}{32} \int\left(4-x^{2}\right) d x \\
&=\frac{6}{32} 4 x-\frac{x^{3}}{3}-0 \\
&=\frac{6}{32} 4-\frac{1}{3}=\frac{6}{32} \times \frac{11}{3}=\frac{11}{16}
\end{aligned}
$$
$$
\begin{aligned}
& \int_{-2}^{2} k\left(4-x^{2}\right) d x=1 \\
\therefore & 2 \int_{0}^{2} k\left(4-x^{2}\right) d x=1 \Rightarrow 2 k\left[4 x-\frac{x^{3}}{3}\right]_{-0}^{2}=1 \\
\therefore & 2 k\left(8-\frac{8}{3}\right)=1 \Rightarrow 2 k\left(\frac{16}{3}\right)=1 \Rightarrow k=\frac{3}{32} \\
\therefore & P[-1 < x < 1] \\
&\left.=\int_{-1}^{1}\left(\frac{3}{32}\right)-4-x^{2}\right] d x=\frac{6}{32} \int\left(4-x^{2}\right) d x \\
&=\frac{6}{32} 4 x-\frac{x^{3}}{3}-0 \\
&=\frac{6}{32} 4-\frac{1}{3}=\frac{6}{32} \times \frac{11}{3}=\frac{11}{16}
\end{aligned}
$$
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