MHT CET · Maths · Probability
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, \(\mathrm{P}[-1 < \mathrm{X} < 1]=\)
- A \(\frac{13}{16}\)
- B \(\frac{1}{2}\)
- C \(\frac{1}{3}\)
- D \(\frac{11}{16}\)
Answer & Solution
Correct Answer
(D) \(\frac{11}{16}\)
Step-by-step Solution
Detailed explanation
Since the function represents a p.d.f.
\(\int_{-2}^{2} k\left(4-x^{2}\right) d x=1 \)
\( \therefore 2 \int_{0}^{2} k\left(4-x^{2}\right) d x=1 \Rightarrow 2 k\left[4 x-\frac{x^{3}}{3}\right]_{-0}^{2}=1 \)
\( \therefore 2 k\left(8-\frac{8}{3}\right)=1 \Rightarrow 2 k\left(\frac{16}{3}\right)=1 \Rightarrow k=\frac{3}{32} \)
\( \therefore P[-1 < x < 1] \)
\( \left.=\int_{-1}^{1}\left(\frac{3}{32}\right)-4-x^{2}\right] d x=\frac{6}{32} \int\left(4-x^{2}\right) d x \)
\( =\frac{6}{32} 4 x-\frac{x^{3}}{3}-0 \)
\( =\frac{6}{32} 4-\frac{1}{3}=\frac{6}{32} \times \frac{11}{3}=\frac{11}{16}\)
\(\int_{-2}^{2} k\left(4-x^{2}\right) d x=1 \)
\( \therefore 2 \int_{0}^{2} k\left(4-x^{2}\right) d x=1 \Rightarrow 2 k\left[4 x-\frac{x^{3}}{3}\right]_{-0}^{2}=1 \)
\( \therefore 2 k\left(8-\frac{8}{3}\right)=1 \Rightarrow 2 k\left(\frac{16}{3}\right)=1 \Rightarrow k=\frac{3}{32} \)
\( \therefore P[-1 < x < 1] \)
\( \left.=\int_{-1}^{1}\left(\frac{3}{32}\right)-4-x^{2}\right] d x=\frac{6}{32} \int\left(4-x^{2}\right) d x \)
\( =\frac{6}{32} 4 x-\frac{x^{3}}{3}-0 \)
\( =\frac{6}{32} 4-\frac{1}{3}=\frac{6}{32} \times \frac{11}{3}=\frac{11}{16}\)
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