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If $\{\mathrm{x}\}$ is a continuous, real valued random variable defined over the interval $(-\infty,+\infty)$ and $\mathrm{its}$ occurrence is defined by the density function given as: $f(x)=\frac{1}{\sqrt{2 \pi} * b} e^{-\frac{1}{2}\left(\frac{x-a}{b}\right)^{2}}$ where ' $a$ ' and 'b' are the statistical attributes of the random variable $\{x\}$. The value of the integral $\int_{-\infty}^{a} \frac{1}{\sqrt{2 \pi} * b} e^{-\frac{1}{2}\left(\frac{x-a}{b}\right)^{2}} d x$ is
(A) 1
(B) 0.5
(C) $\pi$
(D) $\frac{\pi}{2}$



Question ID - 156350 | SaraNextGen Top Answer

If $\{\mathrm{x}\}$ is a continuous, real valued random variable defined over the interval $(-\infty,+\infty)$ and $\mathrm{its}$ occurrence is defined by the density function given as: $f(x)=\frac{1}{\sqrt{2 \pi} * b} e^{-\frac{1}{2}\left(\frac{x-a}{b}\right)^{2}}$ where ' $a$ ' and 'b' are the statistical attributes of the random variable $\{x\}$. The value of the integral $\int_{-\infty}^{a} \frac{1}{\sqrt{2 \pi} * b} e^{-\frac{1}{2}\left(\frac{x-a}{b}\right)^{2}} d x$ is
(A) 1
(B) 0.5
(C) $\pi$
(D) $\frac{\pi}{2}$

1 Answer
127 votes
Answer Key / Explanation : (B) -

0.5

127 votes


127