The random variable X follows normal distribution $f(x) = ce^{-\frac{1}{2}\frac{(x-100)^2}{25}}$. Then the value of c is : (a) $\sqrt{2\pi}$ (b) $\dfrac{1}{\sqrt{2\pi}}$ (c) $5\sqrt{2\pi}$ (d) $\dfrac{1}{5\sqrt{2\pi}}$

> [!TLDR] > Matching the given pdf to the standard normal form $\dfrac{1}{\sigma\sqrt{2\pi}}e^{-\frac{(x-\mu)^2}{2\sigma^2}}$ gives $\sigma=5$, so $c=\dfrac{1}{5\sqrt{2\pi}}$. > > 1. Standard normal density: $f(x)=\dfrac{1}{\sigma\sqrt{2\pi}}\exp\left(-\dfrac{(x-\mu)^2}{2\sigma^2}\right)$. > 2. Given: $f(x)=c\exp\left(-\dfrac12\cdot\dfrac{(x-100)^2}{25}\right)=c\exp\left(-\dfrac{(x-100)^2}{50}\right)$. …

The random variable X follows normal distribution f(x) = ce-(1)/(2)((x-100)^2)/(25). Then the value of c is :

Question 63 of 105

f(x) = ce^{-\frac{1}{2}\frac{(x-100)^2}{25}}

(a)

\sqrt{2\pi}

(b)

\dfrac{1}{\sqrt{2\pi}}

(c)

5\sqrt{2\pi}

(d)

\dfrac{1}{5\sqrt{2\pi}}

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Matching the given pdf to the standard normal form $\dfrac{1}{\sigma\sqrt{2\pi}}e^{-\frac{(x-\mu)^2}{2\sigma^2}}$ gives $\sigma=5$ , so $c=\dfrac{1}{5\sqrt{2\pi}}$ .

Standard normal density: $f(x)=\dfrac{1}{\sigma\sqrt{2\pi}}\exp\left(-\dfrac{(x-\mu)^2}{2\sigma^2}\right)$ .
Given: $f(x)=c\exp\left(-\dfrac12\cdot\dfrac{(x-100)^2}{25}\right)=c\exp\left(-\dfrac{(x-100)^2}{50}\right)$ . …

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Q.The random variable X follows normal distribution $f(x) = ce^{-\frac{1}{2}\frac{(x-100)^2}{25}}$ . Then the value of c is :

Concept understanding — Normal Distribution

Normal Distribution

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Q.The random variable X follows normal distribution $f(x) = ce^{-\frac{1}{2}\frac{(x-100)^2}{25}}$ . Then the value of c is :

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