GATE Overflow Test Series | Probability | Test 1 | Question: 17

Question

Assume that a random variable $X$ is Normal with mean $\mu = 4$ and variance $\sigma^2 = 16.$ The probability that $X$ is between $2$ and $6$ is ____. (Rounded to $2$ decimal points)
$\left(\phi(0.5) =\displaystyle \dfrac{1}{\sqrt{2\pi}} \int_{-\infty}^{0.5} e^{-x^2/2}dx = 0.6915\right)$

Answer 1

$\mu = 4,\sigma = 4$

$P(2 \leq X \leq 6) = P\left(\frac{2-4}{4} \leq \frac{X-4}{4}\leq \frac{6-4}{4} \right)$

$\quad = P\left(-0.5 \leq Z\leq 0.5 \right)$

$\quad = P\left( Z\leq 0.5 \right) - P(Z \leq -0.5)$

$\quad = P\left( Z\leq 0.5 \right) - (1 - P(Z \leq 0.5)$

$\quad = 2.P\left( Z\leq 0.5 \right) - 1$

$\quad = 2\times 0.6915 -1 = 0.383$

Comments

How P(Z <= -0.5) = 1 – P(Z <= 0.5)?

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Because Normal distribution is symmetric about the mean. 

https://en.wikipedia.org/wiki/Normal_distribution#Symmetries_and_derivatives

https://www.mathsisfun.com/data/standard-normal-distribution.html

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Z is standard normal random variable. So,

$P(Z <= -0.5) \equiv  P(Z >= 0.5) \equiv 1 – P(Z <= 0.5)$

Answer 2

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