GATE (TF) Textile 2016 Question Paper Solution | GATE/2016/TF/03

Question 03 (Textile Engineering & Fibre Science)

Let ܺ $X$ be a normally distributed random variable with mean 2 and variance 4. Then, the mean of $\frac{X-2}{2}$ is equal to __-0.01:0.01___.

[Show Answer]

Write Here

X is normally distributed random variable.
Mean(P)=2
np=2 ,p=1-q
and variance( $\sigma$ )=4
npq=4
Now, $\frac{np}{npq}=\frac{2}{4}$
q=2 and p=-1
Now,
Mean of $f(X)=\frac{X-2}{2}$
$Mean=\int_{p}^{q} Xf(X) dx$
$Mean=\int_{-1}^{2} X\frac{X-2}{2} dx$
$Mean=\frac{1}{2}\int_{-1}^{2} X(X-2) dx$
$Mean=\frac{1}{2}[\int_{-1}^{2} (X^2) dx-\int_{-1}^{2} (2X) dx]$
$Mean=\frac{1}{2}[\frac{X^3}{3}]_{-1}^{2}-\frac{1}{2}[X^2]_{-1}^{2}$
$Mean=\frac{1}{2}[\frac{2^3}{3}-\frac{-1^3}{3}]-\frac{1}{2}[2^2-(-1)^2]$
$Mean=0$ (Ans)

Previous Question

Next Question

Frequently Asked Questions | FAQs

GATE Textile Engineering and Fibre Science (TF) Question Papers | GATE Textile Question Answer | GATE Textile Solved Question Papers | GATE Textile Papers | GATE Textile Answer Key