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DSSSB TGT Maths Female Subject Concerned - 18 Nov 2018 Shift 3

Option 1 : \(\dfrac{1}{20}\)

__Concept:__

The **Variance **of X of probability density function f(x) is given as

\(Var(x)=\displaystyle\int_{-\infty}^{\infty}(x-\mu)^2f(x)dx\)

Where \(\mu\) is **mean** given by

\(\mu=\displaystyle\int_{-\infty}^{\infty}x.f(x)dx\)

__Calculation:__

**Mean ** of above given probability density function is given as

\(\mu=\displaystyle\int_{-\infty}^{\infty}x.f(x)dx=\int_{0}^{\infty}x.f(x)dx+\int_{0}^{1}x.f(x)dx+\int_{1}^{\infty}x.f(x)dx\)

\(\mu=\displaystyle\int_{0}^{1}x.(6x-6x^2)dx\)

\(\mu=\displaystyle\int_{0}^{1}(6x^2-6x^3)dx\)

\(\mu=\displaystyle[\frac{-3x^4}{2}+2x^3]^{^1}_{_0}=\frac{1}{2}\)

Now,

**Variance ** of X is given by

\(Var(x)=\displaystyle\int_{0}^{1}(x-\frac{1}{2})^2(6x-6x^2)dx\)

\(Var(X)=\displaystyle[\frac{-6x^5}{5}+3x^4-\frac{5x^3}{2}+\frac{3x^2}{4}]^{^1}_{_0}=\frac{1}{20}\)

**Hence, Variance of the given probability density function is \(\displaystyle\frac{1}{20}\)**