# A pdf is given by f(x) = 6x (1-x) for 0 < x < 1. Find the variance of this distribution?

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• Probability density function P(x) = 6x(1 - x), (x from 0 to 1) inverted parabola)

https://www.wolframalpha.com/input/?i=P%28x%29+%3D...

is already normalised, as confirmed by {x: 0 to 1}∫[6x(1 - x)]dx = 1

By inspection, the mean is x = 1/2 ...in agreement with

μ = {x: 0 to 1}∫[x*6x(1 - x)]dx = 1/2

The Variance V of a Probability Density Function is produced from

V(x) = E(x^2) - μ ^2 where the expectation value of x^2 is given by

E(x^2) = ∫x^2*f(x)dx = {x: 0 to 1}∫[x^2*6x(1 - x)]dx = 3/10

V(x) = 3/10 – 1/4 = 1/20 = 0.05

S.D. = √(1/20) = √(5)/10 ~ 0.2236

• Let f be the probability distribution function for random variable X.

E[X]

= ∫[0,1] xf(x) dx

= ∫[0,1] (6x² - 6x³) dx

= [2x³ - (3/2)x⁴] | [0,1]

= 1/2

E[X²]= ∫[0,1] x²f(x) dx= ∫[0,1] (6x³ - 6x⁴) dx= [(3/2)x⁴ - (6/5)x⁵] | [0,1]

= 3/10

Var(X)

= E[X²] - (E[X])²

= 3/10 - (1/2)²

= 1/20

I am a little out of practice with this, so here I will check it using another formula.

Var(X)

= E[(X - μ)²]

= E[(X - 1/2)²] ... μ = 1/2, by the symmetry of the distribution

= ∫[0,1] (x - 1/2)²f(x) dx

= ∫[0,1] (x² - x + 1/4)(6x - 6x²) dx

= ∫[0,1]  [-6x⁴ + 12x³ - (15/2)x² + (3/2)x] dx

= [(-6/5)x⁵ + 3x⁴ - (5/2)x³ + (3/4)x²] | [0,1]

= (-6/5) + 3 - (5/2) + (3/4)

= 1/20

• Hint: You saw a formula to find the variance, given the pdf. What is this formula?

Answer that as an update to your post and then we will take it from there.

Hopefully no one will spoil you the answer thereby depriving you from your personal enhancement; that would be very inconsiderate of them.

Dont forget to vote me best answer for being the first to correctly walk you through and especially for not spoiling out the answer!