# Thread: probability density function question

1. ## probability density function question

hey,

question

Suppose the probability density function of the length of computer cables is
f(x) = 0.1 from 1200 to 1210 millimeters.

(1.) Determine the mean and standard deviation of the cable length.

I understand how to do p.d.f if you have to variance and so forth but I dont understand this one.

Thanks any help appreciated

2. Originally Posted by Alicia
hey,

question

Suppose the probability density function of the length of computer cables is
f(x) = 0.1 from 1200 to 1210 millimeters.

(1.) Determine the mean and standard deviation of the cable length.

I understand how to do p.d.f if you have to variance and so forth but I dont understand this one.

Thanks any help appreciated
Read this: Uniform distribution (continuous - Wikipedia, the free encyclopedia)

3. Originally Posted by mr fantastic
Here's a derivation for the general uniform distribution:

Let X be a random variable with pdf $\displaystyle f(x) = \frac{1}{b-a}$ for $\displaystyle a \leq x \leq b$ and zero elsewhere.

By definition: $\displaystyle \bar{X} = \int_{a}^{b} \frac{x}{b - a} \, dx = \left( \frac{b^2 - a^2}{2} \right) \, \frac{1}{b - a} = \frac{b + a}{2}$.

By definition: $\displaystyle E(X^2) = \int_{a}^{b} \frac{x^2}{b - a} \, dx = \left( \frac{b^3 - a^3}{3} \right) \, \frac{1}{b - a} = \frac{b^2 + ab + a^2}{3}$.

By definition: $\displaystyle Var(X) = E(X^2) - \bar{X}^2$

$\displaystyle = \left( \frac{b^2 +ab + a^2}{3} \right) - \frac{(b + a)^2}{4} = \frac{4(b^2 + ab + a^2) - 3(b^2 + 2ab + a^2)}{12}$

$\displaystyle = \frac{b^2 - a^2 - 2ab}{12} = \frac{(b - a)^2}{12}$.

In your problem, a = 1200 and b = 1210 .....