# Probability Density Function- Normal distribution

The current in MA in ap iece of copper wire is known to follow a continuous distrubtion over the interval [0,25]. Write down the formula probability density f(x) of the random variable x representing the current. Calculate the mean and variance distribution and find the cumulative distribution function of F(x)

If the function is uniform on [0,25] then the cumulative density function $F(x)=\frac{x-0}{25}$

mean= E[X]=$\int_0^{25} \frac{x}{25}$

$= \frac{x^2}{50} \Big\vert_0^{25}= \frac{625}{50} = 12.5-0=12.5$

I am not sure how to calculate the variance. I missed class, so I really need some good feedback. I took a picture of tmy classmates to quiz to practice these problems and my ansers are way different according to his work:

$$f(x)=\begin{cases} \frac{1}{25} & 0 \leq x \leq 25 \\ 0 & else \end{cases}$$

If he is correct how on earth did he come up with this answer? Since the cumulative distribution function is the integral of the density would it be:

$\int_0^{x} \frac{1}{25}dt$

$\int_0^{x} \frac{t}{25}$

$\Big\vert_0^x = \frac{x}{25}$

Could someone provide info on cumulative and density functions of over normal distribution?

• Where did you get those integration bounds? Integrate over all reals, and you will immediately get $1$, unless I misunderstand... – The Count Mar 10 '17 at 18:07
• What does this have to do with a "normal distribution"? – David G. Stork Mar 10 '17 at 18:13

## 1 Answer

Your question is not entirely clear on whether it is stated that the current is uniformly distributed on $[0,25]$. Your title says "normal" distribution; the statement of the problem does not explicitly state it is uniform (just continuous), and then your answer assumes a uniform distribution. So I have also made the assumption that $X$ is uniform.

You already calculated the cumulative distribution function $$F_X(x) = \begin{cases} 0, & x < 0 \\ \frac{x}{25}, & 0 \le x \le 25 \\ 1, & 25 < x \end{cases}$$ and you also calculated the expectation $$\operatorname{E}[X] = 12.5 = \mu.$$ These are correct. The variance is defined by $$\operatorname{Var}[X] = \operatorname{E}[(X-\mu)^2] = \int_{x=0}^{25} (x - 25)^2 f_X(x) \, dx,$$ where $f_X(x) = \frac{1}{25}$ is the density on $x \in [0,25]$. So what you need to calculate is $$\int_{x=0}^{25} \frac{(x - 25)^2}{25} \, dx.$$ Can you do this?