# Help with Uniform Probability Function from 0 to 10?

I am having trouble understanding this for my high school stats class. I need to find the probabilities for the following (an explanation on how to get the probability would be appreciated).

P(0.8 < X < 8.0)

P(0.2 < X < 4.0 or 5.6 < X < 7.2)

P(X = 2)

P(-1 < X < 9.0)

Again it just says consider a continuous probability distribution from 0 to 10, what are the following probabilities?

### 1 Answer

- MichaelLv 76 years agoFavorite Answer
Well,

just have a look at this notion of "indication of set" notion

a random variable X that is uniformely distributed

over an interval J = [a , b] has a pdf

----------------> f_x (x) = [1/ (b-a) ] . 1I_J (x)

where : 1I_J is the indication function of interval J = [a,b]

defined by :

1I_J (x) = 1 , if x € J and

1I_J (x) = 0 else.

therefore :

for any measurable subset A of R :

P( X € A ) = Integral ( x € A ) f_x (x) dx

and

here,

according to the definition of f_x,

we obviously get (considering the notion of 1I_J )

P( X € A ) = Integral ( x € A n J) dx

finally, if you have understood all this (this is what i hope !!)

we have : a = 0 and b = 10, so 1/(b-a) = 1/10

we get :

for each question the associated A subset :

P(0.8 < X < 8.0) :

A = (0.8 , 8) so A n J = (0.8 , 8) n [0 , 10] = (0.8 , 8)

therefore :

P(0.8 < X < 8.0) = Int (x € (0.8 , 8) ) (1/10) dx = 7.2/10 = 0.72

now a bit faster A = (0.2 , 4) U (5.6 , 7.2)

P(0.2 < X < 4.0 or 5.6 < X < 7.2) = (1/10)[ (4- 0.2) + (7.2 - 5.6)]

now this one is important for your understanding well,

here, A = {2} is a set of measure (length...) = 0 within R

therefore, the integral on it is = 0

and the important notion to understand is :

in continuous probabilities, the probability that a continuous variable takes a precise value (or a set of measure = 0)

is = 0,

so

P(X = 2) = 0

here A = (-1 , 9)

A n J = (0 , 9)

P(-1 < X < 9.0) = 9/10 = 0.9

et voilà !

cheers from southern France !

hope it' ll help !!