# Thread: X is uniformly distributed over

1. ## X is uniformly distributed over

Question : X is uniformly distributed over (0,10) calculate the probability that
i) X < 3
ii) X > 7
iii) 1 < X < 6
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Solution :

$\displaystyle P(x<3) = \frac{1}{3}$

$\displaystyle P(X>7) = \frac{1}{3}$

$\displaystyle P(1<X<6) = \frac{1}{5}$

2. Originally Posted by zorro
Question : X is uniformly distributed over (0,10) calculate the probability that
i) X < 3
ii) X > 7
iii) 1 < X < 6
-------------------------------------------------------------------
Solution :

$\displaystyle P(x<3) = \frac{1}{3}$

$\displaystyle P(X>7) = \frac{1}{3}$

$\displaystyle P(1<X<6) = \frac{1}{5}$

Unfortunately, none of them are correct.

Draw the distribution (note, and I am assuming that X is continuous, that f(x) = 1/10 for $\displaystyle 0 \leq x \leq 10$ and zero otherwise) and shade the area represented by each question. The questions boil down to finding areas of rectangles.

3. ## My work

Originally Posted by mr fantastic
Unfortunately, none of them are correct.

Draw the distribution (note, and I am assuming that X is continuous, that f(x) = 1/10 for $\displaystyle 0 \leq x \leq 10$ and zero otherwise) and shade the area represented by each question. The questions boil down to finding areas of rectangles.

-----------------------------My work---------------------------------

Pdf of uniform distribution

$\displaystyle Pr(x;a,b) = \begin{cases} \frac{1}{b-a} & a \le x < b \\ 0 & otherwise \end{cases}$

Pr(X< 3) = Pr(0 $\displaystyle \le$ X < 3) = $\displaystyle \frac{1}{3-0 } = \frac{1}{3}$

Pr(X>7 ) = Pr(7 < X < 10) = $\displaystyle \frac{1}{10 - 7} = \frac{1}{3}$

Pr(1 < X < 6) = $\displaystyle \frac{1}{6 -1} = \frac{1}{5}$

Please let be know what i have done wrong

4. Originally Posted by zorro
-----------------------------My work---------------------------------

Pdf of uniform distribution

$\displaystyle Pr(x;a,b) = \begin{cases} \frac{1}{b-a} & a \le x < b \\ 0 & otherwise \end{cases}$

Pr(X< 3) = Pr(0 $\displaystyle \le$ X < 3) = $\displaystyle \frac{1}{3-0 } = \frac{1}{3}$

Pr(X>7 ) = Pr(7 < X < 10) = $\displaystyle \frac{1}{10 - 7} = \frac{1}{3}$

Pr(1 < X < 6) = $\displaystyle \frac{1}{6 -1} = \frac{1}{5}$

Please let be know what i have done wrong
You need to review - very, very thoroughly - what the definitions of a and b are in that formula for the pdf.

b = 10 and a = 0.

5. i know b=10 and a=0 but what should i do with that there is no question asking for the interval between 0-10

6. Originally Posted by zorro
i know b=10 and a=0 but what should i do with that there is no question asking for the interval between 0-10
I gave the pdf in my first post (post #3). And you should know how to calculate probability from a pdf.

7. ## Is this correct

Originally Posted by mr fantastic
I gave the pdf in my first post (post #3). And you should know how to calculate probability from a pdf.

$\displaystyle Pr(X<3)= \int_{0}^{3} \frac{1}{10} dx = \frac{3}{10}$

$\displaystyle Pr(X>7) = \int_{7}^{10} \frac{1}{10} dx = \frac{3}{10}$

$\displaystyle Pr(1<X<6) = \int_{1}^{6} \frac{1}{10} dx = \frac{1}{2}$

Are these correct ????

8. Yes they are correct.