## anonymous 3 years ago Fuel mileage is uniformly distributed between 5 km/L to 12 km/L. What is the probability that on the next trip, fuel mileage is between 6 to 9 km/L

1. amistre64

12-5 = 7 |dw:1350993412929:dw|

2. anonymous

ahh finally...someone who understands

3. amistre64

iwould take a gander and say$\frac{1}{7}{(9-6)}$

4. anonymous

what's that 1/7 by the way?

5. anonymous

|dw:1350993820431:dw|

6. amistre64

a distribution curve has an area of 1 underneath it. since this is a uniform distribution. the "curve" is just a rectangle box with an area of 1 since the width is 7 units wide; the height of the box would have to be 1/7 to get an area of 1

7. anonymous

hmm...that's an intuitive way of solving it....and here i thought i have to do $\huge \int \limits_a^b f(x)dx$

8. amistre64

when it gets to normal distributions with a bell curve; they were nice enough to write down tables of for the integration :)

9. anonymous

^?

10. amistre64

a z table is a table of probabilities; it is written up for the same reason that sin and cos tables are written up. So that we dont have to suffer thru the integration of some ungodly looking integral

11. anonymous

...you have succeeded in confusing and scaring me at the same time....

12. amistre64

yay!! happy halloween :)

13. anonymous

it's already christmas from where i am

14. amistre64

hmmm, happy hanakah? ;)

15. anonymous

anyway...this would just be $\int \limits_a^b \frac 1{b-a} dx$ yes?

16. anonymous

wait...that's wrong

17. anonymous

$\int \limits_a^x \frac 1{b-a}dx$ yes?

18. amistre64

if you wanted to do the $\int f(x)dx$then a uniform distribution is a constant function $\int_{5}^{12}k~dx=1$ $12k-5k=1$ $7k=1$ $k=\frac17$ therefore $\int_{6}^{9}\frac17~dx$

19. anonymous

so...i am right...yes?

20. amistre64

1/(b-a) does generalize it yes. but the interval [a,x] seems a little off. that should simply be the interval across which you are integrating and should prolly not be confused with the other parts