Here's the question you clicked on:
lgbasallote
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
12-5 = 7 |dw:1350993412929:dw|
ahh finally...someone who understands
iwould take a gander and say\[\frac{1}{7}{(9-6)}\]
what's that 1/7 by the way?
|dw:1350993820431:dw|
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
hmm...that's an intuitive way of solving it....and here i thought i have to do \[\huge \int \limits_a^b f(x)dx\]
when it gets to normal distributions with a bell curve; they were nice enough to write down tables of for the integration :)
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
...you have succeeded in confusing and scaring me at the same time....
yay!! happy halloween :)
it's already christmas from where i am
hmmm, happy hanakah? ;)
anyway...this would just be \[\int \limits_a^b \frac 1{b-a} dx\] yes?
wait...that's wrong
\[\int \limits_a^x \frac 1{b-a}dx\] yes?
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 \[\int_{6}^{9}\frac17~dx\]
so...i am right...yes?
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