@satellite73
@satellite73 can you help me
hello again
same book?
hi thank you for replying fast :)
yup
just the next lesson
4 and 14?
yup
ok did you do the first 3? height of distribution?
yes
did you get \(\frac{1}{3}\)?
shouldnt it be 2/3?
for the height of the distribution? (actually that is question two)
the height should be \(\frac{1}{3}\) since the base is \(3\) units long and the area is \(1\)
oh ok thank you
but actually for the uniform distribution, although you can use the height (that is why they asked it first) you don't really need to what is the length from \(1.5\) to \(3\)? i .e. what is \(3-.15\)
oops i meant \(3-1.5\)
ok so 3/2 then?
that is the answer to the question i asked yes, but that is not the answer to the question in the book
oh ok yeah I was wondering that because it didnt seem correct since the bigger number is on top
first of 3/2 is not a probability, it is bigger than one the total length from 1 to 4 is 3
so your answer is \[\frac{1.5}{3}\]
so 0.5 or 1/2?
just the length of the line segment divided by the total length of the line
yes \(\frac{1}{2}\)
which just says the distance from \(1.5\) to \(3\) is half of the distance from \(1\) to \(4\)
you can do it in terms of areas as well, but since the height is the same, the area will be one half the total area
height is \(\frac{1}{3}\) base is \(\frac{3}{2}\) area is \[\frac{1}{3}\times \frac{3}{2}=\frac{1}{2}\]
so 0.5/2?
no just one half
or \(.5\) if you prefer
oh ok thank you :)
i think i might have confused you it is a) the length of the interval divided by the total length or b) the area of the region they are the same
for the next one, ignore the part about when the animal is weighed, it makes no difference
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