see attachment!!!!!!!!!!!
A variable X has a uniform distribution on the interval from 7 to 21. (a) What is the height of the density curve for X? (b) What is P(13.23 X 16.15)? (c) What is P(X > 13.43)? (d) What is the value b such that P(b < X < 16.41) = 0.4334? (e) What is the interquartile range of the distribution of X?
can you help?
sorry dude, i have no idea
uniform distribution means all points are equally likely
can you explain c and d
length from 7 to 21 is 14 so the height of the "curve" will be \(\frac{1}{14}\)
but really you are just computing lengths forget the height it is silly
B should be 13.23 less than or equal to x less than or equal to 16.15
sure length from \(13.23\) so \(21\) is \(21-13.23=7.77\) so \[P(X>13.23)=\frac{7.77}{14}\] whatever that is
makes no difference
\(P(X\geq b)=P(X>b)\) for the continuous uniform distribution
in simple english, compute the length of the favorable interval, divide by the total length, which in your caseis 14
in other words, for B it is \[\frac{16.15-13.23}{14}\]
for C it is \[\frac{21-13.23}{14}\]
what about d
actually i think i made a typo, for C it should be \[\frac{21-13.43}{14}\] but you get the idea right?
yes
for D you know know \(b\) but you know \[\frac{16.41-b}{14}=0.4334\]so you can solve for it
oh , I got it . ` for the E , do I have to find median first?
now i cannot help, because i don't know what "interquartile range" means, sorry maybe i can look it up
median is pretty clearly 14 as it is half way between 7 and 21
we have to find q1 which 25 % of it and q 3 is 75% of it
then Interquartile range = q3 - q1
oh i think inter quartile range is a fancy word for the middle part right? middle half
yeah what you wrote
length is 14, divide in to 4 get 3.5
is it 13?
it is a range, not a number
\(7+3.5=10.5\) and \(21-3.5=17.5\) so the middle half of the interval is \((10.5,17.5)\)
|dw:1361331267112:dw|
yeah, but if you wrote the number in order like 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21
at least i think that is right
obivously the median is 14
this is a continuous distribution, not discrete
then we look at the half from 7 to 14
10 is going to be first median? which 25 %
18 is going to be 75 %
am I right?
\(X\) can take on any value decimals, fractons, irrational number etc
no it is the entire interval from 7 to 21, which has length 14, divided in to 4 equal parts not integers
so it's going to be 7
this question can be interpreted as follows: divide the interval in to 4 equal parts, take the middle two
17.5-10.5
don't forget it is asking for an inter quartile Range not a number
yes
i would say 10.5 to 17.5
ok got it
good! good luck
wait for the c i got 0.555, but its wrong
I got the answer, thanks `
Join our real-time social learning platform and learn together with your friends!