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Mathematics 12 Online
OpenStudy (anonymous):

see attachment!!!!!!!!!!!

OpenStudy (anonymous):

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?

OpenStudy (anonymous):

can you help?

OpenStudy (aaronq):

sorry dude, i have no idea

OpenStudy (anonymous):

uniform distribution means all points are equally likely

OpenStudy (anonymous):

can you explain c and d

OpenStudy (anonymous):

length from 7 to 21 is 14 so the height of the "curve" will be \(\frac{1}{14}\)

OpenStudy (anonymous):

but really you are just computing lengths forget the height it is silly

OpenStudy (anonymous):

B should be 13.23 less than or equal to x less than or equal to 16.15

OpenStudy (anonymous):

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

OpenStudy (anonymous):

makes no difference

OpenStudy (anonymous):

\(P(X\geq b)=P(X>b)\) for the continuous uniform distribution

OpenStudy (anonymous):

in simple english, compute the length of the favorable interval, divide by the total length, which in your caseis 14

OpenStudy (anonymous):

in other words, for B it is \[\frac{16.15-13.23}{14}\]

OpenStudy (anonymous):

for C it is \[\frac{21-13.23}{14}\]

OpenStudy (anonymous):

what about d

OpenStudy (anonymous):

actually i think i made a typo, for C it should be \[\frac{21-13.43}{14}\] but you get the idea right?

OpenStudy (anonymous):

yes

OpenStudy (anonymous):

for D you know know \(b\) but you know \[\frac{16.41-b}{14}=0.4334\]so you can solve for it

OpenStudy (anonymous):

oh , I got it . ` for the E , do I have to find median first?

OpenStudy (anonymous):

now i cannot help, because i don't know what "interquartile range" means, sorry maybe i can look it up

OpenStudy (anonymous):

median is pretty clearly 14 as it is half way between 7 and 21

OpenStudy (anonymous):

we have to find q1 which 25 % of it and q 3 is 75% of it

OpenStudy (anonymous):

then Interquartile range = q3 - q1

OpenStudy (anonymous):

oh i think inter quartile range is a fancy word for the middle part right? middle half

OpenStudy (anonymous):

yeah what you wrote

OpenStudy (anonymous):

length is 14, divide in to 4 get 3.5

OpenStudy (anonymous):

is it 13?

OpenStudy (anonymous):

it is a range, not a number

OpenStudy (anonymous):

\(7+3.5=10.5\) and \(21-3.5=17.5\) so the middle half of the interval is \((10.5,17.5)\)

OpenStudy (anonymous):

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OpenStudy (anonymous):

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

OpenStudy (anonymous):

at least i think that is right

OpenStudy (anonymous):

obivously the median is 14

OpenStudy (anonymous):

this is a continuous distribution, not discrete

OpenStudy (anonymous):

then we look at the half from 7 to 14

OpenStudy (anonymous):

10 is going to be first median? which 25 %

OpenStudy (anonymous):

18 is going to be 75 %

OpenStudy (anonymous):

am I right?

OpenStudy (anonymous):

\(X\) can take on any value decimals, fractons, irrational number etc

OpenStudy (anonymous):

no it is the entire interval from 7 to 21, which has length 14, divided in to 4 equal parts not integers

OpenStudy (anonymous):

so it's going to be 7

OpenStudy (anonymous):

this question can be interpreted as follows: divide the interval in to 4 equal parts, take the middle two

OpenStudy (anonymous):

17.5-10.5

OpenStudy (anonymous):

don't forget it is asking for an inter quartile Range not a number

OpenStudy (anonymous):

yes

OpenStudy (anonymous):

i would say 10.5 to 17.5

OpenStudy (anonymous):

ok got it

OpenStudy (anonymous):

good! good luck

OpenStudy (anonymous):

wait for the c i got 0.555, but its wrong

OpenStudy (anonymous):

I got the answer, thanks `

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