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I need to know if I use the Poisson Distribution for problem 5 (AC).
 one year ago
 one year ago
I need to know if I use the Poisson Distribution for problem 5 (AC).
 one year ago
 one year ago

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dumbcowBest ResponseYou've already chosen the best response.2
No use the Binomial distribution p= 1/3 for tie p=1/6 for "00" tie
 one year ago

mariomintchevBest ResponseYou've already chosen the best response.0
so i would use this formula: P^k * (1p)^nk and for part A, I would plug in 10 for n and 3 for k???
 one year ago

mariomintchevBest ResponseYou've already chosen the best response.0
@jim_thompson5910 @karatechopper @JuanitaM
 one year ago

dumbcowBest ResponseYou've already chosen the best response.2
http://en.wikipedia.org/wiki/Binomial_distribution#Probability_mass_function
 one year ago

mariomintchevBest ResponseYou've already chosen the best response.0
so im correct. woohoo! :D
 one year ago

mariomintchevBest ResponseYou've already chosen the best response.0
ok, so i get .0021676912. is that the final answer or is there anything else i needa do?
 one year ago

dumbcowBest ResponseYou've already chosen the best response.2
for part A ? i get about 0.26
 one year ago

mariomintchevBest ResponseYou've already chosen the best response.0
yeah, how'd u get .26 ??
 one year ago

dumbcowBest ResponseYou've already chosen the best response.2
just plugged numbers in formula ... well i cheated and used Excel :) p = 1/3 n=10 k=3 \[\rightarrow \left(\begin{matrix}10 \\ 3\end{matrix}\right)*(\frac{1}{3})^{3} *(\frac{2}{3})^{7} = 120*\frac{128}{3^{10}} = 0.2601\]
 one year ago

mariomintchevBest ResponseYou've already chosen the best response.0
im confused about the second half... after the first equal sign
 one year ago

dumbcowBest ResponseYou've already chosen the best response.2
\[\left(\begin{matrix}10 \\ 3\end{matrix}\right) = \frac{10!}{7! 3!} = 120\] \[(\frac{1}{3})^{3} *(\frac{2}{3})^{7} = \frac{1}{3^{3}}*\frac{2^{7}}{3^{7}} = \frac{2^{7}}{3^{10}} = \frac{128}{59,049}\]
 one year ago

mariomintchevBest ResponseYou've already chosen the best response.0
ok. so i was getting .0021676912 because i didn't multiply it by 120. makes sense.
 one year ago

mariomintchevBest ResponseYou've already chosen the best response.0
so for part B i would just change p to 1/6 ??? (why 1/6 again?) and for part C i would do 1part B ???
 one year ago

dumbcowBest ResponseYou've already chosen the best response.2
its 1/6 because its half of 1/3 > 1/3 * 1/2 = 1/6 part C is looking at chance you tie but Not 00 > 1/3  1/6 = 1/6
 one year ago

mariomintchevBest ResponseYou've already chosen the best response.0
o because half would equal zerozero ??
 one year ago

mariomintchevBest ResponseYou've already chosen the best response.0
and part Cs answer will basically be the same as part B since they're both p=1/6?
 one year ago

dumbcowBest ResponseYou've already chosen the best response.2
umm yep thats what it looks like
 one year ago

mariomintchevBest ResponseYou've already chosen the best response.0
alright, so im on #7 and it's wanting the same info as #6 but normally distributed... would i use : f(x)= ( 1 / [sigma * sqrt(2pi)] ) * e ^ (xmu/sigma) to plug in the same info from 6???
 one year ago

mariomintchevBest ResponseYou've already chosen the best response.0
here are the answers to #6: a) picture b) 1/(ba) = 1/(210150) = 1/60 c) (160150)/(210150) = 10/60 = 1/6 d) 1/6 * 1/6 = 1/36 e) (185165)/(210150) = 20/60 = 2/6 f) (210200)/(210150) = 10/60 = 1/6
 one year ago

mariomintchevBest ResponseYou've already chosen the best response.0
dw:1360115375877:dw
 one year ago

dumbcowBest ResponseYou've already chosen the best response.2
yes that is the function for normal distribution but since its complex....you need to use probability table for normal distribution problems you have to determine zscore (standardized value) first anyway, check your answers from #6 again first though the mean has to be a number between 150 and 210 .... not 1/60
 one year ago

mariomintchevBest ResponseYou've already chosen the best response.0
i got 1/60 by using the formula but i guess thats not correct since it makes no sense to have a mean of 1/60. i guess the mean is 180 ( 150+210 / 2 )
 one year ago

mariomintchevBest ResponseYou've already chosen the best response.0
@satellite73 can you continue where @dumbcow left off??? they are no longer online. :(
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
4 is poisson, 5 is binomial
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
what number are you on?
 one year ago

mariomintchevBest ResponseYou've already chosen the best response.0
it's asking me to repeat the steps of #6 but idk what to really do
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
damn i don't know squat about the normal distribution sorry
 one year ago

mariomintchevBest ResponseYou've already chosen the best response.0
its like the picture i drew above ^^^
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
i know, but you need a table to calculate these
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
the mean is 180 right?
 one year ago

mariomintchevBest ResponseYou've already chosen the best response.0
yes and the standard deviation is 20
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
then you need to convert to a z score or something right? so you can look these up in a table
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
for example \[\frac{165180}{20}=.75\]
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
and \[\frac{185180}{20}=.25\] so you need to use a table to find the area under the normal curve between \(.75\) and \(.25\)
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
you can use this http://www.danielsoper.com/statcalc3/calc.aspx?id=2
 one year ago

mariomintchevBest ResponseYou've already chosen the best response.0
i dont think thats how it works... are both sides supposed to be equal but opposite?
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
not unless they happen to be your question is between 165 and 185 which is not symmetric about 180 it is 5 up and 10 down
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
so your job is to find the area under the curve from .75 to .25 this takes a couple steps
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
from the table , to the left of .25 is 0.59870633 and to the left of .75 is 0.2266273 so you use \[0.598706330.2266273\]
 one year ago

mariomintchevBest ResponseYou've already chosen the best response.0
im soaking it all in
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
the normal curve is symmetric the way the tables work is that they give you the area under the curve to the left of your z score so if you want the area between two z scores you take the larger z score, find the area in a table, then take the smaller area, and subtract
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
dw:1360121214382:dw
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
dw:1360121361823:dw
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
area between .75 and .25 is what you want it is the area to the left of .25 minus the area to the left of .75
 one year ago

mariomintchevBest ResponseYou've already chosen the best response.0
yeah i see what you are doing
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
site is freezing up on me, but you can use the calculator i linked to find the z scores, then the area you want
 one year ago

mariomintchevBest ResponseYou've already chosen the best response.0
ok so we get 0.37207903 or 37% chance ?
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
i am having trouble scrolling up, but i did find the numbers you need, i just didn't subtract
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
yeah that looks right
 one year ago

mariomintchevBest ResponseYou've already chosen the best response.0
can you assist me with #8 too?
 one year ago

mariomintchevBest ResponseYou've already chosen the best response.0
dw:1360121910130:dw
 one year ago

mariomintchevBest ResponseYou've already chosen the best response.0
where do i go from here??
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
we got to look in a table and see what the right hand point give 97.5 % of the area (so the left hand point will have 2.5% and the area between them will be 95%
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
do you have a table?
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
oh forget what i wrote above you just need to go to 95% short people can sit anyway
 one year ago

mariomintchevBest ResponseYou've already chosen the best response.0
im confused. can you reword.
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
i think 95% of the people lie within two standard deviations of the mean, meaning if you look at the table you should see for .95 a z score of 2 you have to check this because i do not have a table if this is true, then two standard deviations is 40 and 180 + 40 = 120 is the max height
 one year ago

mariomintchevBest ResponseYou've already chosen the best response.0
are talking about this table? http://assets.openstudy.com/updates/attachments/5061f687e4b0583d5cd2e54cmahlatse11348604740357standardnormaltable.pdf
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
yeah we can use that
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
so we find .95 in the table, and see what the corresponding z score is
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
looks like about 1.64 or 1.65 half way between them actually lets pick one
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
for the sake of argument, lets say the z score is 1.64 then we convert back to height \[1.64\times 20+180\] will give what we want
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
remember we computed the z score for say \(x\) by \[\frac{x180}{20}\] so if \[\frac{x180}{20}=1.64\] then \[x=1.64\times 20+180\]
 one year ago

mariomintchevBest ResponseYou've already chosen the best response.0
thats the final answer??
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
yeah i forgot what the question was can you repost so i don't have to keep scrolling down?
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
nvm the limit of the height adjustment is what you wanted so yes, that should be the answer
 one year ago

mariomintchevBest ResponseYou've already chosen the best response.0
should i solve or leave as is
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
of course they want a number i would say about 213 or so
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
the next one is similar find .98 in the body of the table, find the corresponding z score and work backwards to find the height
 one year ago

mariomintchevBest ResponseYou've already chosen the best response.0
i got 221 as my final answer
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
looks like a z score of about 2.05 there is nothing really profound about this, it just says that 98% are less than 2.05 standard deviations above the mean since the standard deviation is 20 and the mean is 180 that means 98% are under \(2.05\times 20+180=221\) cm tall
 one year ago

mariomintchevBest ResponseYou've already chosen the best response.0
for the 98 % curve
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
yeah i go the same thing gotta run good luck!
 one year ago
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