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mariomintchev
I need to know if I use the Poisson Distribution for problem 5 (A-C).
No use the Binomial distribution p= 1/3 for tie p=1/6 for "0-0" tie
so i would use this formula: P^k * (1-p)^n-k and for part A, I would plug in 10 for n and 3 for k???
@jim_thompson5910 @karatechopper @JuanitaM
http://en.wikipedia.org/wiki/Binomial_distribution#Probability_mass_function
so im correct. woohoo! :D
ok, so i get .0021676912. is that the final answer or is there anything else i needa do?
for part A ? i get about 0.26
yeah, how'd u get .26 ??
just plugged numbers in formula ... well i cheated and used Excel :) p = 1/3 n=10 k=3 \[\rightarrow \left(\begin{matrix}10 \\ 3\end{matrix}\right)*(\frac{1}{3})^{3} *(\frac{2}{3})^{7} = 120*\frac{128}{3^{10}} = 0.2601\]
im confused about the second half... after the first equal sign
\[\left(\begin{matrix}10 \\ 3\end{matrix}\right) = \frac{10!}{7! 3!} = 120\] \[(\frac{1}{3})^{3} *(\frac{2}{3})^{7} = \frac{1}{3^{3}}*\frac{2^{7}}{3^{7}} = \frac{2^{7}}{3^{10}} = \frac{128}{59,049}\]
ok. so i was getting .0021676912 because i didn't multiply it by 120. makes sense.
so for part B i would just change p to 1/6 ??? (why 1/6 again?) and for part C i would do 1-part B ???
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 0-0 --> 1/3 - 1/6 = 1/6
o because half would equal zero-zero ??
and part Cs answer will basically be the same as part B since they're both p=1/6?
umm yep thats what it looks like
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 ^ (x-mu/sigma) to plug in the same info from 6???
here are the answers to #6: a) picture b) 1/(b-a) = 1/(210-150) = 1/60 c) (160-150)/(210-150) = 10/60 = 1/6 d) 1/6 * 1/6 = 1/36 e) (185-165)/(210-150) = 20/60 = 2/6 f) (210-200)/(210-150) = 10/60 = 1/6
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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 z-score (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
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 )
@satellite73 can you continue where @dumbcow left off??? they are no longer online. :(
4 is poisson, 5 is binomial
what number are you on?
it's asking me to repeat the steps of #6 but idk what to really do
damn i don't know squat about the normal distribution sorry
its like the picture i drew above ^^^
i know, but you need a table to calculate these
the mean is 180 right?
yes and the standard deviation is 20
then you need to convert to a z score or something right? so you can look these up in a table
for example \[\frac{165-180}{20}=-.75\]
and \[\frac{185-180}{20}=.25\] so you need to use a table to find the area under the normal curve between \(-.75\) and \(.25\)
you can use this http://www.danielsoper.com/statcalc3/calc.aspx?id=2
i dont think thats how it works... are both sides supposed to be equal but opposite?
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
so your job is to find the area under the curve from -.75 to .25 this takes a couple steps
from the table , to the left of .25 is 0.59870633 and to the left of -.75 is 0.2266273 so you use \[0.59870633-0.2266273\]
im soaking it all in
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
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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
yeah i see what you are doing
site is freezing up on me, but you can use the calculator i linked to find the z scores, then the area you want
ok so we get 0.37207903 or 37% chance ?
i am having trouble scrolling up, but i did find the numbers you need, i just didn't subtract
yeah that looks right
can you assist me with #8 too?
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where do i go from here??
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%
oh forget what i wrote above you just need to go to 95% short people can sit anyway
im confused. can you reword.
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
are talking about this table? http://assets.openstudy.com/updates/attachments/5061f687e4b0583d5cd2e54c-mahlatse1-1348604740357-standardnormaltable.pdf
so we find .95 in the table, and see what the corresponding z score is
looks like about 1.64 or 1.65 half way between them actually lets pick one
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
remember we computed the z score for say \(x\) by \[\frac{x-180}{20}\] so if \[\frac{x-180}{20}=1.64\] then \[x=1.64\times 20+180\]
thats the final answer??
yeah i forgot what the question was can you repost so i don't have to keep scrolling down?
nvm the limit of the height adjustment is what you wanted so yes, that should be the answer
should i solve or leave as is
of course they want a number i would say about 213 or so
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
i got 221 as my final answer
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
for the 98 % curve
yeah i go the same thing gotta run good luck!