anonymous
  • anonymous
I need to know if I use the Poisson Distribution for problem 5 (A-C).
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
1 Attachment
dumbcow
  • dumbcow
No use the Binomial distribution p= 1/3 for tie p=1/6 for "0-0" tie
anonymous
  • anonymous
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???

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anonymous
  • anonymous
@jim_thompson5910 @karatechopper @JuanitaM
dumbcow
  • dumbcow
http://en.wikipedia.org/wiki/Binomial_distribution#Probability_mass_function
dumbcow
  • dumbcow
n=10 and k=3
anonymous
  • anonymous
so im correct. woohoo! :D
dumbcow
  • dumbcow
haha :)
anonymous
  • anonymous
ok, so i get .0021676912. is that the final answer or is there anything else i needa do?
dumbcow
  • dumbcow
for part A ? i get about 0.26
anonymous
  • anonymous
yeah, how'd u get .26 ??
dumbcow
  • dumbcow
just plugged numbers in formula ... well i cheated and used Excel :) p = 1/3 n=10 k=3 \[\rightarrow \left(\begin{matrix}10 \\ 3\end{matrix}\right)*(\frac{1}{3})^{3} *(\frac{2}{3})^{7} = 120*\frac{128}{3^{10}} = 0.2601\]
anonymous
  • anonymous
im confused about the second half... after the first equal sign
dumbcow
  • dumbcow
\[\left(\begin{matrix}10 \\ 3\end{matrix}\right) = \frac{10!}{7! 3!} = 120\] \[(\frac{1}{3})^{3} *(\frac{2}{3})^{7} = \frac{1}{3^{3}}*\frac{2^{7}}{3^{7}} = \frac{2^{7}}{3^{10}} = \frac{128}{59,049}\]
anonymous
  • anonymous
ok. so i was getting .0021676912 because i didn't multiply it by 120. makes sense.
anonymous
  • anonymous
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 ???
dumbcow
  • dumbcow
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
anonymous
  • anonymous
o because half would equal zero-zero ??
dumbcow
  • dumbcow
right
anonymous
  • anonymous
and part Cs answer will basically be the same as part B since they're both p=1/6?
dumbcow
  • dumbcow
umm yep thats what it looks like
anonymous
  • anonymous
ok coo ! :)
anonymous
  • anonymous
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???
anonymous
  • anonymous
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
anonymous
  • anonymous
|dw:1360115375877:dw|
dumbcow
  • dumbcow
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
anonymous
  • anonymous
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 )
anonymous
  • anonymous
@satellite73 can you continue where @dumbcow left off??? they are no longer online. :(
anonymous
  • anonymous
4 is poisson, 5 is binomial
anonymous
  • anonymous
what number are you on?
anonymous
  • anonymous
number 7
anonymous
  • anonymous
it's asking me to repeat the steps of #6 but idk what to really do
anonymous
  • anonymous
damn i don't know squat about the normal distribution sorry
anonymous
  • anonymous
its like the picture i drew above ^^^
anonymous
  • anonymous
i know, but you need a table to calculate these
anonymous
  • anonymous
the mean is 180 right?
anonymous
  • anonymous
yes and the standard deviation is 20
anonymous
  • anonymous
then you need to convert to a z score or something right? so you can look these up in a table
anonymous
  • anonymous
ok yeah
anonymous
  • anonymous
http://assets.openstudy.com/updates/attachments/5061f687e4b0583d5cd2e54c-mahlatse1-1348604740357-standardnormaltable.pdf
anonymous
  • anonymous
for example \[\frac{165-180}{20}=-.75\]
anonymous
  • anonymous
and \[\frac{185-180}{20}=.25\] so you need to use a table to find the area under the normal curve between \(-.75\) and \(.25\)
anonymous
  • anonymous
you can use this http://www.danielsoper.com/statcalc3/calc.aspx?id=2
anonymous
  • anonymous
i dont think thats how it works... are both sides supposed to be equal but opposite?
anonymous
  • anonymous
no
anonymous
  • anonymous
*arent
anonymous
  • anonymous
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
anonymous
  • anonymous
so your job is to find the area under the curve from -.75 to .25 this takes a couple steps
anonymous
  • anonymous
from the table , to the left of .25 is 0.59870633 and to the left of -.75 is 0.2266273 so you use \[0.59870633-0.2266273\]
anonymous
  • anonymous
you got that?
anonymous
  • anonymous
im soaking it all in
anonymous
  • anonymous
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
anonymous
  • anonymous
|dw:1360121214382:dw|
anonymous
  • anonymous
|dw:1360121361823:dw|
anonymous
  • anonymous
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
anonymous
  • anonymous
yeah i see what you are doing
anonymous
  • anonymous
ok good
anonymous
  • anonymous
site is freezing up on me, but you can use the calculator i linked to find the z scores, then the area you want
anonymous
  • anonymous
ok so we get 0.37207903 or 37% chance ?
anonymous
  • anonymous
i am having trouble scrolling up, but i did find the numbers you need, i just didn't subtract
anonymous
  • anonymous
yeah that looks right
anonymous
  • anonymous
can you assist me with #8 too?
anonymous
  • anonymous
|dw:1360121910130:dw|
anonymous
  • anonymous
where do i go from here??
anonymous
  • anonymous
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%
anonymous
  • anonymous
do you have a table?
anonymous
  • anonymous
oh forget what i wrote above you just need to go to 95% short people can sit anyway
anonymous
  • anonymous
im confused. can you reword.
anonymous
  • anonymous
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
anonymous
  • anonymous
are talking about this table? http://assets.openstudy.com/updates/attachments/5061f687e4b0583d5cd2e54c-mahlatse1-1348604740357-standardnormaltable.pdf
anonymous
  • anonymous
yeah we can use that
anonymous
  • anonymous
.82894
anonymous
  • anonymous
so we find .95 in the table, and see what the corresponding z score is
anonymous
  • anonymous
looks like about 1.64 or 1.65 half way between them actually lets pick one
anonymous
  • anonymous
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
anonymous
  • anonymous
remember we computed the z score for say \(x\) by \[\frac{x-180}{20}\] so if \[\frac{x-180}{20}=1.64\] then \[x=1.64\times 20+180\]
anonymous
  • anonymous
...ok
anonymous
  • anonymous
and then what?
anonymous
  • anonymous
thats the final answer??
anonymous
  • anonymous
yeah i forgot what the question was can you repost so i don't have to keep scrolling down?
anonymous
  • anonymous
1 Attachment
anonymous
  • anonymous
nvm the limit of the height adjustment is what you wanted so yes, that should be the answer
anonymous
  • anonymous
should i solve or leave as is
anonymous
  • anonymous
of course they want a number i would say about 213 or so
anonymous
  • anonymous
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
anonymous
  • anonymous
i got 221 as my final answer
anonymous
  • anonymous
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
anonymous
  • anonymous
for the 98 % curve
anonymous
  • anonymous
yeah i go the same thing gotta run good luck!
anonymous
  • anonymous
thanks a lot!

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