I need to know if I use the Poisson Distribution for problem 5 (A-C).

- anonymous

I need to know if I use the Poisson Distribution for problem 5 (A-C).

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- schrodinger

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- anonymous

##### 1 Attachment

- dumbcow

No use the Binomial distribution
p= 1/3 for tie
p=1/6 for "0-0" tie

- 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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## More answers

- anonymous

@jim_thompson5910 @karatechopper @JuanitaM

- dumbcow

http://en.wikipedia.org/wiki/Binomial_distribution#Probability_mass_function

- dumbcow

n=10 and k=3

- anonymous

so im correct. woohoo! :D

- dumbcow

haha :)

- anonymous

ok, so i get .0021676912. is that the final answer or is there anything else i needa do?

- dumbcow

for part A ?
i get about 0.26

- anonymous

yeah, how'd u get .26 ??

- 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

im confused about the second half...
after the first equal sign

- 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

ok. so i was getting .0021676912 because i didn't multiply it by 120. makes sense.

- 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

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

o because half would equal zero-zero ??

- dumbcow

right

- anonymous

and part Cs answer will basically be the same as part B since they're both p=1/6?

- dumbcow

umm yep thats what it looks like

- anonymous

ok coo ! :)

- 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

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

|dw:1360115375877:dw|

- 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

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

@satellite73
can you continue where @dumbcow left off??? they are no longer online. :(

- anonymous

4 is poisson, 5 is binomial

- anonymous

what number are you on?

- anonymous

number 7

- anonymous

it's asking me to repeat the steps of #6 but idk what to really do

- anonymous

damn i don't know squat about the normal distribution
sorry

- anonymous

its like the picture i drew above ^^^

- anonymous

i know, but you need a table to calculate these

- anonymous

the mean is 180 right?

- anonymous

yes and the standard deviation is 20

- anonymous

then you need to convert to a z score or something right? so you can look these up in a table

- anonymous

ok yeah

- anonymous

http://assets.openstudy.com/updates/attachments/5061f687e4b0583d5cd2e54c-mahlatse1-1348604740357-standardnormaltable.pdf

- anonymous

for example
\[\frac{165-180}{20}=-.75\]

- 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

you can use this
http://www.danielsoper.com/statcalc3/calc.aspx?id=2

- anonymous

i dont think thats how it works...
are both sides supposed to be equal but opposite?

- anonymous

no

- anonymous

*arent

- 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

so your job is to find the area under the curve from -.75 to .25
this takes a couple steps

- 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

you got that?

- anonymous

im soaking it all in

- 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

|dw:1360121214382:dw|

- anonymous

|dw:1360121361823:dw|

- 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

yeah i see what you are doing

- anonymous

ok good

- 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

ok so we get 0.37207903 or 37% chance ?

- anonymous

i am having trouble scrolling up, but i did find the numbers you need, i just didn't subtract

- anonymous

yeah that looks right

- anonymous

can you assist me with #8 too?

- anonymous

|dw:1360121910130:dw|

- anonymous

where do i go from here??

- 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

do you have a table?

- anonymous

oh forget what i wrote above
you just need to go to 95% short people can sit anyway

- anonymous

im confused. can you reword.

- 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

are talking about this table?
http://assets.openstudy.com/updates/attachments/5061f687e4b0583d5cd2e54c-mahlatse1-1348604740357-standardnormaltable.pdf

- anonymous

yeah we can use that

- anonymous

.82894

- anonymous

so we find .95 in the table, and see what the corresponding z score is

- anonymous

looks like about 1.64 or 1.65 half way between them actually
lets pick one

- 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

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

...ok

- anonymous

and then what?

- anonymous

thats the final answer??

- anonymous

yeah i forgot what the question was
can you repost so i don't have to keep scrolling down?

- anonymous

##### 1 Attachment

- anonymous

nvm
the limit of the height adjustment is what you wanted
so yes, that should be the answer

- anonymous

should i solve or leave as is

- anonymous

of course they want a number
i would say about 213 or so

- 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

i got 221 as my final answer

- 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

for the 98 % curve

- anonymous

yeah i go the same thing
gotta run
good luck!

- anonymous

thanks a lot!

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