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mariomintchev
 3 years ago
I need to know if I use the Poisson Distribution for problem 5 (AC).
mariomintchev
 3 years ago
I need to know if I use the Poisson Distribution for problem 5 (AC).

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

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

mariomintchev
 3 years ago
Best ResponseYou've already chosen the best response.0@jim_thompson5910 @karatechopper @JuanitaM

dumbcow
 3 years ago
Best ResponseYou've already chosen the best response.2http://en.wikipedia.org/wiki/Binomial_distribution#Probability_mass_function

mariomintchev
 3 years ago
Best ResponseYou've already chosen the best response.0so im correct. woohoo! :D

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

dumbcow
 3 years ago
Best ResponseYou've already chosen the best response.2for part A ? i get about 0.26

mariomintchev
 3 years ago
Best ResponseYou've already chosen the best response.0yeah, how'd u get .26 ??

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

mariomintchev
 3 years ago
Best ResponseYou've already chosen the best response.0im confused about the second half... after the first equal sign

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

mariomintchev
 3 years ago
Best ResponseYou've already chosen the best response.0ok. so i was getting .0021676912 because i didn't multiply it by 120. makes sense.

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

dumbcow
 3 years ago
Best ResponseYou've already chosen the best response.2its 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

mariomintchev
 3 years ago
Best ResponseYou've already chosen the best response.0o because half would equal zerozero ??

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

dumbcow
 3 years ago
Best ResponseYou've already chosen the best response.2umm yep thats what it looks like

mariomintchev
 3 years ago
Best ResponseYou've already chosen the best response.0alright, 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???

mariomintchev
 3 years ago
Best ResponseYou've already chosen the best response.0here 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

mariomintchev
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1360115375877:dw

dumbcow
 3 years ago
Best ResponseYou've already chosen the best response.2yes 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

mariomintchev
 3 years ago
Best ResponseYou've already chosen the best response.0i 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 )

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

satellite73
 3 years ago
Best ResponseYou've already chosen the best response.14 is poisson, 5 is binomial

satellite73
 3 years ago
Best ResponseYou've already chosen the best response.1what number are you on?

mariomintchev
 3 years ago
Best ResponseYou've already chosen the best response.0it's asking me to repeat the steps of #6 but idk what to really do

satellite73
 3 years ago
Best ResponseYou've already chosen the best response.1damn i don't know squat about the normal distribution sorry

mariomintchev
 3 years ago
Best ResponseYou've already chosen the best response.0its like the picture i drew above ^^^

satellite73
 3 years ago
Best ResponseYou've already chosen the best response.1i know, but you need a table to calculate these

satellite73
 3 years ago
Best ResponseYou've already chosen the best response.1the mean is 180 right?

mariomintchev
 3 years ago
Best ResponseYou've already chosen the best response.0yes and the standard deviation is 20

satellite73
 3 years ago
Best ResponseYou've already chosen the best response.1then you need to convert to a z score or something right? so you can look these up in a table

satellite73
 3 years ago
Best ResponseYou've already chosen the best response.1for example \[\frac{165180}{20}=.75\]

satellite73
 3 years ago
Best ResponseYou've already chosen the best response.1and \[\frac{185180}{20}=.25\] so you need to use a table to find the area under the normal curve between \(.75\) and \(.25\)

satellite73
 3 years ago
Best ResponseYou've already chosen the best response.1you can use this http://www.danielsoper.com/statcalc3/calc.aspx?id=2

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

satellite73
 3 years ago
Best ResponseYou've already chosen the best response.1not 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

satellite73
 3 years ago
Best ResponseYou've already chosen the best response.1so your job is to find the area under the curve from .75 to .25 this takes a couple steps

satellite73
 3 years ago
Best ResponseYou've already chosen the best response.1from the table , to the left of .25 is 0.59870633 and to the left of .75 is 0.2266273 so you use \[0.598706330.2266273\]

mariomintchev
 3 years ago
Best ResponseYou've already chosen the best response.0im soaking it all in

satellite73
 3 years ago
Best ResponseYou've already chosen the best response.1the 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

satellite73
 3 years ago
Best ResponseYou've already chosen the best response.1dw:1360121214382:dw

satellite73
 3 years ago
Best ResponseYou've already chosen the best response.1dw:1360121361823:dw

satellite73
 3 years ago
Best ResponseYou've already chosen the best response.1area 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

mariomintchev
 3 years ago
Best ResponseYou've already chosen the best response.0yeah i see what you are doing

satellite73
 3 years ago
Best ResponseYou've already chosen the best response.1site is freezing up on me, but you can use the calculator i linked to find the z scores, then the area you want

mariomintchev
 3 years ago
Best ResponseYou've already chosen the best response.0ok so we get 0.37207903 or 37% chance ?

satellite73
 3 years ago
Best ResponseYou've already chosen the best response.1i am having trouble scrolling up, but i did find the numbers you need, i just didn't subtract

satellite73
 3 years ago
Best ResponseYou've already chosen the best response.1yeah that looks right

mariomintchev
 3 years ago
Best ResponseYou've already chosen the best response.0can you assist me with #8 too?

mariomintchev
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1360121910130:dw

mariomintchev
 3 years ago
Best ResponseYou've already chosen the best response.0where do i go from here??

satellite73
 3 years ago
Best ResponseYou've already chosen the best response.1we 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%

satellite73
 3 years ago
Best ResponseYou've already chosen the best response.1do you have a table?

satellite73
 3 years ago
Best ResponseYou've already chosen the best response.1oh forget what i wrote above you just need to go to 95% short people can sit anyway

mariomintchev
 3 years ago
Best ResponseYou've already chosen the best response.0im confused. can you reword.

satellite73
 3 years ago
Best ResponseYou've already chosen the best response.1i 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

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

satellite73
 3 years ago
Best ResponseYou've already chosen the best response.1yeah we can use that

satellite73
 3 years ago
Best ResponseYou've already chosen the best response.1so we find .95 in the table, and see what the corresponding z score is

satellite73
 3 years ago
Best ResponseYou've already chosen the best response.1looks like about 1.64 or 1.65 half way between them actually lets pick one

satellite73
 3 years ago
Best ResponseYou've already chosen the best response.1for 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

satellite73
 3 years ago
Best ResponseYou've already chosen the best response.1remember we computed the z score for say \(x\) by \[\frac{x180}{20}\] so if \[\frac{x180}{20}=1.64\] then \[x=1.64\times 20+180\]

mariomintchev
 3 years ago
Best ResponseYou've already chosen the best response.0thats the final answer??

satellite73
 3 years ago
Best ResponseYou've already chosen the best response.1yeah i forgot what the question was can you repost so i don't have to keep scrolling down?

satellite73
 3 years ago
Best ResponseYou've already chosen the best response.1nvm the limit of the height adjustment is what you wanted so yes, that should be the answer

mariomintchev
 3 years ago
Best ResponseYou've already chosen the best response.0should i solve or leave as is

satellite73
 3 years ago
Best ResponseYou've already chosen the best response.1of course they want a number i would say about 213 or so

satellite73
 3 years ago
Best ResponseYou've already chosen the best response.1the next one is similar find .98 in the body of the table, find the corresponding z score and work backwards to find the height

mariomintchev
 3 years ago
Best ResponseYou've already chosen the best response.0i got 221 as my final answer

satellite73
 3 years ago
Best ResponseYou've already chosen the best response.1looks 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

mariomintchev
 3 years ago
Best ResponseYou've already chosen the best response.0for the 98 % curve

satellite73
 3 years ago
Best ResponseYou've already chosen the best response.1yeah i go the same thing gotta run good luck!
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