According to a national survey, 92% of U.S. Dog owners consider their pets to be members of the family. In a random survey of 150 dog owners, what is the probability that at most 132 of them consider their pets to be members of the family?

Question

anonymous · Answer

Help is greatly appreciated, will fan and medal!

amistre64 · Answer

what are your considerations?

amistre64 · Answer

seems like we could do a normal approximation to a binomial distribution ... its a thought.

this site is glitchy at the moment fo rme ...

amistre64 · Answer

or we could just do the binomial stuff ...

amistre64 · Answer

but without a binomCDF function the numbers are going to be rather daunting to play with

anonymous · Answer

Im not sure, thats all it says.

mimi_x3 · Answer

DAng it .... os sucks
If it doesnt state that its normally distributed then it would be safer to use the binomial distribution

anonymous · Answer

Ok could you help me solve it, i dont have a calculator.

anonymous · Answer

Any help @Mimi_x3 ?

mimi_x3 · Answer

Os is acting up really bad But here is a link of a binomial calc http://stattrek.com/online-calculator/binomial.aspx

anonymous · Answer

i dont really know how to use that.

mimi_x3 · Answer

Ohh an amistre is right we can do a normal approximation .... read that incorrectly earlier on

amistre64 · Answer

132/150 - .95(150)
 ----------------
sqrt(.95(.05)(150))

should be our test statistic, our z score

amistre64 · Answer

.95 on top,  not .95(150)

amistre64 · Answer

get my thoghts organized

x - mean
-------
    sd

binom mean = np
sd = sqrt (npq)

adjustment to x by -.5 due to the nature of a binomial shape ...

131.5 - .95(150)
--------------
sqrt(150(.95)(.05))

the left tail of that z score should approximate us good

only thing is, there might be some caveat as the size of p and q ... but i cant remember them at the moment.  if its too high, a poisson is called for

anonymous · Answer

ok so how do i solve that?

amistre64 · Answer

by calculating the values ... its just arithmetic now

amistre64 · Answer

the normal approx should be fine

150(.95)  and 150(.05)  are both larger than 5

anonymous · Answer

so its 131.5-(.95*150) / sqrt of 150*.95*.5?

amistre64 · Answer

thats not the probability, thats the z score .... what do you have to wrok this with anyways?  ti83 maybe?

anonymous · Answer

Nope, just my phone.

amistre64 · Answer

so for a test what do they expect you to use?  im assuming this is for a class.  they give you a table?

anonymous · Answer

they give me a calculator i jsut dont have my own.

amistre64 · Answer

what kind of calculator?  its best to work these in the same manner that youll have to use on a test.

anonymous · Answer

ti83

amistre64 · Answer

ti83 makes this real simple then

hit 2nd, VARS, and pick the binomCDF function and input the values

binomCDF(n,p,x)  and it gives you the left tail area.

anonymous · Answer

ok, i dont have the calculator, so can you show me how to solve it and then give me the answer as well?

amistre64 · Answer

well, im not going to give you the answer ... the wolf can accomplish that.

itll do no good to approximate it and have your solution be in error.

amistre64 · Answer

the z score that we formulate above .... it gives a good indication of the solution, but if you are expected to be more exacting then its no good

anonymous · Answer

my teacher said its fine if its approximated.

amistre64 · Answer

then lets work out the z score

what is  132.5 - .95(150) ??

what is  sqrt(150(.95)(.05)) ?

amistre64 · Answer

did it again ...

amistre64 · Answer

what is  132/150 - .95(150) ??

what is  sqrt(150(.95)(.05)) ?

anonymous · Answer

-141.62 and 2.669

amistre64 · Answer

i need to focus ...

132.5  compares to 95% of 150

132.5 - 150(.95)

132.5 - 142.5 = -11

this is our top value ... of a normal approximation

amistre64 · Answer

out bottom value is the standard deviation which amounts

sqrt(150(.95)(.05))

sqrt(142.5(.05))

sqrt(7.125) = 2.6693

amistre64 · Answer

do these make sense?   or do you know where i go the information from?

anonymous · Answer

yes thats exactly what i got for the bottom

amistre64 · Answer

z = -11/2.6693 = -4.1209...

do you know how to find this value on a z table?

anonymous · Answer

nope

amistre64 · Answer

well, if you dont have a calculator then youll need a table ...

anonymous · Answer

hang on i got a calculator

anonymous · Answer

ti83

amistre64 · Answer

2nd, vars, binomcdf  (150,.95,132)

might be 133, i cant recall of its up to or at most for that terms value

anonymous · Answer

i got 5.413508541 e-4

amistre64 · Answer

one of the more direct approaches is just to wolf it as:
$$\sum_{n=0}^{132}\binom{150}{n}(.95)^{n}~(.05)^{150-n}$$
th sum of all the terms

yep .. and that practically 0

amistre64 · Answer

http://www.wolframalpha.com/input/?i=sum%28n%3D0+to+132%29%28150+choose+n%29+%28.95%29%5En%28.05%29%5E%28150-n%29

anonymous · Answer

ok so the answer is 0?

amistre64 · Answer

i know its rather anticlimatic .. but yeah.  0

amistre64 · Answer

0.05% $\approx$ 0

anonymous · Answer

that doesnt make any sense though, if 92% consider their pets family

amistre64 · Answer

92?  i thought it said 95

anonymous · Answer

92%

amistre64 · Answer

plug in .92 then see what we get

anonymous · Answer

.0551140501

amistre64 · Answer

we would expect:  150(.92) = 138 people to like pets

what is the probability that at most only 132 like pets?

5.51 %

is correct, srry i read 95 to start with

anonymous · Answer

ok thank you very much!