please help.... statistics
 

stuck on this question for 3 hours.... help please..

topic: distribution of the sample means

Tropicana orange juice buys all of its oranges from citrus orange grove in Florida. The amount of juice squeezed from each of these oranges is approximately normally distributed with a mean of 4.70 ounces and a standard deviation of 0.40 ounces.



Tangerine farms claims that only 4% of their oranges will contain less than 5 ounces of juice. tropicana decides to verify their claim by taking random sample of 500 oranges and counting how many contain less than 5 ounces of juice. 30 of the 500 oranges sampled contained less than 5 ounces. If tangerine's claim is true, that only 4% of their oranges contain less than 5 ounces of juice, what is the probability of getting 30 or more oranges containing less than 5 ounces in a sample of 500 oranges?

ANS: 0.0150 = 1.5%

i just pulled my last piece of hair out of my head... and still couldn't figure out answer...

BCIT business statistics ompt 1197

this is where you post your homework questions?

Lol stats, passed that last semester and glad i never have to look at it again.

i normally don't post my hws here, i'll figure out the answer most of the time... but i'm very very stuck on this one..

:fulloffuck:


just take a break and go back to the question and separate it into categories

then take another break and come back to it hopefully it'll seem simple :D

don't you have friends in the class you can ask..?

Any online statistics site related to your question in mind?

It's been a while and need a refresher to answer such question lol.

This question can't be worth much marks right?

does these help.??


z score = x-m/standard deviation

z score = standard / square root of N

and the normal distribution table..

Haven't done these types of stat problems in a while but here's the solution I came up with

Since 4% of oranges contain less than 5 oz of juice, and you have a sample of 500, it's pretty much binomial with n = 500, p = 0.04

so letting X = number of oranges having < 5oz of juice, X ~ Bin(500, 0.04)

so E(X) = 20
Var(X) = 19.2 and Standard deviation is sqrt(19.2)

You're asked to find Pr(X >= 30) (greater or equal to)
and since it's binomial which is discrete whereas normal is continuous you use the normal approximation and apply the continuity correction

Pr(X >=30) = 1 - Pr (X < 30) = 1 - Pr ( X <=29) = 1 - Pr (X <= 29.5) because of the continuity correction.

So standardize this crap to get 1 - Pr (Z <= (29.5 - 20) / sqrt(19.2)) to get Z value of 2.17.

Then use this bad boy here:



to see that the cdf of Z-value 2.17 = 0.985

so 1 - 0.985 = 0.015

:fuckyea:

:toot:

I think that's pretty much dead on...

Beginning of the question was totally unnecessary and didn't even make sense.. If the amount of juice is normally distributed with a mean of 4.7 ounces, how could the percentange of oranges with with <5 ounces be only 5%?

^Tangerine farms are not part of the citrus orange grove in Florida.:troll::lawl:

I'd suggest you never take any stats courses beyond this.

man... I'm looking at the solution and it's not ringing any bells~ All I remember from my 8am Stats 200 was that I either didn't wake up in time for it or that I ended up sleeping through it. In the end, they used some statistical standardization curve to allow me to pass :fuckyea:

Props to T.T :spin:

Maybe there should be a homework help section in RS :lawl:

THANKYOU SOOO MUCHHH!!! =) !!! stats is making my life miserable. Glade i'll be done in 3 weeks :fuckyea:

This is the only stats course i need lol

based on this thread, I've come up with the tentative solution that statistics is gay

^ i concur

trust me .. better to draw out the normal distribution curve...that how I did those questions at BCIT stat. course

we do need one ! good idea! :thumbs:

lol np glad to help :thumbs:

Please show your work.

This is why I like RevScene. Probably more reliable than my classmates.

This just reminded me that I need to take Buec 333

fuuuuuuukk