Anybody that knows statistics/probability..

M

mdeclipse03

New Member
I'm not sure if this is posted in the right place but I could sure use some help with a statistic problem:

Problem:
The random variable x has a normal distribution with u 1000 and o 10.

u = 1,000
o = 10

A. Find the probability that x assumes a value more than 2 standard deviations from its mean. More than 3 standard deviations from u

B. Find the probability that x assumes a value within 1 standard deviation of its mean. Within 2 standard deviation of u

C. Find the value of x that represents the 80th percentile of this distribution. The 10th percentile.



At first I used the formula:

z= x-u/o, plugged in the numbers but it was wrong.

The teacher responded with this:
P(z>2) = .5-.4772 = .0228 & P(Z<-2) =P(Z>2) by symmetry. This is the area of the upper tail plus that of the lower tail.
So, P(Z<-2)+P(Z>2)= 2P(Z>2) = 2(.0228)

So after obtaining that information, I resubmitted using the above formula:

4.94 A) P(Z<-2)+P(Z>2)= 2P(Z>2)
.5 - .4772 (from the back of my book)= 2(.0228) = .0456

P(Z<-3)+P(Z>3)= 2P(Z>3)
.5 - .4987 (number from the back of my book) = 2(.0013) = .0026



B) P(Z<-1)+P(Z>1)= 2P(Z>1)
.5 - .0398 (number from the back of my book) = .4602 = 2(.4602) = .9204


P(Z<-2)+P(Z>2)= 2P(Z>2)
.5 - .4772 (number from the back of my book) = 2(.0228) = .0456



C) P(Z<- .8 )+P(Z> .8 )= 2P(Z>.8 )
.5 - .2281 (number from the back of my book) = .2719 = 2(.2719) = .5438

P(Z<-.1)+P(Z>.1)= 2P(Z>.1)
.5 - .0398 (number from the back of my book) = .4602 = 2(.4602) = .9204



I'm not sure if these even makes sense to anybody but I've been working on this problem for a week now and I have one more week to get it right. Teacher gives us two weeks to make corrections and I am just about lost. I dislike statistics.
 
vraiblonde

vraiblonde

Board Mommy
PREMO Member
Patron
Actually, MD, we do have a statistician onboard. He'll see this probably tomorrow or Monday and most likely answer you.
 
M

mdeclipse03

New Member
well if i turn in the colors of the rainbow, the professor may think I swing his way and that surely is not the case..lol
 
D

dmd196

New Member
STAT HELP - for real!

I don't have a stat book with me, so i'm going to just guide you.

A) is correct

B) part one is correct; part two is incorrect. the number should be higher than the answer from B) part one because it includes two standards of deviation.

C) ur answers are less than 1, so i'm guessing that you didn't understand the question. the question is asking for what value of x is at 80th and 10th percentile. your mean is 1000. that means the value at the 50th percentile is 1000. so your 80th percentile value should be higher than 1000 and and the 10th percentile will be less than 1000. yo
 
M

mdeclipse03

New Member
dmd196 said:
I don't have a stat book with me, so i'm going to just guide you.

A) is correct

B) part one is correct; part two is incorrect. the number should be higher than the answer from B) part one because it includes two standards of deviation.

C) ur answers are less than 1, so i'm guessing that you didn't understand the question. the question is asking for what value of x is at 80th and 10th percentile. your mean is 1000. that means the value at the 50th percentile is 1000. so your 80th percentile value should be higher than 1000 and and the 10th percentile will be less than 1000. yo
Click to expand...

I checked my class and he said for Part B, use Example 4.15: This is what Example 4.15 tells me:

P(-1.33<z<1.33)=P(-1.33<z<0) + P(0<z<1.33)
= A1 + A2 = .4082 + .4082 = .8164

So..for the first part of the question for Part be in regards to 1, based upon that example, I should have:

P(-1<z<1)=P(-1<z<0) + P(0<z<1)
= .0398 + .0398 = .0796

From that, it should be the answer based up what the teacher said to go by. Part C is still confusing to me. For part see, would the value be .5000 since 1000 is divided in half?
 
Tilted

Tilted

..
mdeclipse03 said:
From that, it should be the answer based up what the teacher said to go by. Part C is still confusing to me. For part see, would the value be .5000 since 1000 is divided in half?
Click to expand...

For part C you first need to find the Z that corresponds to the 80th percentile of x's. So find the Z in the standard normal distribution table that corresponds to .3000 (that gives you the 30% of x's that are above u but below the 80th percentile in addition to the 50% that are below u, making up 80%).

Then plug that Z into your equation and solve for x:

z=(x-u)/o

For the 10th percentile you would find the Z on the table that corresponds to .4000 (the 50% of x's that are below u minus the 40% of x's that are below u but above the 10th percentile, giving you 10%). For this one though you have to use the negative of that Z. Then solve the equation for x again.


One caveat - when you look in the table, it probably won't show .3000 or .4000 exactly. One number might be .2996 and the one next to it might be .3015. Depending on how precise your teacher wants you to be, you could either select the Z that corresponds to the one that is closest or you can roughly calculate a more precise Z somewhere in between them. (I hope that makes since)
 
Tilted

Tilted

..
mdeclipse03 said:
P(-1<z<1)=P(-1<z<0) + P(0<z<1)
= .0398 + .0398 = .0796
Click to expand...

You'll have to double check but I'm pretty sure this should be:

.3413 + .3413 = .6826

That is, the probability of falling within either one standard deviation above or one standard deviation below the mean should be .3413, not .0398.

EDIT: I double checked it and the .3413 is correct. The .0398 you used was actually for a Z =.10, not 1.00.
 
Last edited:
Tilted

Tilted

..
mdeclipse03 said:
B. Find the probability that x assumes a value within 1 standard deviation of its mean. Within 2 standard deviation of u

B) P(Z<-1)+P(Z>1)= 2P(Z>1)
.5 - .0398 (number from the back of my book) = .4602 = 2(.4602) = .9204


P(Z<-2)+P(Z>2)= 2P(Z>2)
.5 - .4772 (number from the back of my book) = 2(.0228) = .0456
Click to expand...

dmd196 said:
B) part one is correct; part two is incorrect. the number should be higher than the answer from B) part one because it includes two standards of deviation.
Click to expand...

I think both parts were actually wrong. His formula was calculating the probability that x would lie outside of 1 standard deviation, not within 1. Also for part one I think he misread the probability value for one standard deviation. The answers should have been .6826 and .9544 respectively.
 
PsyOps

PsyOps

Pixelated
mdeclipse03 said:
I checked my class and he said for Part B, use Example 4.15: This is what Example 4.15 tells me:

P(-1.33<z<1.33)=P(-1.33<z<0) + P(0<z<1.33)
= A1 + A2 = .4082 + .4082 = .8164

So..for the first part of the question for Part be in regards to 1, based upon that example, I should have:

P(-1<z<1)=P(-1<z<0) + P(0<z<1)
= .0398 + .0398 = .0796

From that, it should be the answer based up what the teacher said to go by. Part C is still confusing to me. For part see, would the value be .5000 since 1000 is divided in half?
Click to expand...

Ah... so this is for some sort of test. You're trying to cheat.

CHEATER!!!!!!!!! :tantrum
 
M

mdeclipse03

New Member
Tilted said:
For part C you first need to find the Z that corresponds to the 80th percentile of x's. So find the Z in the standard normal distribution table that corresponds to .3000 (that gives you the 30% of x's that are above u but below the 80th percentile in addition to the 50% that are below u, making up 80%).

Then plug that Z into your equation and solve for x:

z=(x-u)/o

For the 10th percentile you would find the Z on the table that corresponds to .4000 (the 50% of x's that are below u minus the 40% of x's that are below u but above the 10th percentile, giving you 10%). For this one though you have to use the negative of that Z. Then solve the equation for x again.


One caveat - when you look in the table, it probably won't show .3000 or .4000 exactly. One number might be .2996 and the one next to it might be .3015. Depending on how precise your teacher wants you to be, you could either select the Z that corresponds to the one that is closest or you can roughly calculate a more precise Z somewhere in between them. (I hope that makes since)
Click to expand...

Ok, I understand Parts A and B. Part C this is what i found.

The Z value of .3000 from the back of my book is .8 (left hand column) and .04 (top column). So if I combine those to I get .804

The Z value of .4000 from the book is 1.2 (left column) and .08 (top column). I combine them which is 1.28

So to solve for X I would use the formula X=u+ZO? If so that would mean 1 + .804(10) for the first part. The second part would be 1 - 1.28(10)?
 
M

mdeclipse03

New Member
PsyOps said:
Ah... so this is for some sort of test. You're trying to cheat.

CHEATER!!!!!!!!! :tantrum
Click to expand...

No its for a homework problem, each we we our assigned homework problems and after the initial submission we can correct them. The book material is not great and I am having a hard time understanding stats. I'm not asking for someone to give me the answer, I'm looking for help to understand what I have to do
 
PsyOps

PsyOps

Pixelated
mdeclipse03 said:
No its for a homework problem, each we we our assigned homework problems and after the initial submission we can correct them. The book material is not great and I am having a hard time understanding stats. I'm not asking for someone to give me the answer, I'm looking for help to understand what I have to do
Click to expand...

Cheating on homework

CHEATER!!!!!!! :tantrum















Guess I better include this :sarcasm:
 
You must log in or register to reply here.
Top