#1

Dear Pit,

I have a maths test tomorrow on probability, and I was wondering if anyone here could help me understand Inverse Normal Problems using bell-curves. I understand Normal Problems, all I have to do is find the upper and lower values, and we are told the standard deviation and mean so that is not a problem, but I suck at the Inverse's. Instead of finding a probability, with the Inverse Normal Problems you've got to find the value of a percentage. So you have to find the area, standard deviation and mean. I have no idea how to do these, I can't find the values of a probability I can only find the probability.

Here's an example of a question I'm trying to answer:

"Find the random variable 'x' for the normal distribution (mean= 63.4 and standard deviation= 15.6)"

P(X>x) = 0.2177"

Please help, it would be dearly appreciated .

I have a maths test tomorrow on probability, and I was wondering if anyone here could help me understand Inverse Normal Problems using bell-curves. I understand Normal Problems, all I have to do is find the upper and lower values, and we are told the standard deviation and mean so that is not a problem, but I suck at the Inverse's. Instead of finding a probability, with the Inverse Normal Problems you've got to find the value of a percentage. So you have to find the area, standard deviation and mean. I have no idea how to do these, I can't find the values of a probability I can only find the probability.

Here's an example of a question I'm trying to answer:

"Find the random variable 'x' for the normal distribution (mean= 63.4 and standard deviation= 15.6)"

P(X>x) = 0.2177"

Please help, it would be dearly appreciated .

#2

Isn't it just 1-what the value on the normal distribution table is?

My stats is a bit rusty, but as i remember, first you wuld standardize the mean and SD, and then find the value of 1-0.2177 on the table, then find out how that P relates to your mean and SD. Like, the value with P(1-0.2177) would be mean+(SD.y) and that would be x.

Yea?

My stats is a bit rusty, but as i remember, first you wuld standardize the mean and SD, and then find the value of 1-0.2177 on the table, then find out how that P relates to your mean and SD. Like, the value with P(1-0.2177) would be mean+(SD.y) and that would be x.

Yea?

#3

Isn't it just 1-what the value on the normal distribution table is?

My stats is a bit rusty, but as i remember, first you wuld standardize the mean and SD, and then find the value of 1-0.2177 on the table, then find out how that P relates to your mean and SD. Like, the value with P(1-0.2177) would be mean+(SD.y) and that would be x.

Yea?

Thanks for the quick response.

How do I standardize the mean and SD? Sorry if that's a stupid question, but I'm really confused.

#4

Bump.

Sorry if there are rules against bumping or w/e, but I really need to know this.

Sorry if there are rules against bumping or w/e, but I really need to know this.

#5

Thanks for the quick response.

How do I standardize the mean and SD? Sorry if that's a stupid question, but I'm really confused.

Ok i just re-read your first question and by 'Normal Problems' do you mean the normal distribution? If yes, then to standardize a x value you have to perform a Z-transformation: Z = (X-mean)/SD

So if you had a normal distribution but didn't know that values on a table, you would standardize the X data value then read the probability off the standard normal distribution table which almost all exams give you.

The trick with your question is that you aren't told the X data value, so you have to work backwards.

Okay, so you know what a normal curve looks like yea? And in the Q it asks for a value that has a probability of 0.2177 above that value. So, logically that value has 1-0.2177 = 0.7823 P below it.

So, your standardized value Z has to equal something with a standard normal probability of 0.7823. Look at the table, it's 0.78.

So, (X-mean)/SD has to equal 0.78.

Plug in the numbers, X = 75.568

Yea?

#6

Ok i just re-read your first question and by 'Normal Problems' do you mean the normal distribution? If yes, then to standardize a x value you have to perform a Z-transformation: Z = (X-mean)/SD

Oops, yeah sorry, I meant Normal Distribution. So by the Z-transformation, you're saying that I divide the mean by the SD and then what?

So if you had a normal distribution but didn't know that values on a table, you would standardize the X data value then read the probability off the standard normal distribution table which almost all exams give you.

The trick with your question is that you aren't told the X data value, so you have to work backwards.

Okay, so you know what a normal curve looks like yea? And in the Q it asks for a value that has a probability of 0.2177 above that value. So, logically that value has 1-0.2177 = 0.7823 P below it.

So, your standardized value Z has to equal something with a standard normal probability of 0.7823. Look at the table, it's 0.78.

So, (X-mean)/SD has to equal 0.78.

Plug in the numbers, X = 75.568

Yea?

Yeah, that's the right answer, I just don't get that part about the Z-transformation. Thanks for the help so far, you're a lifesaver.

#7

Oops, yeah sorry, I meant Normal Distribution. So by the Z-transformation, you're saying that I divide the mean by the SD and then what?

Yeah, that's the right answer, I just don't get that part about the Z-transformation. Thanks for the help so far, you're a lifesaver.

Ach should have made that clearer. Take the X data value, subtract the mean, divide by SD. Z = (X - [mean])/SD

The big trick in that question was working backwards. Usually they'll be like, what is the probability of X>x given mean = 325 and SD = 24. But in this case it was a matter of working back from the table.

But seriously, you should have already learnt all of this quite thoroughly if you have a test on it tomorrow? Slackness on your part to practice methinks

Anyway, no problem

#8

Ach should have made that clearer. Take the X data value, subtract the mean, divide by SD. Z = (X - [mean])/SD

The big trick in that question was working backwards. Usually they'll be like, what is the probability of X>x given mean = 325 and SD = 24. But in this case it was a matter of working back from the table.

But seriously, you should have already learnt all of this quite thoroughly if you have a test on it tomorrow? Slackness on your part to practice methinks

Anyway, no problem

Ok thanks. Yeah, a bit of slackness, but also I don't really need to know these things, the teacher said it's very unlikely that these questions will come up, but I decided to learn them just in case.

Thanks for all your help.