Solving for probabiliy

P

plong

New member
Joined
Oct 5, 2012
Messages
4
p(92<X<108) I know p stands for probability, am I correct in assuming 92 and 108 are the range ? And what does x stand for and how do I find it ? And is there an easy step by step instruction on how do solve these with excel as the prof. wants ? I have been out of school 30 years and only had basic algebra then, Would rather you show me where each number goes. Previous info was mean=100, SD =8, and if I get a z score for 92 and 108 then p(-1<X<1) is this right so far? How do I find x and what from there? Thanks
 
Last edited:
plong said:
p(92<X<108) I know p stands for probability, am I correct in assuming 92 and 108 are the range ? And what does x stand for and how do I find it ? And is there an easy step by step instruction on how do solve these with excel as the prof. wants ? I have been out of school 30 years and only had basic algebra then, Would rather you show me where each number goes. Previous info was mean=100, SD =8. Thanks
Click to expand...
Think of X as a numerical code that shows the result of a single experiment. Its technical name is a "random variable."

P(X = a) = b MEANS b is the probability that the code = a or, alternatively, b is the probability that the result coded as a will happen.

Let's use an example. I flip a fair coin. I code heads as 1 and tails as 0.

P(X = 0) = 0.5. It means that the probability that the code will be 0 is one half, which is the same as saying that the probability of tails from flipping a fair coin is one half. It is just a very concise way of expressing a probability. Of course, there is nothing magic in using X; you could use Y or Z just as well. Furthermore, in different problems, X can be a code for different things, a flip of a coin in one case or a roll of a die in another.

In many, many cases, the code seems quite natural. I can code the result of a 1 showing up on the roll of a die as 1, of a 2 showing up as 2, etc.

So P(92 < X < 108) MEANS the probability that the code X is bigger than 92 but less than 108. More information is required to say what that probability is. Does the problem say anything about a "distribution"?
 
Last edited:
plong said:
p(92<X<108) I know p stands for probability, am I correct in assuming 92 and 108 are the range ? And what does x stand for and how do I find it ? And is there an easy step by step instruction on how do solve these with excel as the prof. wants ? I have been out of school 30 years and only had basic algebra then, Would rather you show me where each number goes. Previous info was mean=100, SD =8, and if I get a z score for 92 and 108 then p(-1<X<1) is this right so far? How do I find x and what from there? Thanks
Click to expand...
This notation is somewhat confusing at first. You are trying to find P, not X.

Any statement in the problem about a normal distribution?
 
JeffM said:
Think of X as a numerical code that shows the result of a single experiment. It's technical name is a "random variable."

P(X = a) = b MEANS b is the probability that the code = a or, alternatively, b is the probability that the result coded as a will happen.

Let's use an example. I flip a fair coin. I code heads as 1 and tails as 0.

P(X = 0) = 0.5. It means that the probability that the code will be 0 is one half, which is the same as saying that the probability of tails from flipping a fair coin is one half. It is just a very concise way of expressing a probability. Of course, there is nothing magic in using X; you could use Y or Z just as well. Furthermore, in different problems, X can be a code for different things, a flip of a coin in one case or a roll of a die in another.

In many, many cases, the code seems quite natural. I can code the result of a 1 showing up on the roll of a die as 1, of a 2 showing up as 2, etc.

So P(92 < X < 108) MEANS the probability that the code X is bigger than 92 but less than 108. More information is required to say what that probability is. Does the problem say anything about a "distribution"?
Click to expand...

No I gave the whole question and all information provided. If I take the z score of 92 and 108 then the problem would now read p(.1587<x<.8413) but I still don't know how to solve for x.
 
You are not asked to solve for symbol X.

You are asked to determine the probability that symbol X has a value between 92 and 108.

Whatever this probability is, the symbol p(92<X<108) stands for it. In other words, find the value of symbol p(92<X<108).

If we assume that the values of X are distributed normally, then the probability graph is a bell curve, and the value you seek is the area (expressed as a percent of the whole area) under this curve between the critical z values that you already calculated (-1σ and 1σ).

You need to look-up this area (stated as a percent of the whole), using your software or a table.

Here is an interactive graph at which you may enter your mean and standard deviation, followed by dragging the shaded area to encompass -1σ to 1σ (as close as you can, anyway) for an estimate of the probability you seek. Use it as a check.

Cheers :cool:
 
Last edited:
plong said:
No I gave the whole question and all information provided. If I take the z score of 92 and 108 then the problem would now read p(.1587<x<.8413) but I still don't know how to solve for x.
Click to expand...

\(\displaystyle Given\ \mu = 100\ and\ \sigma = 8,\ P(92 < X < 108) = P\left(\dfrac{92 - 100}{8} < z < \dfrac{108 - 100}{8}\right) = P(- 1 < z < + 1).\)

You were right on the z scores the first time.

You are solving for P, not X.

If the distribution is normal, then you can look up this probability in a table of the normal distribution. That table is built into excel, but you can use it only if you know that the distribution is normal. However, you do not need a table or excel for this particular situation, you can just memorize the values

\(\displaystyle Given\ normally\ distributed\ z\ scores,\ P(-1 < z < 1) \approx 68\%,\ P(-2 < z < 2) \approx 95\%,\ and\ P(-3 < z < 3) \approx 99\%.\)

But you first have to know that the distribution is normal. You do not seem to have that information so no answer is possible.

EDIT: If the distribution is normal: \(\displaystyle Your\ 0.1587 \approx P(z \le - 1)\ and\ your\ 0.8413 \approx P(z < 1).\)

So, if the distribution is normal: \(\displaystyle P(- 1 < z < 1) = P(z < 1) - P(z < - 1) \approx 0.8413 - 0.1587 = 0.6826 \approx 68\%.\)
 
Last edited:
And, after you understand what's going down, here's some general info (i.e., some rules to estimate percentages within 1, 2 or 3 standard deviations of the mean) to memorize. See the second graph. :cool:
 
mmm4444bot said:
You are not asked to solve for symbol X.

You are asked to determine the probability that symbol X has a value between 92 and 108.

Whatever this probability is, the symbol p(92<X<108) stands for it. In other words, find the value of symbol p(92<X<108).

If we assume that the values of X are distributed normally, then the probability graph is a bell curve, and the value you seek is the area (expressed as a percent of the whole area) under this curve between the critical z values that you already calculated (-1σ and 1σ).

You need to look-up this area (stated as a percent of the whole), using your software or a table.

Here is an interactive graph at which you may enter your mean and standard deviation, followed by dragging the shaded area to encompass -1σ to 1σ (as close as you can, anyway) for an estimate of the probability you seek. Use it as a check.

Cheers :cool:
Click to expand...

Thank you I will try this, I know I am solving for p but I have to know what x is to get the answer for p.
 
plong said:
Thank you I will try this, I know I am solving for p but I have to know what x is to get the answer for p.
Click to expand...
You are told that 92 < X < 108. That is ALL you know about X, and all you NEED to know. You cannot "solve" for X in the sense that you learned in algebra.

In continuous distributions such as the normal distribution, the probability that X is some specific number is always 0. The measures of probability are that

\(\displaystyle P(X < a) = P(X \le a)\ or\ P(X > b) = P(X \ge b)\ or\ P(c < X < d) = P(c \le X < d) = P(c < X \le D) = P(c \le X \le d).\)

You are always dealing with ranges of the random variable in a continuous distribution, and here the range is given. Nothing to find about the random variable.
 
Last edited:
plong said:
I have to know what x is to get the answer
Click to expand...

No, you don't. This is one of the nice things about normally-distributed data; we don't need to know any particular value from the data (i.e., any particular value of X) in order to estimate how much of the data lie within one standard deviation. We do not even need to know what symbol X represents in the real world.

Simply knowing that X is distributed normally is sufficient for us to say that there's a 68% chance (roughly) of picking a particular X with a value within one standard deviation of the average. That is a general rule for estimating because all normally-distributed data fit the same bell curve.

You understand that 92 is one standard deviation below the mean and 108 is one standard deviation above the mean, yes?

Some X values are less than 92, some are more than 108. The rest (about 68% of the total) are in between. We express this fact by writing:

p(92<X<108) = 0.68

Me thinks that you need a refresher course of study. :cool:
 
mmm4444bot said:
No, you don't. This is one of the nice things about normally-distributed data; we don't need to know any particular value from the data (i.e., any particular value of X) in order to estimate how much of the data lie within one standard deviation. We do not even need to know what symbol X represents in the real world.

Simply knowing that X is distributed normally is sufficient for us to say that there's a 68% chance (roughly) of picking a particular X with a value within one standard deviation of the average. That is a general rule for estimating because all normally-distributed data fit the same bell curve.

You understand that 92 is one standard deviation below the mean and 108 is one standard deviation above the mean, yes?

Some X values are less than 92, some are more than 108. The rest (about 68% of the total) are in between. We express this fact by writing:

p(92<X<108) = 0.68

Me thinks that you need a refresher course of study. :cool:
Click to expand...

Me thinks me don't need attitude, thanks. I have been out of high school for over 30 yrs, now in a 5 week crash course on statistics. I have been studying thank you. :-|
 
You must log in or register to reply here.
Top