# Math Help - Hi again, new here, I am completely lost on a problem, have an attachment

1. ## Hi again, new here, I am completely lost on a problem, have an attachment

I really hope the attachment went through. I made it in Paint on my computer and while it is the worst Paint effort ever, I was just trying to give you an idea of what I am dealing with. In case the attachment does not go through, I will explain it anyway, hoping you can visualize it; if it does go through, I will explain anyway since it is the crudest of drawings.

I have a normal distribution diagram here, with 5 vertical lines left to right (instead of the standard normal distribution diagram with the single vertical line to mark the location of the mean). The five lines are marked, from left to right:
X1, X2, X3, X4, X5.

And, I am given percentages corresponding to each X, also from left to right, goes like this:

3%, 23%, 20%, 30%, 7%

The X's and the percentages, I could not include in my attachment.

The problem given to me is this: "For a normal population distribution with mean of 76 and standard deviation of 25, find the values of X1 through X5. I am told to draw a diagram for each caculation to illustrate the problem,and write the correct formula used in the calculation."

I am trying to figure out how to get started. I want to say that I should begin with finding the farthest left X, which is X1 which has the corresponding percentage of 3%. So I would go into my z-table and look for a z-score that corresponds to 3%, correct? I really don't want someone to do the whole thing for me, but I guess if the process of finding the value of all five X's is the same, maybe just walk me through getting one? (That is, if the process IS the same for each?) I hope this is not too much to ask. Actually I really hope that drawing I made in Paint works as an attachement here.

2. Originally Posted by peanutbutter

I really hope the attachment went through. I made it in Paint on my computer and while it is the worst Paint effort ever, I was just trying to give you an idea of what I am dealing with. In case the attachment does not go through, I will explain it anyway, hoping you can visualize it; if it does go through, I will explain anyway since it is the crudest of drawings.

I have a normal distribution diagram here, with 5 vertical lines left to right (instead of the standard normal distribution diagram with the single vertical line to mark the location of the mean). The five lines are marked, from left to right:
X1, X2, X3, X4, X5.

And, I am given percentages corresponding to each X, also from left to right, goes like this:

3%, 23%, 20%, 30%, 7%

The X's and the percentages, I could not include in my attachment.

The problem given to me is this: "For a normal population distribution with mean of 76 and standard deviation of 25, find the values of X1 through X5. I am told to draw a diagram for each caculation to illustrate the problem,and write the correct formula used in the calculation."

I am trying to figure out how to get started. I want to say that I should begin with finding the farthest left X, which is X1 which has the corresponding percentage of 3%. So I would go into my z-table and look for a z-score that corresponds to 3%, correct? I really don't want someone to do the whole thing for me, but I guess if the process of finding the value of all five X's is the same, maybe just walk me through getting one? (That is, if the process IS the same for each?) I hope this is not too much to ask. Actually I really hope that drawing I made in Paint works as an attachement here.
Your percentages make no sense. They should be getting larger as x varies from $x_1$, $x_2$, $x_3$, ...... if things are cumulative .....

You have $\Pr(X < x_1) = 0.03$. So you have to use an inverse normal approach. I'd need to see your tables to know how to use them to find $x_1$ .....

For example, on my tables (which are the standard normal distribution) I can only get values of z to the right of the mean ......

So I first have to find the value of z such that $\Pr(Z {\color{red}>} z) = 0.03 \Rightarrow \Pr( Z < z) = 0.97 \Rightarrow z = 1.881$.

So the value of $x_1$ coresponds to -1.881 (because I need to be on the left hand side of the mean, not the right).

Then $Z = \frac{X - \mu}{\sigma} \Rightarrow -1.881 = \frac{x_1 - 76}{25} \Rightarrow x_1 = 28.975$ .......

Once your values of X are to the right hand side of the mean, I can use the value of z read directly from my tables .....

3. ## Mr Fantastic, can I ask...

I got a pop-up that said I had a private message from you with the title Avoid an Infraction but my pop up blocker was off. Now it is on. I didn't get the message

I am also editing this to say that the z-table at the back of my book also shows values 0 to z to the right of the mean

4. Originally Posted by peanutbutter
I got a pop-up that said I had a private message from you with the title Avoid an Infraction but my pop up blocker was off. Now it is on. I didn't get the message

I am also editing this to say that the z-table at the back of my book also shows values 0 to z to the right of the mean
I see that you have now

So you're OK now with how to get the other x's .....? (I'd suspected from your other question that your tables were as you've said).

And do you understand what I was saying about the percentages needing to increase ....?

5. i hope i'm not a bother but in your explanation, what is the little z and big Z?

6. Originally Posted by peanutbutter
i hope i'm not a bother but in your explanation, what is the little z and big Z?
Z is the usual symbol used to represent the standard normal distribution. This is the distribution your tables give you.

z is the value of Z that satisfies Pr(Z > z) = 0.03. In other words, you have to find the value of Z such that the probability of Z being larger than that value is 0.03. I could just as well used a diferent symbol such as z*, a, etc.

7. Ok I think I have my answers now. Hoping I'm right.

P(X<X1)=0.03
P(Z>z)= 0.03 = P(Z<z) = 0.47 z=-1.88
-1.88= x1-75 / 25
(-1.88) 25 + 76=x1
x1= 29

The next line segment has 23%, but I add that to 3% to get 26, so what I am looking for is at what value do 26 percent of observations lie below, not 23, right? So my question is when I write the probability expression on the first line, am I saying at what value of x2 does 26 percent of observations lie below, or do I say P(X1<x<X2)? I just put the first

P(X<X2)= 0.26
P(Z>z) = 0.26 P(Z<z)=0.24 z= -0.65
-0.65= x2-76 / 25= x2
(-0.65)25 + 76= x2
x2=59.75

Now the next line segment is 20%, I add that to the two below, so 3% + 23% + 20%= 46% so I'm looking at this point for P(X<X3)=0.46. Now I am lost, do I subtract that from 0.5 as I did the others, to find the percentage to look for in the normal table? Because I would be looking for 0.04. Or do I just look for 0.46 in the table?

8. Originally Posted by peanutbutter
Ok I think I have my answers now. Hoping I'm right.

P(X<X1)=0.03
P(Z>z)= 0.03 = P(Z<z) = 0.47 z=-1.88
-1.88= x1-75 / 25
(-1.88) 25 + 76=x1
x1= 29

Mr F says: 29 is close enough within the accuracy of your tables. I get 28.98.

The next line segment has 23%, but I add that to 3% to get 26, so what I am looking for is at what value do 26 percent of observations lie below, not 23, right? So my question is when I write the probability expression on the first line, am I saying at what value of x2 does 26 percent of observations lie below,

Mr F says: That would seem a reasonable interpretation of the question as posted.

or do I say P(X1<x<X2)?

Mr F says: I wouldn't think so.

I just put the first

P(X<X2)= 0.26
P(Z>z) = 0.26 P(Z<z)=0.24 z= -0.65
-0.65= x2-76 / 25= x2
(-0.65)25 + 76= x2
x2=59.75

Mr F says: 59.75 is close enough within the accuracy of your tables. I get 59.92.

Now the next line segment is 20%, I add that to the two below, so 3% + 23% + 20%= 46% so I'm looking at this point for P(X<X3)=0.46.

Mr F says: Again, this seems reasonable.

Now I am lost, do I subtract that from 0.5 as I did the others, to find the percentage to look for in the normal table? Because I would be looking for 0.04. Or do I just look for 0.46 in the table?
X3 is still to the left of the mean so you obviously do it like the others.

In particular, I'd advise setting it out like I did in post #2.

9. OK I promise this is the last post for this. It's just taking me forever to get the logic of the distribution into my head. I give the term dumb blonde a whole new meaning LOL

Finding X3

P(X<X3)= 0.46
P(Z>z)=0.46 --> P(Z<z)=0.04 --> z= -0.10
(-0.10) 25 + 76 = X3
X3=73.5

X4 is 3% + 23% + 20% + 30%= 76%, so

P(X<X4)=0.76
P(Z>z)=0.76 -->P(Z<z)= 1-0.76)= 0.24--> z=0.65
(0.65) 25 + 76= X4
X4= 92.25
(Not sure on this X4, I subtracted the prob. from 1 because I know I'm above the mean by now, and if I subtracted 0.76 from 0.5 it wouldn't look pretty.)

X5 is 3% + 23% + 20% + 30% + 7%= 83%, so

P(X<X5)=0.83
P(Z>z)=0.76 --> P(Z<z) = 1-0.83=0.17 --> z= 0.44
(0.44)25 + 76= X5
X5=87

That looks wrong. Then I thought maybe I *should* subtract from 0.5 still. So 0.5- 0.76= -0.26, which has a z-score in my table of 0.71. And I just ignore the negative sign competely. (the negatives confuse me). So to plug that in I would get X4 = (0.71)25+ 76= 93.75

Then for X5 maybe I should just subtract 0.83 from 0.5. So 0.5-0.83= -0.33. Zscore= 0.96. X5=(0.96)25 + 76= 100. I guess that makes more sense.

So like I promised I won't post on this one anymore but do I have the right final answers? *crosses fingers*

10. Originally Posted by peanutbutter
OK I promise this is the last post for this. It's just taking me forever to get the logic of the distribution into my head. I give the term dumb blonde a whole new meaning LOL

Finding X3

P(X<X3)= 0.46
P(Z>z)=0.46 --> P(Z<z)=0.04 --> z= -0.10
(-0.10) 25 + 76 = X3
X3=73.5

Mr F says: Close enough within the accuracy of your tables. I get 73.49.

X4 is 3% + 23% + 20% + 30%= 76%, so

P(X<X4)=0.76
P(Z>z)=0.76 -->P(Z<z)= 1-0.76)= 0.24--> z=0.65
(0.65) 25 + 76= X4
X4= 92.25
(Not sure on this X4, I subtracted the prob. from 1 because I know I'm above the mean by now, and if I subtracted 0.76 from 0.5 it wouldn't look pretty.)

Mr F says: No. Pr (X < X4) = 0.76 means you need to find the z-value, z*, such that Pr(z < z*) = 0.76. Since now you're clearly to the right of the mean, you should be able to get it straight off your tables: I get z = 0.707 ........ Now convert it: 0.707 = (X4 - 76)/25 etc. to get the value of X4.

X5 is 3% + 23% + 20% + 30% + 7%= 83%, so

P(X<X5)=0.83
P(Z>z)=0.76 --> P(Z<z) = 1-0.83=0.17 --> z= 0.44
(0.44)25 + 76= X5
X5=87

That looks wrong.

Mr F says: It is. It's wrong because your z-value is no good. Read my earlier remarks. Pr(X < X5) = 0.76 requires z* such that Pr(Z < z*) = 0.76 ...... you should get z = 0.954. As a check, calculate Pr(z < 0.954) , you should get back 0.83 ..... (Check above the same way).

[snip]

So like I promised I won't post on this one anymore but do I have the right final answers? *crosses fingers*
You should be able to get X5 now. Let me know how it goes (so you'll have to break your promise)

Originally Posted by peanutbutter
OK I promise this is the last post for this. It's just taking me forever to get the logic of the distribution into my head. I give the term dumb blonde a whole new meaning LOL
'Dumb blondes' are my favourite type (as long as you're XX) .... but I always suggest they dye their hair to get artificial intelligence (re: other post ..... A shame that you're crunchy, though )

11. P(X<X4)=0.76
P(Z>z)=0.76 --> P(Z<z)=0.26--> z=0.71
(0.71)25 + 76= X4
X4= 93.75

P(X<X5)=0.83
P(Z>z)=0.83--> P(Z<z)=0.33--> z= 0.95
(0.95)25 + 76= 99.75

There you go, can I go in the hall of fame of stupid newbie posters?? And to think all the time I spent on this one and I have 8 others to do of the same 5-part format. I realize that in my previous post where I was finding X5 that I put P(Z>z)=0.76 when I should have put 0.83- I guess I was looking at X4 when I did that by accident - slippery keys or something. I also realize my problem was that if I get a probability of say 0.83 I go to the normal table thinking I can look up 0.83 but it is only from 0000 to 4998, so I need to do 0.83-0.5 to get 0.33, which I can find in the normal table. I just don't know if I should add the 0.5 back that I subtracted, if not I give up because I'm going to go crawl into a hole now and work on my other problems. There's a reason they say girls aren't as good at math as guys, I am exhibit A lol

12. Originally Posted by peanutbutter
P(X<X4)=0.76
P(Z>z)=0.76 --> P(Z<z)=0.26--> z=0.71 Mr F says: A couple of corrections (they might be typoes you made): You want Pr(z < z) = 0.76. I'm not sure what the other stuff is all about(subtracting the 0.5 etc.) - it seems that I don't know what your tables look like after all.

(0.71)25 + 76= X4
X4= 93.75 Mr F says: Close enough given the accuracy of your tables. I get 93.66.

P(X<X5)=0.83
P(Z>z)=0.83--> P(Z<z)=0.33--> z= 0.95
(0.95)25 + 76= 99.75

Mr F says: *Ahem* Pr(Z < z) < 0.83 => z = 0.95. You final answer is correct (within the accuracy of your tables).

There you go, can I go in the hall of fame of stupid newbie posters??

Mr F says: There might not be enough room .... it's very crowded.

And to think all the time I spent on this one and I have 8 others to do of the same 5-part format.

Mr F says: If you have learned something from this thread then the other questiosn you have might not be so bad.

I realize that in my previous post where I was finding X5 that I put P(Z>z)=0.76 when I should have put 0.83- I guess I was looking at X4 when I did that by accident - slippery keys or something. I also realize my problem was that if I get a probability of say 0.83 I go to the normal table thinking I can look up 0.83 but it is only from 0000 to 4998,

Mr F says: My tables give 0.5000 to 0.9998 so I'm not sure what your tables look like. You got X5 correct though so it looks like you're on the right track (you took my advice about the hair dye, right ....? )

so I need to do 0.83-0.5 to get 0.33, which I can find in the normal table. I just don't know if I should add the 0.5 back that I subtracted, if not I give up because I'm going to go crawl into a hole now and work on my other problems. There's a reason they say girls aren't as good at math as guys, I am exhibit A lol
Perhaps exhibit B (for Blonde ). Most of the exhibits in that Hall of Fame are guys, you know .....

Feel free to post any future problems. You might yet become smooth peanutbutter ......