Normal Distribution

[desert tendon]

This whole question stumped me. I believe that for a) I had to standardise the norma distribution by using the formula (mean - mu) / standard deviation = Z?

For b) I think I just do P(X < 13) or P(X <= 13), plug values of mu and standard deviation into my calculator to find the value and multiply by 100 for percentage

For c) I'm gonna be honest, I'm not sure how to approach that at all

Finally for d) I think I just compare my mean value from c) and see if it's less than 16 minutes?

[velvet yew]

1. Ask your question and show the work you've done so far. If you've posted a screenshot of a question, specify which part you need help with.
2. Wait patiently for a helper to come along.
3. Once someone helps you, say thank you and close the thread with:

   +close
   

4. Feel free to nominate the person for helper of the week in #helper-nominations
5. Do not ping the mods, unless someone is breaking the rules.
6. If you're happy with the help you got here, and the server overall, you can contribute financially as well:

[keen lava]

(a). We are given P(X > 22) = 0.1, where X ~ N(μ = 16, σ^2), so you can find σ from here.
(b). Correct.
(c). We are given a system of equatiions in the following general form (X ~ N(μ, σ^2)):
P(X > x1) = p1
P(X < x2) = p2
You can find μ and σ from the system.
(d). Yes.

[desert tendon]

okay, I understand a), but for c), I'm confused, I have to find the mean and standard deviation for both system of equations? So they have to satisfy both of them, which means I have to form simultaneous equations?

maybe I should standardise both systems of equations into the form X ~ N(0 , 1^2), find the Z values for both and plug them into (mean - mu) / standard deviation = Z and solve for mean and standard deviation simultaneously?

oh wait that won't work because I don't know the actual mean to figure out my Z value... yeah I'm still stuck on c)

[keen lava]

Note that P(X < x) = F(x), and F(x) can be expressed in terms of erf(x) or Φ(x), depending on whether you want to solve this numerically or using a table.
In the end you'll get a system of linear equations.

[desert tendon]

isn't Φ(x) the same as P(Z = z) or something like that?

[keen lava]

No. Φ(x) is Laplace's function: Φ(x) = 1/2 + F((x - μ)/σ).
Here's a table of its values if you need it.

[desert tendon]

we haven't learned that yet, is there not a simpler approach to this? We've only been taught these for Normal Distribution

[keen lava]

Oh. Well, it's basically the same thing, just without adding 1/2. Less comfortable to use, to be honest.

[desert tendon]

okay so, I just find the z value equivalents and solve for µ and standard deviation simultaneously?

they're in the second image so it's quicker to just plug them in the formula on the first image

[keen lava]

Yes. Though, you might want to solve the system generally first.

[desert tendon]

generally?

[keen lava]

Yes, generally. Without plugging in values.

[desert tendon]

oh, so you mean using inverse normal distribution to find the z-value equivalents?

[keen lava]

No, no. Why inverse normal distribution? That's not needed here.

Let me show how you can start.

Do you prefer using error function or Laplace's function?

[desert tendon]

umm, those two things are way out of my knowledge scope

[keen lava]

Oh, you want to do it like this?

So, using the CDF of a RV with standard normal distribution z(p).

[desert tendon]

yeah

what does RV stand for?

[keen lava]

Alright. We have:
P(X > x1) = p1
P(X < x2) = p2
As P(X < x) = F((x - μ)/σ):
F((x1 - μ)/σ) = 1 - p1
F((x2 - μ)/σ) = p2
Taking the inverses, we get:
(x1 - μ)/σ = z(1 - p1)
(x2 - μ)/σ = z(p2)
And it should be easy from here.

[keen lava]

As usual, random variable.

[desert tendon]

oh, yeah

[desert tendon]

okay, I think I get it, I'll solve it now and see what I get

[keen lava]

Note, however, that this table is written not for z(p), but for z(1 - p) for some reason. Maybe that's to avoid negative values, since z(1 - p) = -z(p) for any p between 0 and 1.

[desert tendon]

okay, rounded to 3 s.f. I got σ = 36.2 and µ = -11.6 , I'm not sure about mu though, can it be negative?

[keen lava]

Well, in general - yes, but certainly not in this case.
What expressions did you get for μ and σ?

[desert tendon]

getting rid of the fractions, I got 17 - µ = 0.78814...σ for the first equation and 8 - µ = 0.5398...σ for the second equation

[keen lava]

Oh...

No, you shouldn't do it like that.

Until you derive the expressions for what you want to find, forget about the values.

Find the general formulas first.

[desert tendon]

so something like P(X > 17) = 0.2 first?

[keen lava]

No, you're still using values.

[desert tendon]

oh I get what you mean

[keen lava]

We have the following system:
(x1 - μ)/σ = z(1 - p1)
(x2 - μ)/σ = z(p2)
Express μ and σ from here.

[desert tendon]

yeah

okay, so µ = x1 - z(1-p1)σ and σ = (x1 - µ)/z(1-p1) for the first system, µ = x2 - z(p2)σ and σ = (x2 - µ)/z(p2) for the second system?

[keen lava]

What do you mean "for the first/second system"? There's only one system.

[desert tendon]

oh right, I mean, µ = x1 - z(1-p1)σ and σ = (x1 - µ)/z(1-p1) for the first RV and µ = x2 - z(p2)σ and σ = (x2 - µ)/z(p2) for the second RV?

or is RV not the right term?

[keen lava]

No. I think you're misunderstanding me.

This is a system of two equations.

All you've done is rearrange some terms, but this won't allow you to find μ and σ.

You need to express them.

Expressing means you are only left with constants and variables that you know the value of. In this case that's p1 and p2.

We can multiply both sides by σ and rearrange a bit, giving:
μ + z(1 - p1)σ = x1
μ + z(p2)σ = x2
Now you can easily express σ. Try doing that.

[desert tendon]

okay, I did the first equation - second equation, leaving me with z(1-p1)σ - z(p2)σ = x1 - x2, then I factorised the LHS, which is [z(1-p1) - z(p2)]σ = x1 - x2, and finally I divided both sides by [z(1-p1) - z(p2)], giving me σ = (x1 - x2) / z(1-p1) - z(p2) and this should be right since it's expressed in variables that I know the values of this time

[keen lava]

Yeah, nice! So, σ = (x1 - x2)/(z(1 - p1) - z(p2)).
Now, you can, of course, find μ by substituting σ into one of the equations. But can you see how you can do it in a similar way to how σ has been found?

[desert tendon]

yes

this is weird, I got σ = 36.2, which is fine, but I got µ = -2.57 this time...

rounded to 3 s.f.

[keen lava]

Did you remember that the table is written for z(1 - p), not z(p)?

[desert tendon]

yeah, but isn't z(1 - p), in this case, z(0.8)?

because p1 was 0.2

oh wait, I should've done 1 - 0.8416

[keen lava]

I'm getting the following:
σ = (x1 - x2)/(z(1 - p1) - z(p2)) = (x1 - x2)/(z(1 - p1) + z(1 - p2)) = (17 - 8)/(z(1 - 0.2) + z(1 - 0.1)) ≈ 9/(0.8416 + 1.2816) ≈ 4.24

[desert tendon]

ah, i see where I went wrong

I kept other one as z(0.1)

[keen lava]

What expression did you get for μ?

[desert tendon]

I don't think I got an expression, once I got σ, I substituted back to one of either μ + z(1 - p1)σ = x1 or
μ + z(p2)σ = x2

[desert tendon]

actually, why did the signs change when you had -z(p2) and then you wrote z(1-p2) afterwards?

because when I tried calculating σ, I didn't do that

[keen lava]

Normal distribution is symmetric, so its quantiles are symmetric around 0. As I mentioned above, for any p between 0 and 1 we have z(1 - p) = -z(p).

[keen lava]

μ + z(1 - p1)σ = x1
μ + z(p2)σ = x2
Try dividing the first equation by z(1 - p1) and the second by z(p2). What do you see?

[desert tendon]

I see that (µ + z(1 - p1)σ)/z(1-p1) = x1/z(1 - p1) and (µ + z(p2)σ)/z(p2) = x2/z(p2)

but I'm not sure what that implies

[keen lava]

Well, if you divide fully, you get:
μ/z(1 - p1) + σ = x1/z(1 - p1)
μ/z(p2) + σ = x2/z(p2)

So, what can you do now?

[desert tendon]

ohh, the first equation subtract the second equation and then rearrange to find µ

[keen lava]

Yeah, nice!

Don't forget to apply -z(p2) = z(1 - p2) as before.

[desert tendon]

rounded to 3 s.f. I got µ = 12.6

[keen lava]

Hm, I'm getting a different value.

Can you show how you calculated it?

[keen lava]

One of the quantiles isn't correct. We have:
μ = (x1/z(1 - p1) - x2/z(p2))/(1/z(1 - p1) - 1/z(p2)) = (x1/z(1 - p1) + x2/z(1 - p2))/(1/z(1 - p1) + 1/z(1 - p2)) = (17/z(1 - 0.2) + 8/z(1 - 0.1))/(1/z(1 - 0.2) + 1/z(1 - 0.1)) ≈ (17/0.8416 + 8/1.2816)/(1/0.8416 + 1/1.2816) ≈ 13.4

Also, again, better to express μ first, then substitute the values.

[desert tendon]

yeah, I hesitated to do that because I thought it would look messy

[keen lava]

Oh, also.

Why did you take quantiles with different numbers of significant digits on left and right side?

[desert tendon]

here?

[keen lava]

Yes.

[desert tendon]

I didn't want to write it all out again so I put dots at the end of the other ones lol

I entered everything in my calculator anyways

[keen lava]

Oh.

Well, alright.

[desert tendon]

so I think that's it since you gave me both the mean and standard deviation

but I think I understand everything of what you explained to me

[keen lava]

Nice! You can try changing the signs in the initial equations around to see how that changes things.

[desert tendon]

alright cool, I'll experiment with that a bit later, but thank you so much!

[frosty helm]

@desert tendon has given 1 rep to @keen lava

[keen lava]

You're welcome!

I do recommend learning about error function and Laplace's function, though.

As they are both odd, it may be a bit easier to use them in these problems.

The latter also works if you want to use a table like I showed above.

[desert tendon]

it's just that they look quite advanced than what I'm used to right now, espcially the error function

but I'll learn them anyways

[keen lava]

Oh, by the way.

Hold on, one sec. I have a table that may be useful.

Here you go. This is the relationship between three functions for normal distribution: CDF F(x), Laplace's function Φ(x) and error function erf(x).

As you can see, they are all very similar.

[desert tendon]

+close