Maths/Stats problem!

[Marked 197]

So I got a finance exam coming up. Actually this isn't extremely important, but I think it's probably something I got to do. One of the valuation formulas we gotta learn involves using the standard normal probably table. You know that thing where you got a z value (say 1.12) and you look on the left side for 1.1 then you look along the top for 0.02 and find the p value aka the probability. 

 

 

Well in this exam our z values are greater than 2 decimal places... It's more like 1.1264. The true p value lies somewhere between 2 of the probabilities listed in the column (if its 1.1264 then its something between the 0.02 and 0.03, ie between 0.3686 and 0.3708. We're given this table. 

 

So my question, how do I find the correct the p value using these inputs? In my class it was referred to as a linear approximation, but then we weren't ever told how to do this. Actually its not that important because it will only change the answer by a few decimals and its more about understanding that calculations. But still... this is really bugging me coz i can't figure it out or find the answer on google. I hope someone here can T.T

[black mage 28]

So, you want to find a value of p between some scope of range, right?

 

Let's make an example for a p value between 1.12 and 1.13.

 

Let say that we have a linear function f(x), defined as f(x) = mx + c.

Let x be 1.12 and 1.13.

Thus, for x = 1.12, we have  f(x) = 0.3686 (by looking up to the table given above).

And, for x = 1.13, we have  f(x) = 0.3708 (also by looking up to the table given above).

Substitute the x and f(x) to the equation, hence we got 2 linear equation :

0.3686 = m*1.12 + c and

0.3708 = m*1.13 + c

 

then by multiplying both sides of 0.3686 = m*1.12 + c with (-1) and we'll get -0.3686 = -m*1.12 - c

Add the -0.3686 = -m*1.12 - c into 0.3708 = m*1.13 + c and we get 0.0022 = m*0.01

Thus, we have m = 0.22

 

Now substitute m = 0.22 into our linear equation, thus we'll have

0.3686 = 0.22*1.12 + c and

0.3708 = 0.22*1.13 + c

 

Do the operation, and we'll end up with :

0.3686 = 0.2464 + c and

0.3708 = 0.2486 + c

And we can get the value of c easily, which is 0.1222

 

Substitute the m and c that we've got into the linear function that we define on top of my post ;) and we'll get :

f(x) = 0.22*x + 0.1222

 

Or in this case, p = 0.22*x + 0.1222

So, if you want to find the value of p for x = 1.1264, simply substitute the x into the function that we have above.

p for x = 1.1264 is = f(1.1264) = 0.22*1.1264 + 0.1222 = 0.37008

 

That's how you solve it.

 

Note that the function is work only if the value of x is between 1.12 and 1.13. If you want to find the value for another range, you need to rework everything from the top.

 

If you're in doubt, trust me, I'm a Mathematician. A man that can blow up some heads by just talking :shifty:

Edited June 9, 2015 by black mage

[zahraa 26]

I think there is a problem with the function above (f(x) = 0.22*x + 0.1222)  The normal format for linear function is f(x) = mx + c but if you want to write the function for that table, c is 0 because for the first data, x is 0 and f(x) is also 0 so c must be 0 too.

[black mage 28]

No, it's not a function for that table. It's a function to approximate the p value between 1.12 and 1.13 (looks like I forgot to mention this on my previous post). If you want to create a function that can cover the whole table, we need another method called regression, which I won't talking about since we're gonna stray a bit from the topic.

 

If we can create the function for the whole table by just using 2 value like what I done in the above, what's the point of the whole table then :P

Edited June 9, 2015 by black mage

[zahraa 26]

oh I see! Thanks for the explanation! ^^

[Marked 197]

OMG. I literally love you. TQ  :love: