Easy Math Problem HELP!!!!

[Kyle Barry]

Using normal scoring, how many people out of 800, can be expected to score out of 53 and 57? Normal is Mu-50 Sigma-10, just a problem my friends need solved, but we're idiots so we need cubers' help, thanks.

 

[Lucas Garron]

Between 53 and 57?
Assuming a continuous distribution of scores (discrete would cause a bit of problems depending on inclusive/exclusive):

Bunch o' ways, like:
NormalCDF[(53-50)/10 to (57-50)/10] ≈ 0.1401

Out of 800, 112.

 

[Kyle Barry]

Thanks, but when i put that in the calculator i get .021, not .1401, what am i doing wrong

 

[cmhardw]

Between 53 and 57?
Assuming a continuous distribution of scores (discrete would cause a bit of problems depending on inclusive/exclusive):

Bunch o' ways, like:
NormalCDF[(53-10)/50 to (57-10)/50] ≈ 0.1401

Out of 800, 112.

Lucas I have to ask, because I am learning about this kind of math now for the first time and want to make sure I understand correctly. If we are assuming continuous, and that it's between 53 and 57 wouldn't we do:
(52.5-50)/10 and (57.5-50)/10 getting z values of z=0.25 and z=0.75

which is 0.7734 - .5987 = 0.1747

So out of 800 would be 140? I ask because I am not sure. I only studied pure math in school, so now I am trying to learn about some financial and statistical math to improve my ability to get a job in this field. This is the first time in my life that I've tried to study statistics.

Chris

 

[Lucas Garron]

If we are assuming continuous, and that it's between 53 and 57 wouldn't we do:
(52.5-50)/10 and (57.5-50)/10 getting z values of z=0.25 and z=0.75

which is 0.7734 - .5987 = 0.1747

No, we wouldn't.
Really, it depends on how his distribution actually works. Your calculation finds the probability of getting a score between 52.5 and 57.5 -which might be correct, say, if the distribution is continuous but scores get rounded to the nearest integer (in which cases we want to take everything between your values).

Also, in case Kyle meant "score out of" as scoring beyond/over:
>53: 0.382089
>57: 0.241964
(1-CDF to those values.)

 

[Tim Reynolds]

In any case, Lucas, you should do (53-50)/10 instead of (53-10)/50

 

[Lucas Garron]

In any case, Lucas, you should do (53-50)/10 instead of (53-10)/50

Bah, he has a brain, right?

Uh, sorry, Kyle.
Lemme edit that post...

(See, this is why Mathematica needs to copy better.)

 

[cmhardw]

If we are assuming continuous, and that it's between 53 and 57 wouldn't we do:
(52.5-50)/10 and (57.5-50)/10 getting z values of z=0.25 and z=0.75

which is 0.7734 - .5987 = 0.1747

No, we wouldn't.

I see my error now, we would do something like what I was asking when using a normal distribution to approximate a discrete binomial distribution, but we don't have that in this case.

Gotcha, thanks for helping me clear that up Lucas.

Chris

 

[Kyle Barry]

Thanks, not a problem, we figured it out eventually, with some help, we realized we were supposed to do it with a calculator anyway (well not me, im not even in the class), and used the z score table in our book, thanks for the help though.

 

[Swordsman Kirby]

normalcdf(53, 57, 50, 10) * 800

 

[Lucas Garron]

normalcdf(53, 57, 50, 10) * 800

You look like you use a TI. Bah.

 

[Swordsman Kirby]

normalcdf(53, 57, 50, 10) * 800

You look like you use a TI. Bah.

I'm under the assumption that Kyle doesn't have Mathematica.

 

[Lucas Garron]

normalcdf(53, 57, 50, 10) * 800

You look like you use a TI. Bah.

I'm under the assumption that Kyle doesn't have Mathematica.

So am I.