normal distribution problem: Suppose that SAT scores among

[xxbabezxx233]

I need help approaching the answer or to make sure I am correct.

Suppose that SAT scores among U.S. college students are normally distributed with a mean of 500 and a standard deviation of 100. What is the probability that a randomly selected individual from this population has an SAT score at or below 600?

I would find the z-score, which would be 600-500, and divide by 100. I would get 1.00 now when I look at my z-score chart, I would get 0.8413? Would this be correct? What would I do from then?

Also if I want to figure out what people would be above 600...?

 

[marcmtlca]

Draw a picture of the bell curve Centered at 0. Now put a line at z=1 and shade everything under the curve from -infinity to z. this is the proportion of people who have lower results (.8413). The total area under the curve is 1. Find out a way to obtain the area of the unshaded region using the two things I just mentioned.

 

[xxbabezxx233]

is there a site where I can draw my distribution on the computer. I'm still a bit confused on how to draw it. But i'll try it out thanks. Any more information would be greatly appreciated

 

[galactus]

Here's a close diagram of your distribution curve.

The shaded region is the area under the curve that corresponds to z=1 or a score below 600.

Remember, the hint marc gave you. The area under the entire curve is 1. If the area from \(\displaystyle ({-\infty},1]\) is 0.8413, then the area from \(\displaystyle [1, {\infty})\) would be ?????.

What if the question were 'scored exactly 600'?.

Do you have a nice calculator that does integration?.

You can find the area under the curve by:

\(\displaystyle \L\\\frac{1}{\sqrt{2{\pi}}}\int_{-\infty}^{z}e^{\frac{-x^{2}}{2}}dx\)

For instance, use z=1 and the solution is 0.8413

 

[marcmtlca]

I don't know how to solve that exactly 600 thing. Is there a standard size interval that people use?

 

[galactus]

The way I do it is to use the continuity correction and subtract them.

Like so: \(\displaystyle 600\pm{0.5}\)

\(\displaystyle \L\\\frac{599.5-500}{100}=0.995\)

\(\displaystyle \L\\\frac{600.5-500}{100}=1.005\)

The corresponding z-scores are 0.8401 and 0.8426, respectively.

Subtract: 0.8426-0.8401=0.0024

The probability of getting exactly a 600 is 0.24%.

The the way I have always seen it done.

 

[xxbabezxx233]

so the answer wont be 84%.. and is there a reason why we number the line that way?

 

[marcmtlca]

I see, well it seems like an acceptable way to do it.

 

[galactus]

so the answer wont be 84%.. and is there a reason why we number the line that way?

Look at the posts. That is an example of exactly 600.

You're on the right track.

 

[marcmtlca]

The idea is that the probability of getting exactly 600 in a continuous setting is 0. And so to get an answer we need to create an interval around 600 and calculate the probability that the score lies in that interval.