Does anybody know how to find percentage in a normal curve???

On a normal curve, 68% of the scores fall within 1 standard deviation of the mean, and 95% of the scores fall within 2 standard deviations of the mean. So, if 500 is the mean:

b=68% because 400 and 600 are each 1 standard deviation from the mean, and d=95% since 300 and 700 are each 2 standard deviations from the mean. The formula to figure out the rest of them is (score-mean)/standard deviation. You can convert the standard deviations to percentages easily since the SD for this problem is 100. If you don’t want to use the formula, you can just plot the rest of them on a normal curve, using the information I’ve given you to figure out the percentages.

You need a special table or a calculator to do this. You work out a z score, then you read the area under the normal curve from the table.

z = (Raw score – Mean) / Standard Deviation

a. Area between 500 and 600 = area below 600 – area below 500

z score for 600 = (600 – 500)/100 = 1

That means that 600 is one standard deviation greater than the mean. Since the distribution is normal we can go to the table and read off the fraction of the area beneath the normal curve that is less than or equal to one standard deviation

Area below 600 = 0.841345 (by calculator)

The area below 500 is exactly half the area below the normal curve, because 500 is the mean

Area below 500 = 0.5 (because 500 is the mean)

Therefore the area between 500 and 600 = 0.841345 – 0.5

= 0.341345

= 34% to nearest integer

(b) z score for 400 = -1 and z score for 600 = 1

Area below 400 = 0.158655 (by calculator)

Area below 600 = 0.841345 (by calculator)

Therefore the area between 400 and 600 = 0.68269

= 68% to nearest integer

Make sure you understand why the answer to (b) is exactly double the answer to (a). After that, all the questions are the same.

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Try this:
http://www.caladis.org/compute/?q=%24SATscore&v=SATscore%3Anorm%2C500%2C100&x=off&n=m&h=fd&a=rad