Prove the sum of two rational numbers is rational where a, b, c, and d are integers and b and d cannot be zero.
2 answers:
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The standard deviation of what? Percentiles from any normal distribution look the same, just like the unit normal, so you can't really determine the standard deviation of the original scores. You can determine a z score from a percentile. That tells us the number of standard deviations, positive or negative, a given score is away from the mean score. It's a normalized test result.
Your percentile is (a hundred times) the probability that another score is less than your score. We have a normal distribution, so that probability is the integral of the standard normal from negative infinity to our normalized score.
Let's call the percentile rank
, already scaled between zero and 1.
corresponds to a z score 
because the fiftieth percentile means we got an exactly average score, 0 standard deviations away from the mean.
We know 68% of the probability will be between -1 and +1 standard deviation. So
corresponds to 
and
corresponds to 
Similarly, 95% of the probability will be between -2 and +2 standard deviations. So
corresponds to 
and
corresponds to 
That's about the list I can do off the top of my head. I think three standard deviations is 99.7%. For the rest we just consult a z table or integrated normal table. We find p in the body of the table (maybe |.5-p| depending on the table) and then the column headings tell us our z score.
In this modern age, your computer can do this for you quickly
Your percentile is (a hundred times) the probability that another score is less than your score. We have a normal distribution, so that probability is the integral of the standard normal from negative infinity to our normalized score.
Let's call the percentile rank
We know 68% of the probability will be between -1 and +1 standard deviation. So
Similarly, 95% of the probability will be between -2 and +2 standard deviations. So
That's about the list I can do off the top of my head. I think three standard deviations is 99.7%. For the rest we just consult a z table or integrated normal table. We find p in the body of the table (maybe |.5-p| depending on the table) and then the column headings tell us our z score.
In this modern age, your computer can do this for you quickly