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Vivian earned a score of 940 on a national achievement test.

 The mean test score was 850 with a standard deviation of 100.  What porportion of students had a higher score than Vivian? (Assume that test scores are normally distributed.)

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Bill W. | Looking for help with Excel, SQL, or other business technology?Looking for help with Excel, SQL, or oth...
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Hi Vivian!
To solve this question, you'll need to use a table of the standardized Normal distribution.  Fortunately, you can just use a graphing calculator, or you can find a Normal distribution calculator on the internet.  It'll ask for the three facts from your question.  (If the calculator also asks for cumulative probability, leave that blank because that's what you're trying to solve for.)
             Standard score         940
             Mean                       850
             Standard deviation    100
So, the answer will be the cumulative probability, which will be a number with a few decimals.  However, this represents the students who had a LOWER score than Vivian.  You'll need to subtract this from 1 in order to get the opposite number, the proportion of students who had a higher score than Vivian.  Be sure to multiply the result by 100 to get a percentage.  That's it.
Here's some additional info that might help explain the approach.  The first step to solving this question is to figure out the difference between Vivian's score and the mean score:  940 minus 850 is 90.  This tells us how different her score was from the average.
Since the standard deviation is 100, we know that Vivian's difference (90) is within one standard deviation from the mean.  In other words, if you increase the mean by one standard deviation, 850 + 100, you get 950, and you can see that Vivian's score of 940 doesn't get that high.  Specifically, her difference of 90 is 90% of the standard deviation of 100.  The decimal equivalent of 90% is 0.90, and this is called Vivian's z-score.
In math notation, we've done this:  z = (X - μ) / σ = (940 - 850) / 100 = 0.90
   where z is the z-score
             X is Vivian's score (940)
             µ is the mean (850)
             σ is the standard deviation (100)
As you may know, in a normal distribution it's expected that about 68% of all observations will fall within 1 standard deviation of the mean, 95% will fall within 2 standard deviations, and 99% will fall within 3 standard deviations.  You now have all the info you need to do a rough calculation of the cumulative probability.  Here's how to put it together:
     The upper half of the 68% (the scores between the mean and the first standard deviation) = 34%
     Vivian's 90% of the 34% = 30.6%
     All the test scores lower than the mean = 0.500  (this is 1/2 of all test scores)
  + All the test scores between the mean and Vivian's score  = 0.306  (see above)
   = 0.806, which is not the real cumulative probability, but an approximation.
Hope this helps!