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A testing bureau reports that the mean for the population of Graduate Record Exam (GRE) scores is 500 with a standard deviation of 90.

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3 Answers

Based on an assumption you're using z charts for this (and I could be wrong), you need to get the z scores for the 300 and 400 first.
z = (x - μ) / σ
A z score is saying how many standard deviations it is.
So for instance, taking a point 300 - mean of 500 = -200.  That's the distance from the mean.  Then dividing by 90 tells you how many standard deviations that is.
You'll want to do that for both points, 300 and 400.
Both of those are to the left of the mean, and you need that "slice" between the 2.  That can be the trickiest to do, and unfortunately the hardest to explain, especially without pictures.
z charts can vary, so I don't even know what your specific charts would look like.  However, the slice between those 2 points is one point to some location less the other point to the same location.
For instance, if you have a tail chart, the distance from the 400 to the end of the left tail gives you that whole entire section.  So then you want the area from the 300 to the end of the left tail subtracted off.  (Draw it and think about that a minute.)
------------------- (400)
The area to the 400 minus the area to the 300 leaves you with the area in between, the slice.
You can also do this if you have a chart that shows from the 0 out to a point, and you can even do it if you have a chart that goes from one end and cross over the center.  Either way, it's the difference between the two.  (Of course, you will be using the z scores for those on the charts.)
Hi Nicole!  As you may know, with a normal distribution (using the numbers from your question):
One standard deviation = 68.2% of all observations
-- Lower limit = 500 - 90 = 410              = 34.1% of all observations
-- Upper limit = 500 + 90 = 590             = 34.1% of all observations
Two standard deviations = 95.4% of all observations
-- Lower limit = 500 - 90 - 90 = 320       = 47.7% of all observations
-- Upper limit = 500 + 90 + 90 = 680     = 47.7% of all observations
Since your range of 300 to 400 is roughly between 2 standard deviations below the mean (320) and 1 standard deviation below the mean (410), your answer will be in the neighborhood of 47.7% minus 34.1% (2σ minus 1σ), which is 13.6% of all test scores.  Your answer will likely be within 1 or 2 percentage points of this.
To solve your question, you'll need to use the NORMDIST function in Excel, or find a Normal Distribution Calculator using a search engine (online).  The goal is to find two percentage numbers, one for 300 and one for 400.  (Note, each percentage will show the cumulative probability P(X<x), which is how many scores were between 0 and the number, for example between 0 and 300.)  You just need to plug-in the mean (500), the standard deviation (90), and the random variable (run it first with 300, then again with 400).  For example, the result you get with 300 should be 0.01313, which is 1.313%.
After you have both percentages, you just need to subtract them, to find the proportion of all scores that fall between those two values.
Hope this helps!
            Compute the Z for each condition i.e. Z1=(400-500)/90=-100/90=-1.11 and Z2=(300-500)/90=-2.22
The area under the normal curve between those two limits is .1206 (12.06%) which would be the proportion of the scores between 300 and 400