Question 1184960: The test scores for a very large statistics class have a bell-shaped distribution with a mean of 69 points.
If 95% of the scores are between 47 and 91, what is the standard deviation?
I know I have to use the empirical rule, but I have no clue how to solve for SD. It just doesn't make much sense to me.
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! 95% of the area under the normal distribution curve is between 47 and 91.
find the z-scores that haves 95% of the area under the normal distribution curve.
those z-score are plus minus 1.96 as shown below.
use the z-score formula to solve for s.
the z-score formula is z = (x - m) / s.
z is the z-score
s is the standard deviation
m is the mean
x is the raw score.
with the low z-score, your formula becomes:
-1.96 = (47 - 69) / s
solve for s to get:
s = (47 - 69) / -1.96
simplify to get:
s = -22 / -1.96, resulting in s = 11.22449, rounding to 5 decimal places.
with the high z-score, the formula becomes:
1.96 = (91 - 69) / s
solve for s to get:
s = 1.96 / 22 = 11.22449, round to 5 decimal laces.
since the normal distribution curve is symmetric about the mean, you only needed to do one side to get the standard deviation.
your standard deviation is 11.22449.
here's the analysis that confirms this to be true.
here are the displays from using the online statistical calculator by david lane.
first is to find the z-scores associated with 95% area under the normal distribution curve between them.
second is to confirm the standard deviation calculated is correct.
the same calculator can be used with z-score or with real scores.
with z-scores, the mean is 0 and the standard deviation is 1.
with real scores, the mean is the mean of the scores and the standard deviation is the standard deviation of the scores you just calculated using the z-scoree formula.
the calculator i used can be found on https://davidmlane.com/hyperstat/z_table.html
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