Question 1140214:
If the mean score on a math quiz is 12.8 and 77.0%
of the students in your class scored between 9.2 and 16.4
determine the standard deviation of the quiz scores.
The standard deviation of the quiz scores is
.
(Round to one decimal place as needed.)
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! assuming the distribution of the scores are normal, then you get the following.
z = (x - m) / s
z is the z-score
x is the raw score
m is the mean
s is the standard deviation
if the mean is 12.8 and the high score is 16.4, then the formula becomes:
z = 16.4 - 12.8 / s which becomes z = 3.6 / s
if you solve for s, you will get s = 3.6 / z
you need to find z.
you are given that 77% of the scores are between 9.2 and 16.4.
the 77% is assumed to be in the middle of the normal distribution curve.
this means that the tails on either end are the same.
each tail is therefore (100% - 77%) / 2 = 11.5%.
if you look up the z-score that has 11.5% of the normal distribution curve to the left of it, you will find that the z-score is equal to -1.200358858.
that would be the low z-score.
since the normal distribution is symmetric, then the high z-score will be 1.200358858.
you have 77% of the normal distribution is between z-score of -1.200358858. and 1.200358858.
your high z-score formula becomes z = (16.4 - 12.8) / s which becomes 1.200358858 = 3.6 / s.
solve for s to get s = 3.6/1.200358858 which makes s = 2.999103123
that would be your standard deviation.
your mean is 12.8
your standard deviation is 2.999103123
77% of your scores will be between 9.2 and 16.4
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