Carcass wrote:
The scores for the 500 students who took Ms. Johnson’s final exam have a normal distribution. There are 80 students who scored at least 92 points out of a possible 100 total points and 10 students who scored at or below 56.
Quantity A |
Quantity B |
The average (arithmetic mean) score on the final exam |
87 |
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A little extra background on
standard deviations above and below the mean If, for example, a set has a standard deviation of 4, then:
1 standard deviation = 4
2 standard deviations = 8
3 standard deviations = 12
1.5 standard deviations = 6
0.25 standard deviations = 1
etc
So, if the mean of a set is 9, and the standard deviation is 4, then:
2 standard deviations ABOVE the mean =
17 [since 9 + 2(4) = 17] 1.5 standard deviations BELOW the mean =
3 [since 9 - 1.5(4) = 3] 3 standard deviations ABOVE the mean =
21 [since 9 + 3(4) = 21] etc.
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We're told that 16% scored at least 92 points (80/500 = 10%)
When we examine the normal distribution curve, we see that 16% of all values are more than 1 standard deviation ABOVE the mean.
So, we know that a test score of 92 points is 1 standard deviation ABOVE the mean.
So, if m = the mean of all the scores
and if d = the standard deviation of all the scores...
We can write:
m + d = 92We're also told that 2% scored less than 56 points (10/500 = 2%)
When we examine the normal distribution curve, we see that 2% of all values are more than 2 standard deviations BELOW the mean.
So, we know that a test score of 56 points is 2 standard deviations BELOW the mean.
So, if m = the mean of all the scores
and if d = the standard deviation of all the scores...
We can write:
m - 2d = 56So, we now know two things:
m + d = 92m - 2d = 56Take the TOP equation, and multiply both sides by 2 to get:
2m + 2d = 184m - 2d = 56ADD the two equations to get: 3m = 240
Solve: m = 80
Answer: 80
Cheers,
Brent
_________________
Brent Hanneson - founder of Greenlight Test Prep