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The scores for the 500 students who took Ms. Johnson’s fina
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22 May 2017, 01:45
22 May 2017, 01:45
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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 |
A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.
Re: The scores for the 500 students who took Ms. Johnson’s fina
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18 Sep 2017, 09:57
18 Sep 2017, 09:57
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Quote:
Remember that a normal distribution curve has divisions of 34 percent, 14 percent, and 2 percent on each side of the mean. 80 out of 50 is 16 percent, or 14 percent + 2 percent, and 10 out of 500 is 2 percent. Draw a normal distribution curve and label it. There are three standard deviations between 92 and 56, so 92 - 56 = 36, and 36 ÷ 3 = 12. The mean is 92 - 12 = 80, which is smaller than Quantity B.
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Re: The scores for the 500 students who took Ms. Johnson’s fina
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26 May 2019, 13:18
26 May 2019, 13:18
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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 |
-------ASIDE-----------------------------
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.
-------------------------------------------------
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 = 92
We'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 = 56
So, we now know two things:
m + d = 92
m - 2d = 56
Take the TOP equation, and multiply both sides by 2 to get:
2m + 2d = 184
m - 2d = 56
ADD the two equations to get: 3m = 240
Solve: m = 80
Answer: 80
Cheers,
Brent
