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Retired Moderator
Joined: 10 Apr 2015
Posts: 6218
Given Kudos: 136
In a distribution of grapes, a weight of 19.2 grams is 0.5 u
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Updated on: 26 May 2019, 10:01
Updated on: 26 May 2019, 10:01
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In a distribution of grapes, a weight of 19.2 grams is 0.5 units of standard deviation above the mean, and a weight of 9.6 grams is 0.25 units of standard deviation below the mean.
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.
Quantity A |
Quantity B |
the mean of the distribution (in grams) |
the standard deviation of the distribution (in grams) |
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.
Originally posted by GreenlightTestPrep on 26 May 2019, 07:28.
Last edited by Carcass on 26 May 2019, 10:01, edited 1 time in total.
Last edited by Carcass on 26 May 2019, 10:01, edited 1 time in total.
Edited by Carcass - added the answer choices
Retired Moderator
Joined: 10 Apr 2015
Posts: 6218
Given Kudos: 136
Re: In a distribution of grapes, a weight of 19.2 grams is 0.5 u
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28 May 2019, 07:59
28 May 2019, 07:59
1
GreenlightTestPrep wrote:
In a distribution of grapes, a weight of 19.2 grams is 0.5 units of standard deviation above the mean, and a weight of 9.6 grams is 0.25 units of standard deviation below the mean.
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.
Quantity A |
Quantity B |
the mean of the distribution (in grams) |
the standard deviation of the distribution (in grams) |
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.
----------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.
-----------------------------------
Let m = the mean of the distribution
Let d = the standard deviation of the distribution
A weight of 19.2 grams is 0.5 units of standard deviation above the mean
We can write: m + 0.5d = 19.2
A weight of 9.6 grams is 0.25 units of standard deviation below the mean
We can write: m - 0.25d = 9.6
We now have:
m + 0.5d = 19.2
m - 0.25d = 9.6
Subtract the bottom equation from the top equation to get: 0.75d = 9.6
Divide both sides by 0.75 to get: d = 12.8
Now that we know d = 12.8, we can take any equation and replace d with 12.8.
So, take the equation m + 0.5d = 19.2 and replace d with 12.8
We get: m + 0.5(12.8) = 19.2
Simplify: m + 6.4 = 19.2
Solve: m = 12.8
We get:
QUANTITY A: 12.8
QUANTITY B: 12.8
Answer: C
Cheers,
Brent
Re: In a distribution of grapes, a weight of 19.2 grams is 0.5 u
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27 Jun 2024, 04:26
27 Jun 2024, 04:26
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