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07 May 2026, 19:00
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A community center recorded the number of evening workshops it held in each of the 12 months of last year. What was the standard deviation of the number of evening workshops held per month last year?
(1) The standard deviation of the numbers of evening workshops held per month during the first 6 months of last year was 1.
(2) The standard deviation of the numbers of evening workshops held per month during the last 6 months of last year was 5.
(1) The standard deviation of the numbers of evening workshops held per month during the first 6 months of last year was 1.
(2) The standard deviation of the numbers of evening workshops held per month during the last 6 months of last year was 5.
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10 May 2026, 03:00
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GMAT Club Official Solution:
A community center recorded the number of evening workshops it held in each of the 12 months of last year. What was the standard deviation of the number of evening workshops held per month last year?
The standard deviation is a measure of the variation of the data points from the mean, a measure of how widespread a given set is. When the standard deviation is low, the data points tend to be close to the mean, while a high standard deviation implies that the data is spread out over a broader range of values. In essence, the standard deviation can be thought of as a measure of the distance from the mean, and since distance cannot be negative, the standard deviation cannot be negative, meaning that it must be greater than or equal to zero: \(SD\geq0\).
Next, the standard deviation of a set is zero if and only if the set consists of identical numbers, which is the same as saying that the set consists of only one distinct number.
(1) The standard deviation of the numbers of evening workshops held per month during the first 6 months of last year was 1.
This gives only the standard deviation of the first 6 monthly values. It gives no information about the last 6 monthly values, so it cannot determine the standard deviation of all 12 values. Not sufficient.
(2) The standard deviation of the numbers of evening workshops held per month during the last 6 months of last year was 5.
This gives only the standard deviation of the last 6 monthly values. It gives no information about the first 6 monthly values, so it cannot determine the standard deviation of all 12 values. Not sufficient.
(1)+(2) Even if we know the standard deviation within each half of the year, we still do not know how far apart the average of the first 6 months and the average of the last 6 months are.
For example, the 12 monthly values could be:
4, 4, 4, 6, 6, 6, 0, 0, 0, 10, 10, 10
The first 6 values have standard deviation 1, and the last 6 values have standard deviation 5.
But the 12 monthly values could also be:
4, 4, 4, 6, 6, 6, 100, 100, 100, 110, 110, 110
Again, the first 6 values have standard deviation 1, and the last 6 values have standard deviation 5.
However, the standard deviation of all 12 values is much larger in the second case, because the two halves of the year are centered around very different averages.
So the two statements together still do not determine the standard deviation for the full year.
Not sufficient.
Answer: E.
A community center recorded the number of evening workshops it held in each of the 12 months of last year. What was the standard deviation of the number of evening workshops held per month last year?
The standard deviation is a measure of the variation of the data points from the mean, a measure of how widespread a given set is. When the standard deviation is low, the data points tend to be close to the mean, while a high standard deviation implies that the data is spread out over a broader range of values. In essence, the standard deviation can be thought of as a measure of the distance from the mean, and since distance cannot be negative, the standard deviation cannot be negative, meaning that it must be greater than or equal to zero: \(SD\geq0\).
Next, the standard deviation of a set is zero if and only if the set consists of identical numbers, which is the same as saying that the set consists of only one distinct number.
(1) The standard deviation of the numbers of evening workshops held per month during the first 6 months of last year was 1.
This gives only the standard deviation of the first 6 monthly values. It gives no information about the last 6 monthly values, so it cannot determine the standard deviation of all 12 values. Not sufficient.
(2) The standard deviation of the numbers of evening workshops held per month during the last 6 months of last year was 5.
This gives only the standard deviation of the last 6 monthly values. It gives no information about the first 6 monthly values, so it cannot determine the standard deviation of all 12 values. Not sufficient.
(1)+(2) Even if we know the standard deviation within each half of the year, we still do not know how far apart the average of the first 6 months and the average of the last 6 months are.
For example, the 12 monthly values could be:
4, 4, 4, 6, 6, 6, 0, 0, 0, 10, 10, 10
The first 6 values have standard deviation 1, and the last 6 values have standard deviation 5.
But the 12 monthly values could also be:
4, 4, 4, 6, 6, 6, 100, 100, 100, 110, 110, 110
Again, the first 6 values have standard deviation 1, and the last 6 values have standard deviation 5.
However, the standard deviation of all 12 values is much larger in the second case, because the two halves of the year are centered around very different averages.
So the two statements together still do not determine the standard deviation for the full year.
Not sufficient.
Answer: E.
General Discussion
07 May 2026, 20:40
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We need the standard deviation for all 12 months.
Statement (1):
First 6 months have standard deviation 1.
This gives information only about half the data. We do not know the last 6 months or the overall distribution.
Not sufficient.
Statement (2):
Last 6 months have standard deviation 5.
Again, only partial information.
Not sufficient.
Together:
Knowing the standard deviations of the two halves still does not determine the standard deviation of all 12 months, because the two groups could have very different means.
Example:
Different overall SDs are possible.
Answer: E
Statement (1):
First 6 months have standard deviation 1.
This gives information only about half the data. We do not know the last 6 months or the overall distribution.
Not sufficient.
Statement (2):
Last 6 months have standard deviation 5.
Again, only partial information.
Not sufficient.
Together:
Knowing the standard deviations of the two halves still does not determine the standard deviation of all 12 months, because the two groups could have very different means.
Example:
- First 6 months: all centered near 10
- Last 6 months: all centered near 100
Different overall SDs are possible.
Answer: E
