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805+ (Hard)|   Math Related|         
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GMAT Club Olympics 2025 (Day 6): During a science experiment 07 Jul 2025, 08:00
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Corrected Mean: 6.35 Corrected Standard Deviation: 1.45
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62% (02:07) correct 38% (02:16) wrong based on 371 sessions

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During a science experiment, each measurement result was affected by the same calibration error and had to be adjusted. The corrected result m for each result was calculated as m = -0.5n + 10, where n was the original recorded value. The average (arithmetic mean) of the original measurements was 7.3 and the standard deviation of the original measurements was 2.9.

Select for Corrected Mean the average (arithmetic mean) of the corrected results, and select for Corrected Standard Deviation the standard deviation of the corrected results. Make only two selections, one in each column.
Corrected Mean Corrected Standard Deviation
17.3
8.55
6.35
1.45
-1.45
-3.65
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Corrected Mean: 6.35 Corrected Standard Deviation: 1.45
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Bunuel
During a science experiment, each measurement result was affected by the same calibration error and had to be adjusted. The corrected result m for each result was calculated as m = -0.5n + 10, where n was the original recorded value. The average (arithmetic mean) of the original measurements was 7.3 and the standard deviation of the original measurements was 2.9.

Select for Corrected Mean the average (arithmetic mean) of the corrected results, and select for Corrected Standard Deviation the standard deviation of the corrected results. Make only two selections, one in each column.
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­

GMAT Club Official Explanation:



Since each original measurement n was transformed into a corrected result m = -0.5n + 10, the corrected mean is simply the average of all the m values. But since this transformation is applied identically to each individual value, we can apply the same transformation directly to the original mean.

That is, the corrected mean is:

Corrected mean = -0.5 * (original mean) + 10

Corrected mean = -0.5 * 7.3 + 10 = -3.65 + 10 = 6.35

As for the standard deviation, two key properties are useful:

  • If we multiply (or divide) all the numbers in a set by a non-zero constant, the standard deviation will be respectively multiplied or divided by the absolute value of that constant. For example, if the SD of \(\{a, \ b, \ c\}\) is \(s\), then the SD of \(\{2a, \ 2b, \ 2c\}\) will be \(2s\) and the SD of \(\{-3a, \ -3b, \ -3c\}\) will be \(s*|-3|=3s\).
  • If we add (or subtract) a constant to all the numbers in a set, the standard deviation will NOT change. For example, if the SD of \(\{a, \ b, \ c\}\) is \(s\), then the SD of \(\{a+4, \ b+4, \ c+4\}\) or of \(\{a-1, \ b-1, \ c-1\}\) will also be \(s\).

Since the transformation is m = -0.5n + 10, the scaling factor is -0.5 and the shift is +10. The +10 has no effect on the standard deviation, and the multiplication by -0.5 scales the SD by |-0.5| = 0.5.

Original SD = 2.9
Corrected SD = |-0.5| * 2.9 = 1.45
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GMAT Club Olympics 2025 (Day 6): During a science experiment 07 Jul 2025, 08:31
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Hi,

Mean is affected by both operations (+ or -) and (* or /) and since each term had to go through the correction same will be applied to mean leading to 6.35 as updated mean.

SD is only affected (* or /) operation so applying that part of correction to the given SD lead to 1.45. Remember SD can never be negative. It is a positive quantity and sign of the numbers in the set does not affect the SD.

For Ex: Original Set: 2, 4, 6, 8, 10
Mean = (2 + 4 + 6 + 8 + 10) / 5 = 30 / 5 = 6
SD = √(40 / 5) = √8 = ~2.828

Case 1: Add 5 to each number: 7, 9, 11, 13, 15
Mean = 6 + 5 = 11
SD = unchanged : ~2.828

Case 2: Subtract 3 from each number: −1, 1, 3, 5, 7
Mean = 6 – 3 = 3
SD = still : ~2.828

Case 3: Multiply all numbers by 2 : 1, 2, 3, 4, 5
Mean = 6 × 2 = 12
SD = 2 × original SD = 2 × 2.828 = ~5.656

Case 4: Divide all numbers by 2 : 1, 2, 3, 4, 5
Mean = 6 / 2 = 3
SD = original SD / 2 = 2.828 / 2 = ~1.414


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During a science experiment, each measurement result was affected by the same calibration error and had to be adjusted. The corrected result m for each result was calculated as m = -0.5n + 10, where n was the original recorded value. The average (arithmetic mean) of the original measurements was 7.3 and the standard deviation of the original measurements was 2.9.

m = - 0.5n + 10

Corrected mean = -0.5*7.3 + 10 = - 3.65 + 10 = 6.35
Corrected Standard deviation = |-0.5|*2.9 = 1.45

Constance term affects corrected mean but not the standard deviation. |-0.5| ensures that the standard deviation remains non-negative.

Select for Corrected Mean the average (arithmetic mean) of the corrected results, and select for Corrected Standard Deviation the standard deviation of the corrected results. Make only two selections, one in each column.

Corrected MeanCorrected Standard Deviation
6.351.45
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GMAT Club Olympics 2025 (Day 6): During a science experiment 07 Jul 2025, 08:56
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Information given:
- Each measurement is adjusted by: m = -0.5n + 10, where n is the original value
- Original mean: 7.3
- Original standard deviation: 2.9

Question:
- Find the mean of the corrected results
- Find the standard deviation of the corrected results

- New mean = a * old mean + b
- New mean = -0.5 * 7.3 + 10 = -3.65 + 10 = 6.35

- Standard deviation is unaffected by shift (+10), only scale factor matters
- New SD = 0.5 * 2.9 = 1.45

Answer: Corrected mean 6.35, corrected standard deviation 1.45
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During a science experiment, each measurement result was affected by the same calibration error and had to be adjusted. The corrected result m for each result was calculated as m = -0.5n + 10, where n was the original recorded value. The average (arithmetic mean) of the original measurements was 7.3 and the standard deviation of the original measurements was 2.9.

Select for Corrected Mean the average (arithmetic mean) of the corrected results, and select for Corrected Standard Deviation the standard deviation of the corrected results. Make only two selections, one in each column.
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GMAT Club Olympics 2025 (Day 6): During a science experiment 07 Jul 2025, 09:09
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We have a formula for the corrected measurements: m = -0.5n + 10, where n is the original measurement
For the corrected mean, we need to apply this formula to the original mean:
Original mean = 7.3
Corrected mean = -0.5(7.3) + 10 = -3.65 + 10 = 6.35
For the standard deviation, we need to understand how linear transformations affect the standard deviation. When we have a formula of the form ax+ b:
- The constant (b) doesn't affect the standard deviation
- The coefficient (a) multiplies the standard deviation by its absolute value
Original standard deviation = 2.9
Corrected standard deviation = 1-0.5l × 2.9 = 0.5 × 2.9 = 1.45
Therefore, the corrected mean is 6.35 and the corrected standard deviation is 1.45.
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When you adjust every measurement by multiplying it by –0.5 and then adding 10, the entire set of values is both flipped and scaled, and then shifted upward. Since the average of the original readings was 7.3, multiplying by –0.5 produces an average change of –3.65, and then adding 10 shifts that to a final average of 6.35. In contrast, the standard deviation measures how spread‐out the values are, and adding the same constant to every reading (the +10) doesn’t affect the spread at all. it merely relocates the center. However, multiplying all readings by –0.5 cuts the spread in half (and flips its direction, which doesn’t affect the size of the spread), so the original spread of 2.9 becomes 1.45 in the corrected data.

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During a science experiment, each measurement result was affected by the same calibration error and had to be adjusted. The corrected result m for each result was calculated as m = -0.5n + 10, where n was the original recorded value. The average (arithmetic mean) of the original measurements was 7.3 and the standard deviation of the original measurements was 2.9.

Select for Corrected Mean the average (arithmetic mean) of the corrected results, and select for Corrected Standard Deviation the standard deviation of the corrected results. Make only two selections, one in each column.
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During a science experiment, each measurement result was affected by the same calibration error and had to be adjusted. The corrected result m for each result was calculated as m = -0.5n + 10, where n was the original recorded value. The average (arithmetic mean) of the original measurements was 7.3 and the standard deviation of the original measurements was 2.9.

Select for Corrected Mean the average (arithmetic mean) of the corrected results, and select for Corrected Standard Deviation the standard deviation of the corrected results. Make only two selections, one in each column.

m== -0.5n + 10
m=-0.5*7.3+10 ; 6.35

SD = l-0.5l *2.9 ; 1.45

correct option mean 6.35 & standard deviation 1.45
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The problem requires calculating the new mean and standard deviation of a dataset after each data point `n` undergoes a linear transformation `m = -0.5n + 10`. The original mean was 7.3 and the original standard deviation was 2.9.

The rules for the linear transformation of statistical measures are as follows. For a transformation `y = ax + b`:
1. The new mean is calculated by applying the full transformation to the old mean: `Mean(y) = a * Mean(x) + b`.
2. The new standard deviation is calculated by multiplying the old standard deviation by the absolute value of the scaling factor: `SD(y) = |a| * SD(x)`. The additive constant `b` does not affect the spread of the data.

To find the Corrected Mean, we apply the first rule:
Corrected Mean = (-0.5) * (Original Mean) + 10
Corrected Mean = (-0.5) * 7.3 + 10
Corrected Mean = -3.65 + 10 = 6.35

To find the Corrected Standard Deviation, we apply the second rule:
Corrected Standard Deviation = |-0.5| * (Original Standard Deviation)
Corrected Standard Deviation = 0.5 * 2.9 = 1.45

The Corrected Mean is 6.35, and the Corrected Standard Deviation is 1.45.
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m= -0.5n+10
Original mean= 7.3

Corrected Mean= -0.5*7.3+10=6.35

In standard deviation we only consider the scaling factor,
Original SD=2.9
Corrected SD= |-0.5*2.9|= 1.45
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The formula for corrected result m for each result is calculated as m = -0.5n + 10, n is the original recorded value.

Given, average (arithmetic mean) of the original measurements(n) is 7.3 and the standard deviation of the original measurements(n) is 2.9.

so the given values can be substituted in place of n respectively,

Thus, for Corrected Mean, m = -0.5(7.3)+10 = 6.35
and for Corrected Standard Deviation, m = -0.5(2.9)+10 = 8.55
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Original Mean : 7.3
Original Standard Deviation : 2.9.

Corrected result can be caluclated as : m = -0.5n + 10, where n was the original recorded value.

Mean
Let x be the total number of entries.
Corrected Mean = ((-0.5)*(Orginal Mean)*x + 10*x)/x = -0.5*(Orginal Mean) + 10 = -0.5*7.3 + 10 = -3.65+10 = 6.35

Standard Deviation
Trick : Standard Deviation remains same if a constant k is added to or subtracted from all the entries.
Standard Deviation gets multiplied by k if all entries are multiplied by constant k.
Standard Deviation gets divided by k if all entries are divided by constant k.

Thus, here, m = -0.5n + 10
Adding 10 to all the entries will have no effect on the SD.
But since we are multiplying by -0.5, therefore SD also gets multiplied by -0.5.
Therefore, Corrected Standard Deviation = 2.9*(-0.5) = -1.45
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Given equation,
m=−0.5n+10
Where:

m is the corrected result
n is the original recorded value

The original measurements' statistics are:
Average (arithmetic mean) of original measurements, n' = 7.3

Standard deviation of original measurements, SDn = 2.9

Corrected Mean
For a linear transformation of the form m=an+b, the new mean, m' =an' +b

In this case, a=−0.5 and b=10.

m' =(−0.5)(7.3)+10
m' =−3.65+10
m' =6.35

The average (arithmetic mean) of the corrected results is 6.35.

Corrected Standard Deviation
For a linear transformation of the form m=an+b, the new standard deviation, SDm
=∣a∣SDn
In this case, a=−0.5.

SDm =∣−0.5∣(2.9)
=(0.5)(2.9)
=1.45

The standard deviation of the corrected results is 1.45.
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GMAT Club Olympics 2025 (Day 6): During a science experiment 07 Jul 2025, 09:41
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m = -0.5n + 10
So Corrected Mean => -0.5 * 7.3 + 10 = -3.65 + 10 = 6.35

Corrected Standard Deviation => We know when every data point is multiplied some factor then standard deviation is also multiplied by same factor so => |-0.5| * 2.9 = 1.45

why |-0.5| => because spread is always positive. when u calculate the SD u. will square the diff and then take root which then make the multiplier positive.
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The questions uses inferences form general statistic theory, i will state some below which will help to solve even similar type questions -

For mean :
1. All the numbers of set is multiplied/divided by a constant, the mean also gets multiplied by same constant
2. A constant is added/subtracted to all numbers of set, the ,mean also gets shifted (added/subtracted) by same constant

For standard deviation :
1. All the numbers of set is multiplied by a constant, the standard deviation also gets multiplied by same constant in absolute value
2. A constant is added/subtracted to all numbers of set, the ,standard deviation remain same


Now using the aove pricniples, the question becomes quite straigh forward

1. Given, m=-0.5n+10, hence new mean = -0.5(7.9)+10 = 6.35

2. Given, m=-0.5n+10, hence new SD = half of old SD (consider absolute value) = 1.45
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Bunuel
 


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During a science experiment, each measurement result was affected by the same calibration error and had to be adjusted. The corrected result m for each result was calculated as m = -0.5n + 10, where n was the original recorded value. The average (arithmetic mean) of the original measurements was 7.3 and the standard deviation of the original measurements was 2.9.

Select for Corrected Mean the average (arithmetic mean) of the corrected results, and select for Corrected Standard Deviation the standard deviation of the corrected results. Make only two selections, one in each column.
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Arithmetic mean wass 7.3 sd was 2.9 implies 7.3*7.3/n=2.9n=7.3*7.3/2.9 n= 18 m=19
Substituting we get corrected am and sd
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GMAT Club Olympics 2025 (Day 6): During a science experiment 07 Jul 2025, 09:46
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Fact
Arithmetic mean changes when anything is multiplied or added to the original recorded value.
so just plugin the 7.3 into m = -0.5n + 10 and we get 6.35 as the new mean

However standard deviation changes when something is multiplied and not added to the original recorded value..
So we need to do 0.5 * original recorded value. =1.45
Also remember standard deviation is always positive

Hence for mean = 6.35
for standard deviation =1.45
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During a science experiment, each measurement result was affected by the same calibration error and had to be adjusted. The corrected result m for each result was calculated as m = -0.5n + 10, where n was the original recorded value. The average (arithmetic mean) of the original measurements was 7.3 and the standard deviation of the original measurements was 2.9.

Select for Corrected Mean the average (arithmetic mean) of the corrected results, and select for Corrected Standard Deviation the standard deviation of the corrected results. Make only two selections, one in each column.
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Mean= -0.5*7.3 + 10 =6.35
std. deviation= -0.5*2.9=-1.45 ..no affect of +10

Ans 6.35 & -1.45
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Given:
Original measurement= n
Corrected measurement= m
The transformation applied to each original value n is: m=−0.5n+10
Average of original measurement = 7.3
Standard Deviation of corrected measurement = 2.9

Corrected Average mean : m = -0.5(7.3)+10 = 6.35

Standard Deviation relies on the spread of value and is not affected by constant change.
Therefore, Standard Deviation for corrected data = 0.5* 2.9 = 1.45

Answer: Corrected Mean: Option 3, Corrected Standard Deviation: Option 4



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During a science experiment, each measurement result was affected by the same calibration error and had to be adjusted. The corrected result m for each result was calculated as m = -0.5n + 10, where n was the original recorded value. The average (arithmetic mean) of the original measurements was 7.3 and the standard deviation of the original measurements was 2.9.

Select for Corrected Mean the average (arithmetic mean) of the corrected results, and select for Corrected Standard Deviation the standard deviation of the corrected results. Make only two selections, one in each column.
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Dipan0506
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GMAT Club Olympics 2025 (Day 6): During a science experiment 07 Jul 2025, 09:55
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The corrected m is taken with 7.5 and
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