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GMAT Club Olympics 2025 (Day 6): During a science experiment

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08 Jul 2025, 00:03

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m=-0.5n+10 , n =initial value
Avg(n)=7.3 , SD=2.9
This is a linear transformation:
So, new mean =a(old mean)+ b
Here a=-0.5 , b=10
So , new mean =-0.5(7.3)+10=6.35 (Ans)
And , new corrected SD= |a|SD = 0.5*2.9=1.45 (Ans)

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08 Jul 2025, 01:17

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Ans: mean = 6.35 and Standard Deviation = 1.45

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.

What we are doing here is using the corrected result equation to calculate the new Mean and Standard Deviation.

previous mean = 7.3
corrected equation = m = -0.5n + 10
What happens to the mean of a set of values if we multiply all the values by a constant? = it gets multiplied too.
What happens to the mean of a set of values if we add a constant to all the values = The value also gets added to the mean (mean shifts by that value)
in other words new mean = -0.5 (old mean) + 10 = -0.5 (7.3) + 10 = 6.35

previous Standard deviation: 2.9
When a constant value is added to all the values of the set --> standard deviation remains the same
When a constant value is multiplied with all the values of the set --> standard deviation gets multiplied by the same value.
New standard deviation = 2.9* 0.5 = 1.45
Remember: Standard deviation can not be negative (it's the distance of values of the set from the mean, and distance can not be negative)

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08 Jul 2025, 02:01

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Given:
Correction formula: m=−0.5n+10
Original Mean = 7.3, Original SD = 2.9

Corrected Mean:
Apply the formula to the mean:
−0.5×7.3+10=6.35
Corrected SD:
Standard deviation is only affected by scaling and not shifting, so:
---> 0.5×2.9=1.45

[*]Corrected Mean = 6.35
[*]Corrected SD = 1.45

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08 Jul 2025, 03:02

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Bunuel

This question was provided by GMAT Club
for the GMAT Club Olympics Competition

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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 result m
m = -0.5n + 10

Original Mean = 7.3
Corrected mean = -0.5 * 7.3 + 10 = 6.35

Original SD = 2.9

Here the addition of a number to all the measurements taken does not affect the SD
Also, SD is mod hence |-0.5| = 0.5
Corrected SD = 0.5 * 2.9 = 1.45

Corrected mean = 6.35
Corrected SD = 1.45

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08 Jul 2025, 03:56

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$x =$ number of measurements
$S = n_1+n_2+n_3+.......+n_x=$ sum of original mesaurements
We know the average of the original measurements is $\frac{S}{x} = 7.3$
The corrected measurement is given by $m = -0.5n+10$.
Here, the original measurement is being multiplied by $-0.5$ and then $10$ is being added to it. Let incorporate that into the average.

New Average $= \frac{(-0.5)n_1+10+(-0.5)n_2+10+........+(-0.5)n_x+10}{x}$

$=\frac{(-0.5)(n_1+n_2+n_3+........+n_x)+10x}{x}$

$=\frac{(-0.5)S}{x}+\frac{10x}{x}$

$=(-0.5)(7.3)+10$

$=6.35$

Clearly, the average is going through the same operations as the original mesaurements. The original average is being multiplied by -0.5 and this product is being added to 10 to get the corrected average.

To find the corrected standard deviation, there are two ways ->

1. Adding a constant to all numbers in the set does not change the standard deviation, but multiplying all the numbers in the set by a constant scales the standard deviation by |constant|.
If the original SD was 2.9, then the corrected standard deviation will be $|-0.5|*2.9 = 1.45$ (because the original values in the set were multiplied by constant $-0.5$. Adding $10$ to all the values in the set did not affect the $SD$)

2. The formula for Standard Deviation is $SD = \sqrt{\frac{Σ(x_i - mean)^2}{n}}$, where $x_i$ is each value in the set and n is the number of values in the set. To find the corrected standard deviation, we can just plug in the new corrected values in place of $x_i $ and mean.

New $SD = \sqrt{\frac{Σ[((-0.5)n_i+10) - ((-0.5)mean_{old}+10)]^2}{x}}$

$=(-0.5) \sqrt{\frac{Σ(n_i+10) - (mean_{old}+10)^2}{x}}$

$=(-0.5) \sqrt{\frac{Σ(n_i+10 - mean_{old}-10)^2}{x}}$

$=(-0.5) \sqrt{\frac{Σ(n_i - mean_{old})^2}{x}}$

We know $\sqrt{\frac{Σ(n_i - mean_{old})^2}{x}} = 2.9$

$=(-0.5)(2.9)$

$=-1.45$

But $SD$ cannot be negative, because it's essentially a measure of distance, and only the magnitue matters in distance.

New $SD = |-1.45| = 1.45$

Takeaway ->
1. When adding/subtracting/multiplying/dividing a constant to/from all numbers in the set, do the same operations on the old mean to get the new mean.
2. Adding/subtracting a constant to/from all numbers in the set has no affect on the standard deviation. Multiplying/dividing a constant with all the numbers in the set scales the standard deviation by |constant|.

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08 Jul 2025, 04:10

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Bunuel

This question was provided by GMAT Club
for the GMAT Club Olympics Competition

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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= original recorded value,
n= corrected value

m = -0.5n + 10,

Average n given,

On multiplication of values, average get affected and new average can be derived by multiplying the same number which has been multiplied to all values. same for addition.

Hence, average of m = -0.5*7.3 +10 = -3.65+10 = 6.45.

Standard deviation won't get impacted by addition but multiply only.
Std deviation of m = 0.5*2.9 = 1.45

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08 Jul 2025, 04:31

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The correction only implies a linear transformation, multiply by -0.5 and add 10.
The mean will change in the same way as each one of the measurements:
new mean = -0.5*mean+10 = 6.35

The standard derivation measures how far the measurements are from the mean. The correction multiplies by -1/2 each value, so the new standard derivation will be multiplied by 1/2 too. We don't apply the sign because standard derivation is always non negative. The correction also adds 10, that does not change the standard derivation of a set.
new standard derivation = 1/2 * standard derivation = 1/2 * 2.9 = 1.45

The right answers are:
Corrected Mean=6.35
Corrected Standard Deviation=1.45

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08 Jul 2025, 04:48

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For the corrected mean, we can just substitute the original mean 7.3 into the corrected result equation :
-> (-0.5 * 7.3) + 10 = 6.35

For the corrected SD, we can just multiply |-0.5|, since SD can't be -ve. There is no need to add 10 (multiplication affects SD but addition doesn't).
-> 0.5 × 2.9 = 1.45

Corrected mean = 6.35
Corrected SD = 1.45.

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08 Jul 2025, 05:27

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if a constant is multiplied to the values then both standard deviation and mean are multiplied by it. addition does not have any effect.
however the standard deviation will be positive
therefore we get mean as -3.65 and standard deviation as 1.45

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08 Jul 2025, 05:44

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m = -0.5n + 10
Linear transformation means
New mean = nm= -0.5nn+10 where nm is new mean and nn is the new recorded value of n
So substitute n with the orginal value of n (-0.5)x7.3 + 10 = 6.35
Correlated mean is therefore = 6.35
Given that the STD is also affected the same way linearly,
new STD = (0.5)old STD
0.5x2.9= 1.45

Bunuel

This question was provided by GMAT Club
for the GMAT Club Olympics Competition

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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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08 Jul 2025, 05:45

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Bunuel

This question was provided by GMAT Club
for the GMAT Club Olympics Competition

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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.

n was the original value.
m is the corrected value
m= -0.5n+10

Now note:
For mean:
- If each value in the data increases/decreases by a certain amount, then the mean will increase or decrease by same value.
- If each value in the data is multiplied or divided by a certain amount, then the mean will be multiplied or divided by the same value.

For standard deviation:
- If each value in the data increases/decreases by a certain amount, the standard deviation remains the same.
- If each value in the data is multiplied or divided by a certain amount, then the standard deviation will be multiplied or divided by the same absolute value.

Original mean = 7.3
Each value is computed as:
m= -0.5n +10
Hence, new mean:
= -0.5*7.3+10
= 6.35

Original standard deviation:
New standard deviation:
= 2.9*0.5
= 1.45

Here, the absolute value is taken, because the standard deviation depends on spread and spread doesn't depend on the direction(positive or negative).

ANSWER:
New mean: 6.35
New standard deviation: 1.45

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08 Jul 2025, 05:47

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Original value = n
Corrected value m = −0.5n+10
Mean of original values n' =7.3
Standard deviation of original values σn=2.9

Now we know that,
when transforming data with a linear function
m = a*n + b

New Mean = a*n' + b
New Standard deviation = |a|*(σn)

Considering the transformation is linear since m = -0.5n + 10
a = -0.5

Corrected mean,
m' = -0.5n' + 10
= -0.5*(7.3) + 10
= 6.35

Corrected Standard deviation σn ' = |a|*(σn) = |-0.5 * (2.9) = 1.45

Corrected Mean = 6.35
Corrected Standard deviation = 1.45

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08 Jul 2025, 05:59

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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.

Consider Two Samples n1,n2;

Corrected Result m1 = -0.5n1 + 10; m2 = -0.5n2 + 10;

The average (arithmetic mean) of the original measurements was 7.3 and the standard deviation of the original measurements was 2.9.

Avg of Original = (n1+n2)/2 = 7.3;

SD = 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 = (m1+m2)/2 = ( -0.5n1 + 10 -0.5n2 + 10) / 2 = -0.5(n1+n2)/2 + 10 = -0.5*7.3 + 10 = 6.35

Using the Same method, corrected Standard Deviation can be calculated; however, as we know, adding anything to the original sample value doesn't affect the Standard Deviation, but multiplication or division affects the standard deviation by same multiplier or divisor. Here multiplier is -0.5, however the square will make it +ve, so new SD would be 0.5 * 2.9 = 1.45;

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08 Jul 2025, 06:45

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1. Corrected Mean

let $x_m$ be corrected mean, and $x_n$ be original mean

if m = -0.5n + 10, as this is a linear transformation then,
$x_m$ = -0.5$x_n $+ 10, according to mean theory
= -0.5*7.3 + 10
= 6.35

2. Corrected Standard deviation

For std dev, only the multiplication |-0.5| = 0.5 will be accounted for, not the additive components.
Corrected std. dev. = 0.5* 2.9
= 1.45

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08 Jul 2025, 06:54

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Given:
Original mean n = 7.3
Original SD = 2.9
Correction formula: m = -0.5n + 10

1. Corrected Mean (m):
m = (-0.5) * (7.3) + 10
m = -3.65 + 10 = 6.35

2. Corrected Standard Deviation (S.D. of new mean):

S.D. of m = |-0.5| * (2.9) = 1.45

Therefore, you should select:
Corrected Mean: 6.35
Corrected Standard Deviation: 1.45

Bunuel

This question was provided by GMAT Club
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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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08 Jul 2025, 07:01

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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

Mean in a linear transformation will change to = -0.5*7.3+10=6.35
StD will change to = 0.5*2.9=1.45

Ans C, D

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08 Jul 2025, 07:07

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Mean is modified similarly as the measurements:
Corrected Mean = -0.5*Mean+10 = -0.5*7.3+10 = -3.65+10 = 6.35

Standard derivation doesn't change if some number is added to all the measurements. But it changes if some number multiplies all the measurements. And it changes in the same absolute value as the measurements change.
Corrected Standard Deviation = |-0.5|*Standard Deviation = 0.5*2.9 = 1.45

Correct answers: Corrected Mean=6.35 and Corrected Standard Deviation=1.45

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08 Jul 2025, 07:35

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Each measurement is affected by m = -0.5n + 10, then the average will affected by this formula too.
new_average = -0.5*average + 10 = -0.5*7.3+10 = -3.65+10 = 6.35

Standard derivation is affected only by scaled measurements, not by displacements of this measurements. In the formula, it is affected only by "-0.5*", not by "+10". And standard derivation is always positive or zero.
new_sd = abs(-0.5)*2.9 = 1.45

Answers
Corrected Mean 6.35
Corrected Standard Deviation 1.45

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08 Jul 2025, 07:45

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Corrected Mean: When a linear transformation (m=an+b) is applied to each data point, the new mean (μm ) is simply the transformed value of the original mean (μn).
So, μm = aμn +b.

In this case, a=−0.5, b=10, and μn =7.3, leading to μm =(−0.5)(7.3)+10=−3.65+10=6.35

Corrected Standard Deviation: For a linear transformation (m=an+b), the new standard deviation (σm) is the absolute value of the multiplicative factor 'a' times the original standard deviation (σn).

The additive constant 'b' does not affect the spread (standard deviation) of the data. So, σm = ∣a∣ σn
Here, a=−0.5 and σn =2.9, resulting in σm = ∣−0.5∣ (2.9) = 0.5×2.9 = 1.45.