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Average:
We know that multiplying or shifting all the numbers in a set affects the average exactly the same way.
So if the average for n was 7.3, the average for m will be: \((-0.5*7.3) +10 = 6.35\)

Standard deviation:
We know that multiplying a whole set by a number will result in the standard deviation being multiplied by the absolute value of that number. Shifting a set does not affect the standard deviation. So the standard deviation for m will be: \(0.5*2.9=1.45\)

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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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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.

Just basic statistics:
Therefore, the average can be just calculated by adding the average as "n". This gives us: 10- 3.65 = 6.35
For the standard deviation, we have the case, that the fixed value of 10 is not relevant for us. But the "variable" value is. That leads us to: 0.5*2.9 = 1.45, because the value cant be negative.
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corrected mean: 6.35
corrected sd: 1.45

explanation
mean adjustment
m=-0.5n+10
new mean= 0.5*7.35+10= 6.35

sd adjustment
|-0.5| * 2.9=1.45
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If you multiply and add to all elements, the average does the same.

Average:
7.3 * -0.5 + 10 = 6.35

Multiplying affects the SD (lowering it in this case), but addition doesn't (same dispersion but moved 10 units up):

SD
2.9 * -0.5 = -1.45

YET, SD formula squares the differences, hence, is NEVER NEGATIVE.

-> 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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The Corrected Mean: 6.35
Corrected Standard Deviation: 1.45

To calculated the corrected mean, take the original mean and replace for "n" in the correction equation.
-0.5(7.9) + 10 = 6.35

To calculate the corrected standard deviation, replace the "n" for 2.9, but since it is a unit of how much deviation occurs from the center, one does not have to add the "10" and just need to multiple by the corrective -0.5 to get
-0.5(2.9) = 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.
So, this is very simple. We know that correct value, m = 10-0.5n. so Arithmetic mean of values of m in each measurment will be = 10-0.5x(arithmetic mean of values of incorrecct measurments)

now we know that, arithmetic mean value of incorrect values (original) was 7.3, so upon solving we found that arithmetic mean of correct value will be 6.35.

And we noticed that standard deviation was 2.9. for original data. but our new data is 10-0.5n, but we know standard deviation when multiplied by a value increases by that number of times. so final standard deviation will be 2.9/2 = 1.45. { standard deviation do not changes when a constant is added or subtracted from each term}
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Clearly correction on each no will be directly result correction in the average no
So,Corrected mean
= -0.5*7.3+10= 6.35

Now standard deviation =2.9

Standard deviation = summation √((no-mean)^2
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New Mean = a * Old Mean + b
New Mean = -0.5 * 7.3 + 10
= -3.65 + 10
= 6.35

New SD = |a| * Old SD
New SD = 0.5 * 2.9
= 1.45
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As, for each value, m = -0.5n + 10, corrected mean will match the same formula:
Corrected Mean = -0.5 * Mean + 10 = -0.5 * 7.3 + 10 = -3.65 + 10 = 6.35

Standard deviation is a measure of how spread out a set of values is relative to the mean of that set. Adding 10 does not affect the standard deviation, but multiplying by -0.5 scales the standard deviation by the same factor. Sign does not matter here.
Corrected Standard Deviation = 0.5 * Standard Deviation = 0.5 * 2.9 = 1.45

Corrected Mean: 6.35 and Corrected Standard Deviation: 1.45
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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.
The error is in the equation m = -0.5n + 10,

So let's assume that all the items are of mean values, Hence new items are 10 - 3.65 => 6.35.

Hence the new mean is 6.35.

The standard deviation won't differ by the addition, It might impact with the multilication hence the new SD is 1.45.

Hence IMO 6.35, 1.45.
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Each corrected value m is a linear transformation of the original n: m = -0.5n + 10.

The mean of the transformed data = -0.5 ×7.3 + 10 = 6.35.

The standard deviation scales by the absolute value of the multiplier = |-0.5|×2.9=1.45.

Therefore, corrected mean = 6.35 and corrected standard deviation = 1.45.
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Finding the Corrected Mean:
  • Original average is 7.3
  • Formula: m = -0.5n + 10
  • New average = -0.5(7.3) + 10 = -3.65 + 10 = 6.35

Finding the Corrected Standard Deviation:
  • Original standard deviation is 2.9
  • When we multiply data by a number, the standard deviation gets multiplied by that same number
  • We're multiplying by -0.5, so new standard deviation = 0.5 × 2.9 = 1.45
    (Adding 10 doesn't change the standard deviation)

Answer:
  • Corrected Mean: 6.35
  • Corrected Standard Deviation: 1.45
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1. The average (arithmetic mean) of the corrected results = (-0.5 * original mean + 10) = (-0.5 * 7.3 + 10) = 6.35

2. The standard deviation of the corrected results = 0.5 * original SD = 0.5 * 2.9 = 1.45
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\(m = -0.5n + 10\)

Average \(=7.3\)
Std deviation \(=2.9\)

Corrected Mean:

\(m=(-0.5*7.3)+10=6.35\)

Corrected Standard Deviation:

Since all the values are increased by 10, standard deviation will not be affected by this.

But since all the values are halved by \(-0.5n\), standard deviation will also be halved

\(\frac{2.9}{2}=1.45 \)


Corrected Mean: C

Corrected Standard Deviation: D
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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.
· Average (arithmetic mean) of the original measurements, Am = 7.3
· Standard deviation of the original measurements, Sd = 2.9.

To find
· Corrected Mean
· Corrected Standard Deviation

Solution:
If m = a*n + b
Then, Average corrected mean, Acm = a * Am +b
Acm = -0.5* 7.3 + 10
Acm = 6.35

For standard deviation, only multiplication factor, a will affect, not the conatant part, b:
Corrected standard deviation, CSd = |a| *Sd
CSd = |-0.5| * 2.9
Csd = 1.45

Answer:
· Corrected Mean = 6.35
· Corrected Standard Deviation = 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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Let n be the original recorded value. The corrected result m is given by the equation: m = -0.5n + 10

We are given:
1)Average (arithmetic mean) of the original measurements,\(\mu_n\) = 7.3
2)Standard deviation of the original measurements,\(\sigma_n\) = 2.9

We need to find the Corrected Mean (\(\mu_m\)) and the Corrected Standard Deviation (\(\sigma_m\)).

1. Corrected Mean (\(\mu_m\))
For a linear transformation m = an + b, the new mean is given by \(\mu_m\) = a\(\mu_n\) + b. In this case, a = -0.5 and b = 10.
On substituting the values:
\(\mu_m\) = 6.35

2. Corrected Standard Deviation (\(\sigma_m\))
For a linear transformation m = an + b, the new standard deviation is given by \(\sigma_m\) = |a|\(\sigma_n\).
The constant 'b' (the shift) does not affect the spread (standard deviation) of the data.

\(\sigma_m\) = |-0.5|(2.9)
\(\sigma_m\) = (0.5)(2.9)
\(\sigma_m\) = 1.45

Therefore, the Corrected Mean is 6.35 and the Corrected Standard Deviation is 1.45.
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Correct Answer: Corrected Mean= 6.35 & Corrected Standard Deviation= 1.45

Let’s find corrected mean first,
We have original mean= 7.3
we need to correct using, m = -0.5n + 10
n= 7.3
m= -0.5*7.3 + 10
m= -3.65 + 10
m= 6.35

Now let’s find corrected standard deviation,
We can calculate this using,
Transform formula= m= |-0.5|*n
We have original standard deviation= 2.9
we need to correct using, m = -0.5n + 10
n= 2.9
m= |-0.5|*2.9
m= 1.45
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