Is the standard deviation of the set of measurements x1, x2,

Its D..

1)square root of variance is standard deviation.. so it will be 2.. that is less than 3 ..sufficient

2) s.d is 2 .. that is less than 3.. sufficient..

so D is the answer..

OA?????

Is the standard deviation of the set of measurements x1, x2, x3, x4, ..., x20 less than 3 ?

(1) The variance for the set of measurements is 4.
(2) For each measurement, the difference between the mean and that measurement is 2.

1 Statement:

s.deviation = sqrt(variance) = > thus, the std is 2. Sufficient;


2 Statement:

The standard deviation cannot be greater than the difference of observation and the mean of the population, (take min and max of the population compare to mean). Thus, we can safely assume that the std is less than 3. Sufficient.



Post Kudos, if you like my post.

Learned a new thing today

I am of the opinion that calculation is not needed to prove D is sufficient.

Standard deviation of an observation is how far it is from the mean of the observations.
so by definition of SD, D is sufficient..

Please correct me if my understanding is wrong.
Regards,
Sach

For 2 to hold good,

say mean is 4 and for each of the 20 measurements to have a difference of 2 from the mean, all the measurements have to be either 2 or 6. .

Please correct me if my understanding is not correct.

Nice Official Question.
Here is what i did on this Question=>
We need to see if the SD is less than 3 or not.

Statement 1->
Variance =4
Hence S-D=2

Hence Sufficient

Statement 2=>
Using the equation => S.D = \(\sqrt{\frac{\sum(Mean-x_i)^2}{N}}\)
S.D => 2
Hence Sufficient

Hence D

For the second expression, given m-xi = 2
does the SD= \sqrt{4/20} ?
This does not equal 2, which is the answer stated in the thread

Hi

Each m - xi = 2, so there are twenty such 2's. So first we will find variance, and then find square root of that variance to get SD.
To get variance, we need to square all the values of (m - xi), and then find their average.

Since all (m - xi) are 2, their squares will be '4', and thus their average will also be '4'.
So our variance is 4 - it becomes same as first statement. SD = √variance = √4 = 2.

Bunuel can you please elaborate on second statement you say " So, we have that \(m-x_i=2\), since N=20 then we know everything to calculate the standard deviation. Sufficient."

but the question is asking if SD the set of measurements x1, x2, x3, x4, ..., x20 less than 3 ?[/b]

\(standard \ deviation= \sqrt{variance} = \sqrt{\frac{\sum(m-x_i)^2}{N}}\).

(2) says that \(m-x_i=2\), so \(standard \ deviation= \sqrt{variance} = \sqrt{\frac{2^2+2^2+...+2^2}{20}}=\sqrt{\frac{20*2^2}{20}}=2\).

\(L = \left\{ {{x_1}\,,\,\,{x_2}\,,\,\,{x_3}\,,\,\,\, \ldots \,\,\,,\,\,{x_{20}}} \right\}\)

\({\sigma _L} = \sigma \,\,\mathop < \limits^? \,\,3\)

\(\left( 1 \right)\,\,\,{\sigma ^{\,2}} = 4\,\,\,\,\,\mathop \Rightarrow \limits^{\sigma \,\, \geqslant \,\,0} \,\,\,\,\sigma \,\, = \,\,2\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\left\langle {{\text{YES}}} \right\rangle\)


\(\left( 2 \right)\,\,\,\left| {{x_k} - \mu } \right| = 2\,\,\,,\,\,\,\,1 \leqslant k \leqslant 20\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,{\sigma ^{\,2}} = \,\,\,\frac{1}{{20}}\,\, \cdot \,\,\sum\limits_{k = 1}^{20} {\,\,{{\left( {\left| {{x_k} - \mu } \right|} \right)}^2}\,\,\, = } \,\,\,\frac{1}{{20}}\left( {20 \cdot 4} \right)\,\,\, = 4\)

Hence \(\,\,\, \left( 2 \right)\,\,\,\,\, \Rightarrow \,\,\,\,\left( 1 \right)\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\left\langle {{\text{YES}}} \right\rangle \,\,\,\,\)


The correct answer is therefore (D).


This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.

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