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Re: Is the standard deviation of the set of measurements x1, x2, [#permalink]
25 Jun 2012, 02:02

1

This post received KUDOS

Expert's post

SOLUTION

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

CALCULATING STANDARD DEVIATION OF A SET {x1, x2, ... xn}: 1. Find the mean, m, of the values. 2. For each value x_i calculate its deviation (m-x_i) from the mean. 3. Calculate the squares of these deviations. 4. Find the mean of the squared deviations. This quantity is the variance. 5. Take the square root of the variance. The quantity is th SD.

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

(1) The variance for the set of measurements is 4. The variance is just the square of the standard deviation, so if variance=4 then SD=\sqrt{4}=2. Sufficient.

(2) For each measurement, the difference between the mean and that measurement is 2. So, we have that m-x_i=2, since N=20 then we know everything to calculate the standard deviation. Sufficient.

Re: Is the standard deviation of the set of measurements x1, x2, [#permalink]
29 Jun 2012, 03:28

Expert's post

1

This post was BOOKMARKED

SOLUTION

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

CALCULATING STANDARD DEVIATION OF A SET {x1, x2, ... xn}: 1. Find the mean, m, of the values. 2. For each value x_i calculate its deviation (m-x_i) from the mean. 3. Calculate the squares of these deviations. 4. Find the mean of the squared deviations. This quantity is the variance. 5. Take the square root of the variance. The quantity is th SD.

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

(1) The variance for the set of measurements is 4. The variance is just the square of the standard deviation, so if variance=4 then SD=\sqrt{4}=2. Sufficient.

(2) For each measurement, the difference between the mean and that measurement is 2. So, we have that m-x_i=2, since N=20 then we know everything to calculate the standard deviation. Sufficient.

Re: Is the standard deviation of the set of measurements x1, x2, [#permalink]
29 Jun 2012, 03:55

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.

Re: Is the standard deviation of the set of measurements x1, x2, [#permalink]
19 Sep 2012, 22:23

Bunuel wrote:

SOLUTION

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

CALCULATING STANDARD DEVIATION OF A SET {x1, x2, ... xn}: 1. Find the mean, m, of the values. 2. For each value x_i calculate its deviation (m-x_i) from the mean. 3. Calculate the squares of these deviations. 4. Find the mean of the squared deviations. This quantity is the variance. 5. Take the square root of the variance. The quantity is th SD.

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

(1) The variance for the set of measurements is 4. The variance is just the square of the standard deviation, so if variance=4 then SD=\sqrt{4}=2. Sufficient

(2) For each measurement, the difference between the mean and that measurement is 2. So, we have that m-x_i=2, since N=20 then we know everything to calculate the standard deviation. Sufficient.

Re: Is the standard deviation of the set of measurements x1, x2, [#permalink]
19 Jun 2014, 16:37

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