Each term of set T is a multiple of 5. Is standard deviation

Revised version of this question is below:

Each term of set T is a multiple of 5. Is standard deviation of T positive?

The standard deviation of a set shows how much variation there is from the mean, how widespread a given set is. So, a low standard deviation indicates that the data points tend to be very close to the mean, whereas high standard deviation indicates that the data are spread out over a large range of values. So, basically we can say that it in a sense measures the distance and the distance can not be negative, which means that the standard deviation of any set is greater than or equal to zero: .

Next, the standard deviation of a set is zero if and only the set consists of identical numbers (or which is the same if the set consists of only one number).

(1) Each term of set T is positive --> if T={5} then then SD=0 but if set T={5, 10} then SD>0. Not sufficient.

(2) Set T consists of one term --> any set with only one term has the standard deviation equal to zero. Sufficient.

Answer: B.
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my answer is B..

1.) Let T = { 5,15}
Mean = 10
\(sd =\sqrt{\frac{((5 - 10)^2 + (15-10)^2)}{2}}\)

=>\(sd = 5\) > 0

but, if T = {5,5}
Then Sd = 0 ,, uncertainity,,,hence, insufficient..

2.) T has only one member..
sd = 0 hence, sufficient.. sd is not positive..

Ans is B

1) If pos nos are 5 10 15 the SD>0
If pos nos are same 15 15 15 SD= 0

SD=0 or SD>0

2) Only one number SD= 0
Hence suff to say SD is not greater than zero . SUFFICIENT

SD in layman terms is the distance between each term and the mean of the set

Given: Each term of the set is a multiple of 5
Required: SD of the set > 0?

Statement 1: Each term of the set is positive
There can be various versions of this set
Set 1: {5, 10, 15}
In this case, SD = 5 (Distance of the elements of the set from the mean)
Set 2: {5, 5, 5}
In this case, SD = 0 (Distance of the elements of the set from the mean)
No unique value of SD

INSUFFICIENT

Statement 2: Set T consists of one term
If there is only one element in the set, then SD = 0
Hence we have a unique value of SD

SUFFICIENT

Correct Answer B

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