The standard deviation is most commonly used to get a sense of how far the typical score of a distribution differs from the mean. In computing the standard deviation, why is it necessary to square the deviations from the mean for each score?

so as to spread out the differences

Step-by-step explanation: if we try to take the algebraic sum of each data set, negative and positive number of the same value cancels out.

If we try to take the modulus of the data set to have an absolute value, and sum up the answer, that gives us the mean deviation.

But by taking the square of the deviations (and taking the square root at the end) allows for more spread of the data set which gives a better results.

Let us test this with some number.

Assuming the data set

2, 3, 1 The mean is 2

By taking algebraic sum of deviation, we have

(2-2) + (3-2) + (1-2) = 0 + 1 - 1 = 0

You can see 2 and -2 canceling out each other.

If we take the absolute value of deviation and sum it up, we have the mean deviation ( which is not important for this question, so we won't bother doing it)

For sum of squares of deviation, we have

(2-2)² + (3-2)² + (1-2)² = 1 + 1 = 2

√2 = 1.4142

As we can see from our value it is alot more detailed and spreads better

Answer; squaring helps to spread out the differences and also gives a positive value.

Step-by-step explanation:

squaring helps to spread out the differences and also gives a positive value so that the sum will not be equal to zero. When we take the algebra sum of each of the data set positive and negative numbers of the same value will gives zero. e.g 6-6 =0. Now taking the square and the square roots at end gives a spread of the data set which in turn yield a better results.

for example giving the set of data 2, 4, 6

Mean = (2+4+6)/3 =4

Now

Algebra sum of deviation = (2-4)+(4-4)+(6-4)= 0

While the sum of squares of the deviation will not give zero.

360/29

step-by-step explanation:

for this pair of functions g(x)=x-2 and f(x)=|x| is (g o f)(a)= |a| -2

step-by-step explanation:

g(x)=x-2

f(x)=|x|

(g o f)(a)=?

(g o f)(x)=g(f(x))

(g o f)(x)=g(|x|)

(g o f)(x)=|x|-2

(g o f)(a)=|a|-2