Question 1016582
The range is the maximum value minus the minimum value,
so in both cases the range is {{{14-0=14}}} .
You notice that the middle value (called the median) is {{{7}}} ,
and that happens to also be the mean or average,
because {{{((0+6+7+8+14))/5=35/5=7}}} .
The set 0, 6, 7, 8, 14 has
three scores clustered around the mean in the center of the distribution (6, 7, and 8),
and two extreme values (0, and 14).
When you replace {{{0}}} for {{{6}}} and {{{14}}} for {{{8}}} ,
you get 0,0,7,14,14,
which also has {{{7}}} and the mean, {{{((0+0+7+14+14))/5=35/5=7}}} ,
and also as the median.
All the other values are as far from {{{7}}} as possible,
while still having a range of {{{14}}} .
The values in the set 0,0,7,14,14 are more spread out,
showing more variability.
The range does not show that.
It only takes into account the extreme values,
and so it is the same for the two sets: {{{14-0=14}}} .
The standard deviation does show the different variability,
because it takes into account how different from the mean all the values are.
The standard deviation can be calculated by using a calculator or some computer program.
Otherwise, you can
calculate the differences between each value and the mean,
square them,
calculate the average of those squares,
and then take the square root of the result.
For the set 0,0,7,14,14, the differences from the mean are
{{{0-7=-7}}} , {{{0-7=-7}}} , {{{7-7=0}}} , {{{14-7=7}}} , and {{{14-7=7}}} .
The squares of those differences are
{{{(-7)^2=49}}} , {{{(-7)^2=49}}} , {{{0^2=0}}} , {{{7^2=49}}} , and {{{7^2=49}}} .
Their average is {{{((49+49+0+49+49))/5=4*49/5}}} .
The square root of that is
{{{sqrt(4*49/5)=sqrt(49)sqrt(4)/sqrt(5)=7*2/sqrt(5)=14/sqrt(5)}}} ,
which can be more elegantly written as {{{14sqrt(5)/5}}} .
For the set 0,6,7,8,14, the differences from the mean are
{{{0-7=-7}}} , {{{6-7=-1}}} , {{{7-7=0}}} , {{{8-7=1}}} , and {{{14-7=7}}} .
The squares of those differences are
{{{(-7)^2=49}}} , {{{(-1)^2=1}}} , {{{0^2=0}}} , {{{1^2=1}}} , and {{{7^2=49}}} .
Their average is {{{((49+1+0+1+49))/5=100/5}}} .
The square root of that is
{{{sqrt(100/5)=sqrt(100)/sqrt(5)=10/sqrt(5)}}} ,
which can be more elegantly written as {{{10sqrt(5)/5}}} .
Comparing the two sets' standard deviations, we find that
{{{14/sqrt(5)>10/sqrt(5)}}} , which showa that the set 0,0,7,14,14, has more variability.
Of course we knew that, because we made it so,
by moving the values {{{6}}} and {{{8}}} away from the mean {{{7}}} .