Question 1190761
<font color=black size=3>
Let's say we had the set
{400,500,600}
The mean is 500. We can check as such
(400+500+600)/3 = 1500/3 = 500


Now let's add 10 to each number
{410,510,610}
and repeat the process of finding the mean
(410+510+610)/3 = 1530/3 = 510
The mean has increased by 10.


We can rewrite that equation like so
(410+510+610)/3 = (400+500+600 + 10+10+10)/3 
(410+510+610)/3 = 500 + (10+10+10)/3 
(410+510+610)/3 = 500 + (3*10)/3 
(410+510+610)/3 = 500 + 10
(410+510+610)/3 = 510
If we add 10 to each number in the set, then the mean goes up by 10.


This can be applied in a more general sense, and it doesn't only work with this particular set (nor does it only apply to sets of 3 items)


{x1,x2,x3,...,xn} = original set of values
S = sum of the original items = x1+x2+x3+...+xn
S/n = original mean


Let's say we add on k to each number
This is equivalent to adding nk because there are n copies of k being added on.
So,
new mean = (S+nk)/n
new mean = (S/n)+(nk/n)
new mean = (old mean)+k
Therefore, what happened with the "adding 10" example earlier wasn't just a coincidence.


----------------------


Now let's consider multiplying each item in the set {x1,x2,...,xn} by some constant R
We would get this new set {Rx1,Rx2,...,Rxn}


Let's find the new mean of this scaled set.
new mean = (sum of the x values)/n 
new mean = (Rx1+Rx2+...+Rxn)/n 
new mean = (R*(x1+x2+...+xn))/n 
new mean = (R*S)/n
new mean = R*(S/n) 
new mean = R*(old mean)


So if we were to multiply each value by R = -5, then,
new mean = R*(old mean)
new mean = -5*(500)
new mean = -2500


Or you could go back to the example with 3 items from earlier
{400,500,600}
Multiply everything by -5 to get
{-2000,-2500,-3000}
Then compute the mean
(-2000+(-2500)+(-3000))/3 = -7500/3 = -2500


To summarize:
If you add 10 to each number, then the mean goes up by 10
If you multiply each number by -5, then the mean is multiplied by -5



============================================
Answers:
a) 510
b) -2500
</font>