Question

A list of numbers has an average of 48 and an SD of 12. The list is transformed by subtracting 3 from each entry and then dividing the result by 5.

Answer 1

Find the average of the transformed list.

Answer 2

@phi

Answer 3

the average is sum/N
we don't know the sum or N but we know sum/N= 48
and sum= N*48
if we subtract 3 from each number, and there are N numbers, how much do we subtract?
I think N*3
so the new sum is
Sum_new= N*48-N*3= N(48-3)= N*45
if we divide Sum_new by N we get the new average

Answer 4

if we toss in divide by 5
then
Sum_new= N*45/5 = N*9
Sum_new/N= 9 is the average for this new list

Answer 5

thanks

Answer 6

please help @phi

Answer 7

how do you find the s.d. of a list?

Answer 8

A list of numbers has an average of 48 and an SD of 12. The list is transformed by subtracting 3 from each entry and then dividing the result by 5.

Answer 9

Find the SD of the transformed list.

Answer 10

yes. How do you find the standard deviation of a list ?

Answer 11

please help.

Answer 12

to find the standard deviation squared (the variance) of N numbers:

\[\sigma^2=\frac{ 1 }{ N }\sum_{i=1}^{N}(x_i-\bar{x})^2\]

Answer 13

the original list has s.d.=12 and var= 12^2= 144
We don't know how many numbers we have, but the computation would look like this:
\[ \frac{(a-48)^2}{N} + \frac{(b-48)^2}{N}+...+\frac{(z-48)^2}{N}= 144 \]
we can write this as
\[ (a-48)^2 + (b-48)^2 + ... + (z-48)^2 = 144 N \]

Answer 14

The transformed numbers will each have the form
\[ \frac{a-3}{5} \]
Their variance (that is, standard deviation squared) is computed as
\[ \frac{1}{N}\left(\frac{a-3}{5}-9\right)^2 + \frac{1}{N}\left(\frac{b-3}{5}-9\right)^2 + ... + \frac{1}{N}\left(\frac{z-3}{5}-9\right)^2= \sigma^2\]
notice that we use the mean of the transformed list (the 9) rather than the mean of the original list.

we can simplify this a bit. Multiply both sides by N. Write 9 as 45/5 so we can combine the fractions:
\[ \left(\frac{a-3-45}{5}\right)^2 + \left(\frac{b-3-45}{5}\right)^2+...+ \left(\frac{z-3-45}{5}\right)^2= N\sigma^2\]
we now get
\[ \frac{(a-48)^2}{25}+ \frac{(b-48)^2}{25}+ ... +\frac{(z-48)^2}{25}= N\sigma^2\]
or, after multiplying both sides by 25:
\[ (a-48)^2 + (b-48)^2 + ...+(c-48)^2= 25N\sigma^2 \]
Interestingly, we know the sum on the left side, it is 144N, so we have
\[ 144N= 25N\sigma^2 \]
solving for sigma, we get
\[ \sigma= 12/5 \]

Answer 15

Thanks