Question

Help with 1-3 Exercises

Answer 1

they dont really tell you that smaller sets are easier to work with do they .....

Answer 2

\[mean:~\frac{a+b+c+...+z}{n}\]
\[adjusted~mean:~\frac{(a+b+c+...+z)-kn}{n}=\frac{a+b+c+...+z}{n}-k\]
in other words, for large numbers, you can make it easier on yourself by subtracing a common value from all data set, finding the mean, and adding back the value you subtracted

Answer 3

since to find the standard deviation, you have to subtract the mean to start with, the smaller set is already closer to what you need ..... the standard deviation of the smaller set is equal to the larger set

Answer 4

the decimals can be adjusted in a similar fashion:\[\frac{ar+br+cr+...+zr}{n}=r\frac{a+b+c+...+z}{n}\]

Answer 5

-2,2,-1,6,-3,3,-1,-4?... How to i plug them in?

Answer 6

you have 3 sets of numbers to play with; im not sure which one you are working on with those

Answer 7

Number 1..

Answer 8

ohhh hmm lol i dont understand.... 4, 8, 5, 12, 3, 9, 5, 2 . Could you help me from the beginning ... how i get the mean ?

Answer 9

id suggest subtracting 4 from the given set then:

{ 4, 8, 5, 12, 3, 9, 5, 2} - 4

{ 0, 4, 1, 8, -1, 5, 1, -2} add em up

16

since there are 8 numbers to begin with, we divide this by 8

16/8 = 2, 2 is the mean of the smaller set .... not that we really needed to so this for this set, but its good practice

since we subtracted 4, lets add it back: 2+4 = 6 which is the mean of the larger set

Answer 10

ooohh i see

Answer 11

notice that if we had chosen to subtract 6 to start with; the mean of the smaller set would have been zero:

{ 4, 8, 5, 12, 3, 9, 5, 2} - 6

{ -2, 2, -1, 6, -3, 3, -1, -4}; add em up to get: 0, and 0+6 = 6

Answer 12

the standard deviation is a process of zeroing out the mean and squareing the results

since our mean is 6: { 4, 8, 5, 12, 3, 9, 5, 2} - 6
produces { -2, 2, -1, 6, -3, 3, -1, -4}

so lets square these and add em up

{ -2, 2, -1, 6, -3, 3, -1, -4}

{ 4, 4, 1, 36, 9, 9, 1, 16}

{ 80 }

Answer 13

divide again by the number of data points given, in this case 8

80/8 = 10 ; this value is called the variance

and then take the square root of it: sqrt(10)

Answer 14

this process will become more apparent on the second problem that has larger numbers

Answer 15

okay i get it... so for the second one 102 98 103 86 101 110 it's -102? then add them

Answer 16

the value you subtract by is completely up to you; its only purpose is to make you life simpler. as a rule of thumb, you want it to be as close to the mean as possible, id use about half the range: 80 to 102 is the range; so 100 seems to be a good guess of the mean:

Answer 17

102 98 103 86 101 110
-100 ..............................
-------------------------
2 -2 3 -14 1 10

Answer 18

i think that sums up to zero .... and 0 + 100 = 100

Answer 19

well, 0/6 = 0 for the smaller set ... and add back the 100 :)

Answer 20

you see how in step 2 it tells you to subtract the mean from the data set?

since 100 is the mean of the data set, and we (by luck) already subtracted 100 to get our smaller set .... we dont need to rework it again do we :)

Answer 21

standard deviation is: square, add, divide, then sqrt

2 -2 3 -14 1 10

Answer 22

so we add these now?? then divide the added numbers

Answer 23

yep, think of it like finding the mean of the squares

Answer 24

square these first, then add em

Answer 25

2 -2 3 -14 1 10
4 4 9 196 1 100 <-- now add

Answer 26

ahhh 314 then divide

Answer 27

is 52.3333

Answer 28

correct, since this is a "population" of 6 data points ... divide by 6 to determine the variance

52.333333 is good

standard deviation is the square root of variance; so sqrt that

Answer 29

7.23417813807024

Answer 30

correct, and clean that up as needed, prolly rounded to like 7.23

Answer 31

do you know the difference between a population, and a sample?

Answer 32

Thanks so much for helping me...

Answer 33

no i do not

Answer 34

if you are getting further into statistics with this, which is the whole reason for learning this to begin with .... a sample is a subset of a population. the standard deviation of a "sample" will be 1 less then the dataset

for example, in this case, 6-1 = 5, we would have divided the variance by 5 and sqrted for the "sample" standard deviation

Answer 35

if you are not having to do these things by hand, then my "smaller set" work is not needed. A calculator will addem all up just as easily

Answer 36

\[mean=\frac{sum(data~set)}{\#~data~set}\]

Answer 37

okay thanks. so for the #3. 8.2, 11.6, 8.7, 10.6, 9.4, 10.1, 9.3

Answer 38

oh no , 2.25,3.61,1,0.810000000000001,0.0899999999999994,0.16,0.159999999999999

Answer 39

sum {8.2, 11.6, 8.7, 10.6, 9.4, 10.1, 9.3}

http://www.wolframalpha.com/input/?i=sum+%7B8.2%2C+11.6%2C+8.7%2C+10.6%2C+9.4%2C+10.1%2C+9.3%7D

Answer 40

if by calculator, just sum up the dataset as is ..... makes for less work

Answer 41

i wanted to see how to work was completed cause the calculator just give answers.

Answer 42

Your awsome though thanks for helping :)

Answer 43

:) youre welcome

Answer 44

if by hand, i would have considered multiplying by 10 and finding the mean

{8.2, 11.6, 8.7, 10.6, 9.4, 10.1, 9.3} * 10

{82, 116, 87, 106, 94, 101, 93}

80 to 120 range, so about 100 guess on the mean is fine

{82, 116, 87, 106, 94, 101, 93} - 100

{-18, 16, -13, 6, -6, 1, -7}, add em up

{-21}

there are 7 numbers in the dataset: -21/7 = -3

the mean of this smaller set is -3, add back 100 to get 97 .... but thats not working out like a want it to so i will have to consider why .... and adjust it appropriately

Answer 45

\[\frac{(d_1-k)r+(d_2-k)r+(d_3-k)r+...+(d_n-k)r}{n}\]
\[r~\frac{d_1-k+d_2-k+d_3-k+...+d_n-k}{n}\]
\[r~(\frac{d_1+d_2+d_3+...+d_n}{n}-k)\]
i see, i cant divide out the decimals first, i would have to reduce the set and then divide out 10 .... bummer

Answer 46

sorry, i was doing fine; 9.7 is the mean

-3 + 100 = 97

97/10 = 9.7

i entered the wrong values to begin with :)

Answer 47

k How do i show the work ?