Question

Thomas surveyed his class to find the number of pets that each student had. He complied his results into a probability distribution which is shown. What is the expected number of pets for a student in his class?

Answer 1

Here is the probability distribution that goes along with it http://screencast.com/t/LGJmj0Wl7Pn

Answer 2

So far i have 0(.04) + .3(1) + .25(2) + .005(3) = .95, However i can't remember what operation to use to find the expected outcome of all students.

Answer 3

that 3 or more part has me overthinking it ... but otherwise your process is fine,

\[E(X)=\sum x~ p(x)\]

Answer 4

well the asnwer doesnt show .95
its just ~1, ~2, ~3, ~4

Answer 5

unless they want me to round up?

Answer 6

.95 is approx 1 isnt it?

Answer 7

yes here ill see if it is correct

Answer 8

yes it was

Answer 9

Would you mind helping me with 1 more question?

Answer 10

i spose they wanted a more realistic value ... kind of hard to have .95 pets :)

Answer 11

i can try

Answer 12

here it is http://screencast.com/t/w08FenBTPr

Answer 13

is the average 80%?

Answer 14

100+80+60+0 = 240 240/4 =60, well 60%

Answer 15

what is the probability of each outcome?

Answer 16

.5^2 ...

Answer 17

P(0 correct) = (3 0) .5^0 .5^3
P(1 correct) = (3 1) .5^1 .5^2
P(2 correct) = (3 2) .5^2 .5^1
P(3 correct) = (3 3) .5^3 .5^0

Answer 18

E(X)= 0 P(0)+ 1P(1)+ 2P(2)+ 3P(3)

you agree?

Answer 19

hmm, maybe better as 0 60 80 and 100 .... not 0 1 2 3

Answer 20

or, find expected number of points and determine it from the grading rubric

Answer 21

how are you getting 5^2, im getting 5.2

Answer 22

i just took 3(.1)+2(.8)+1(.6)

Answer 23

each question is a 50/50 chance of being right,

so for any given test (this is a binomial approach) we have:

rrr
rrw
rwr
wrr
rww
wrw
wwr
www

Answer 24

0 right = 3 wrong

there is 1 www, and w = .5

1 right is 3 cases, rww, wrw, wwr, r=.5 and w=.5

etc

Answer 25

.5^3 is the result that the probability of right = probability of wrong when guessing between 2 options. .5 and .5

Answer 26

does that make sense?

Answer 27

yes

Answer 28

then each permutation result has a .5^3 probability

0 correct = 1 case of .5^3
1 correct = 3 cases of .5^3
2 correct = 3 cases of .5^3
3 correct = 1 case of .5^3

it matter not if we use the number of points scored or the number of questions correct as a basis .... they have equal weight regardless, but you might feel safer finding the expected (mean) of number correct.

Answer 29

0(1*.5^3)+1(3*.5^3)+2(3*.5^3)+3(1*.5^3)

3*.5^3+6*.5^3+3*.5^3

.5^3(3+6+3)

12*.5^3

Answer 30

i was just doing that... in my calculator lol

Answer 31

i just get 1.5

Answer 32

so our average number correct is 1.5 ... rnd to 2?

Answer 33

yep

Answer 34

try it using point values instead and you should get the same result but in point format instead of number correct format

Answer 35

0(1*.5^3)+60(3*.5^3)+80(3*.5^3)+100(1*.5^3)

we get 65 as the average points earned for guessing

Answer 36

so do we round to 80 or stay at 60 ... but thats up to how your material wants you to proceed

Answer 37

just stay but here

Answer 38

There are eight possible ways that the questions on the quiz can be answered.
If R stands for a right answer and W stands for a wrong answer, the eight ways are: RRR, WRR, RWR, RRW, WWR, WRW, RWW, WWW.
This means that the probability of 3 right answers is 18
, the probability of 2 right answers is 38
, the probability of 1 right answer is 38
, and the probability of 0 right answers is 18
. To find the average number of points a student is likely to be awarded on the quiz,
find the expected value:(18)(100) + (38)(80) + (38)(60) + (18)(0) = 65.

Answer 39

thanks for the help, relly appreciate it. :P gave you best

Answer 40

youre welcome :)