Question

Find the probability that at least 4 of the 10 randomly selected parts weigh greater than 41 pounds.
Mean 40 pounds
SD 1.5 pounds

Answer 1

Is that all the information given?

Answer 2

yes

Answer 3

Let's assume we are given the probability distribution of selecting a part. The mean of this distribution is 40 , and standard deviation is 1.5 pounds.

Answer 4

It looks like there is more to this question. it says 'the parts' what parts?

Answer 5

"Find the probability that at least 4 of the 10 randomly selected parts weigh greater than 41 pounds. "
Was there a first part to this question

Answer 6

Parts refers to what they are choosing not anything else. its like saying randomly selected fruits.

Answer 7

The problem is, we need to know the probability of choosing one part that is greater than 41 pounds in order to answer the question.
Can we assume the parts are normally distributed with a mean of 40 pounds and st. dev of 1.5?

Answer 8

the original population of parts

Answer 9

yes we can assume that

Answer 10

i found the number for that was around 0.2525

Answer 11

hmm, does it state that in the question (that's why i wasnt sure)
also is this a multiple choice question, what are the choices .

Answer 12

no it does not and its not multiple choice

Answer 13

P ( X > 41) = normalcdf ( 41, 1e99, 40 , 1.5) = .25249 ~ .2525

Answer 14

yeah i dont like to make assumptions that are not given explicitly :)
this problem is a bit vague, but we can just impose our assumptions i guess

Answer 15

ok so we know the probability of choosing 1 part that is greater than 41 pounds.
now lets answer the question, suppose we sample 10 parts from the population. what is probability at least 4 is greater than 41 pounds

Answer 16

yeah at this point I'm very ok with assumptions this has been giving me lots of trouble

Answer 17

so we are looking at samples of ten (instead of samples of one)

Answer 18

lets define a variable Y
Y = number of parts which are success , or greater than 41 pounds
note that Y can equal to 0,1,2,3,... 9,10

Answer 19

ok

Answer 20

the question is asking
find P( Y>= 4)
this is a binomial distribution problem

Answer 21

it is easier to find the complement
P( Y>= 4) = 1 - P( Y < 4 )

Answer 22

ok

Answer 23

1 - P( Y<4 ) = 1 - [ P( Y=0) + P( Y= 1) + P( Y = 2) + P( Y = 3 ) ]

Answer 24

we need to find
P( Y=0) + P( Y= 1) + P( Y = 2) + P( Y = 3 )

lets do this in separate calculations

P(Y=0) = (1-.2525)^10

P(Y=1) = (10 choose 1 ) * (.2525)^1 * (1-.2525)^9

P(Y=2) = (10 choose 2 ) * (.2525)^2 *(1-.2525)^8

P(Y=3) = (10 choose 3 ) * (.2525)^3 *(1-.2525)^7

Answer 25

if you want to be consistent you can write
P (Y = 0) = (10 choose 0 ) * (.2525)^0 * ( 1-.2525)^10 ,
= 1 * 1 * (1-.2525)^10

Answer 26

ok

Answer 27

also you can do this on a TI 83 calculator ,
1- P ( Y < 4) = 1 - P( Y <= 3 ) = 1 - binomcdf( 10, .2525, 3 ) = .22999

Answer 28

can you elaborate on the calculator way??

Answer 29

There is a built-in command binomcdf (binomial cumulative density function) that can be used to quickly determine "at most". Because this is a "cumulative" function, it will find the sum of all of the probabilities up to, and including.

the command is
2ND -> VARS ->(press down until you see binomcdf (

binomcdf (number of trials, probability of occurrence, number of specific events)

or shortened we have

binomcdf (n, p, r)

Answer 30

P(Y <= 3) , p = .2525, n = 10

is equal to
binomcdf ( 10, .2525, 3 )

Answer 31

also I used the fact that P ( Y<4 ) = P ( Y <= 3 ) , which makes sense since 0,1,2,3 are less than 4

Answer 32

Woah i had no idea this was possible on the calc

Answer 33

yes its nice

Answer 34

there is a limitation though, for example if you want P ( 2<= Y <= 5 ) , you have to do
P ( Y < = 5) - P ( Y <= 1 )
since
P( Y<= r ) = binomcdf ( n, p , r )

Answer 35

ah ok

Answer 36

so whats next to solve this completely

Answer 37

so you subtract that from 1 to get your answer

Answer 38

1 - binomcdf( 10, .2525, 3 ) = .22999

Answer 39

oh ok that makes sense

Answer 40

wait does this answer the original question??

Answer 41

yes

Answer 42

there were two parts to answering the question. first we had to find the probability of picking 1 item that had a weight greater than 41 pounds (ie picking samples of one).

Then the second part, suppose instead of sampling one item you sampled ten at a time, what is probability that at least 4 of the items have weight greater than 41 pounds.

we needed the first part in order to find p = .2525 for the binomial distribution

Answer 43

do u still need help @rowerguy508 ?

Answer 44

do you agree with this? so far

Answer 45

who?

Answer 46

you Noland

Answer 47

oh yes i do

Answer 48

I mean if you have a different solution,

Answer 49

great work guys

Answer 50

i wasnt sure because the directions were a bit vague