Question

Find P(X <= k) in each case:

a) n=20, p=0.05, k=2
b) n=15, p=0.7, k=8
c) n=10, p=0.9, k=9

Please help! I don't need the answers, I just need to know how to do the questions.

Answer 1

plug in the given values into the appropriate place

Answer 2

what given space?

Answer 3

given values

Answer 4

im confused, plug them into what?

Answer 5

what type of calculator do you have?

Answer 6

a scientific one

Answer 7

sharp EL-W535X

Answer 8

okay well whats the formula i would use on paper if i was to do it by hand?

Answer 9

what class is this for?

Answer 10

statistics for the sciences

Answer 11

yeah

Answer 12

so what do i do?

Answer 13

P(X≤k) = P(X=0) + P(X=1) + ...+ P(X=k)

Answer 14

does that look familiar?

Answer 15

yeah i know how to do that but am i really supposed to do that all the way up to 20?

Answer 16

well im doing this on paper
i guess i can show my work for the first one and write P(x=2) +--->P(x=20) = ?

Answer 17

you think that would be okay?

Answer 18

i have no idea, depends on what your teacher wants

Answer 19

im not sure tbh
well if theres a faster way i think it would be better to do it that way bc atleast all my work will be shown

Answer 20

do you have a binomcdf function on your calculator?

Answer 21

i have no idea

Answer 22

Just a quick tip, for the last one, instead of doing P(x=0) all the way up to P(x=9), you can just do 1 - P(x = 10). I think you can use a similar method for the 2nd part. Either do that or just use your calculator like you've been saying :P

Answer 23

i dont know how to calculate this question

Answer 24

JE so for part a can i do 1 - P(x=20)? to get the answer?

Answer 25

No, for part a you should just do P(x=0) + P(x=1) + P(x=2)

Answer 26

If you find for any question like this you have to sum a load of things then it might be better to do 1 - the things that aren't in the sum, just to simplify calculations.

Answer 27

but why cant i do this for part a?

Answer 28

You could, but you'd be doing 1 - P(x=20) - P(x=19) - P(x=18) - P(x=17) - ... - P(x=3) which is a lot of calculation. It's much easier in this case to do P(x=0) + P(x=1) + P(x=2).

Answer 29

all the way up to 20?

Answer 30

No, just up to 2. If you're calculating P(x<=k) then you sum from 0 to k, since x can only take the values 0, 1, 2, 3, ..., k if it is <=k.

Answer 31

Like on that other question, you can use P(x<=k) = 1-P(x>k) if it makes the calculations easier, which is very useful for the last part!

Answer 32

okay so i do P(x=0) + P(x=1) + P(x=2) and then - 1?

Answer 33

You don't need the -1 but yes. Essentially you're doing P(x <= 2), so this means x can only take the values 0, 1, 2. You'd only need to do 1 - (stuff) if you were trying to calculate P(x > 2).

Answer 34

okay so i just add up P(x=0) + P(x=1) + P(x=2) to get the answer for a?

Answer 35

Yes :)

Answer 36

okay I got 0.924 for a)

Answer 37

how about for b and c the k values are pretty high

Answer 38

For b I think you're just going to have to add up all the stuff again; but as billj5 was saying your calculator might have a function to calculate P(x<=8) for you, if you know about it.

Answer 39

i dont :(

Answer 40

Ah well you may just have to do the sum :P Maybe ask someone who you can show your calculator to if it has this function?

Answer 41

okay well thank you again JE!

Answer 42

You're welcome :)