anonymous
  • anonymous
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.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
plug in the given values into the appropriate place
anonymous
  • anonymous
what given space?
anonymous
  • anonymous
given values

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anonymous
  • anonymous
im confused, plug them into what?
anonymous
  • anonymous
what type of calculator do you have?
anonymous
  • anonymous
a scientific one
anonymous
  • anonymous
sharp EL-W535X
anonymous
  • anonymous
okay well whats the formula i would use on paper if i was to do it by hand?
anonymous
  • anonymous
what class is this for?
anonymous
  • anonymous
statistics for the sciences
anonymous
  • anonymous
yeah
anonymous
  • anonymous
so what do i do?
anonymous
  • anonymous
P(X≤k) = P(X=0) + P(X=1) + ...+ P(X=k)
anonymous
  • anonymous
does that look familiar?
anonymous
  • anonymous
yeah i know how to do that but am i really supposed to do that all the way up to 20?
anonymous
  • anonymous
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) = ?
anonymous
  • anonymous
you think that would be okay?
anonymous
  • anonymous
i have no idea, depends on what your teacher wants
anonymous
  • anonymous
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
anonymous
  • anonymous
do you have a binomcdf function on your calculator?
anonymous
  • anonymous
i have no idea
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anonymous
  • anonymous
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
anonymous
  • anonymous
i dont know how to calculate this question
anonymous
  • anonymous
JE so for part a can i do 1 - P(x=20)? to get the answer?
anonymous
  • anonymous
No, for part a you should just do P(x=0) + P(x=1) + P(x=2)
anonymous
  • anonymous
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.
anonymous
  • anonymous
but why cant i do this for part a?
anonymous
  • anonymous
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).
anonymous
  • anonymous
all the way up to 20?
anonymous
  • anonymous
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.
anonymous
  • anonymous
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!
anonymous
  • anonymous
okay so i do P(x=0) + P(x=1) + P(x=2) and then - 1?
anonymous
  • anonymous
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).
anonymous
  • anonymous
okay so i just add up P(x=0) + P(x=1) + P(x=2) to get the answer for a?
anonymous
  • anonymous
Yes :)
anonymous
  • anonymous
okay I got 0.924 for a)
anonymous
  • anonymous
how about for b and c the k values are pretty high
anonymous
  • anonymous
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.
anonymous
  • anonymous
i dont :(
anonymous
  • anonymous
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?
anonymous
  • anonymous
okay well thank you again JE!
anonymous
  • anonymous
You're welcome :)

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