anonymous
  • anonymous
An experiment consists of 48 binomial trials, each having probability 3/4 of success. Use an approximating normal curve to estimate the following probabilities. (a) between 35 and 39 successes, inclusive.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
use the fact that when np>5, then a normal approximation with mean np and std. dev with sqrt(np(1-p)) allows you to solve this. So, the normal curve with mean 36 and std. dev. of 3 will be what you need to use
anonymous
  • anonymous
yea i found all those already. n= 48 p = .75 mean = 36 st. dev. = 3 I tried 36.5-36/3 but the answer I got was wrong.... I don't know what I'm doing wrong.
anonymous
  • anonymous
oops not 36.5-36 that's for another question. Ignore the bottom part.

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anonymous
  • anonymous
I guess I should say, you should then use 35
anonymous
  • anonymous
I thought I would do 35 - 36/3 and 39-36/3 but that's wrong too. I'm not sure what I'm doing.
anonymous
  • anonymous
35-36=-1, then divide by 3 is -1/3 39-36=3 then divide by 3 is 1
anonymous
  • anonymous
those are the z-values you want to use...Is that still giving you the wrong answer?
anonymous
  • anonymous
should be around 47% or so yes?
anonymous
  • anonymous
I know I got those answers, but when I subtracted the area of those...it was still wrong. .8413-.3821 = .4592
anonymous
  • anonymous
Still not right....
anonymous
  • anonymous
what is the answer? maybe they want you to use a continuity correction
anonymous
  • anonymous
I don't know the answer, they won't give it to me. I have to keep on trying until I get it right.
anonymous
  • anonymous
hmm, that is strange. if you get really desperate you could always sum up the bernoulli cases 35, 36, 37, 38, and 39 if you have access to technology that computes combinations
anonymous
  • anonymous
Yes I have a TI-83. How would I do that? 35C0 + 36C.....those?
anonymous
  • anonymous
48C35*p^35*(1-p)^13+48C36*p^36*(1-p)^12+...
anonymous
  • anonymous
.5795 is what I got
anonymous
  • anonymous
if you use continuity corrections, then you get this answer. I had forgot to do that. when you go from a discrete to a continuous probability, this can sometimes greatly affect your answer, especially when bounded on both sides
anonymous
  • anonymous
.5795 is incorrect too... Thanks for trying to help me though!
anonymous
  • anonymous
man, that is crazy,I'm a stats man and this is embarrassing ;( is it a computer system? then maybe it is rejecting due to rounding, you think?
anonymous
  • anonymous
with continuity correction i got around .57. other than that, I got nothing...though this problem will bother me for a while!
anonymous
  • anonymous
I tried .5795, .57, and .58. None of them worked. I have been working on this problem for a while now haha so it's okay. Thank you for trying to help me with this though! I have another question though if you still want to help me? The third part of the question asks me to find "more than 36 successes" I did 36.5-36 / 3 which = .1666, so the area would be .5636 but that's incorrect. What am I doing wrong?
anonymous
  • anonymous
greater than means you have to take 1-minus the value from the z-score
anonymous
  • anonymous
the z-table gives you less than percentage, not greater than percentage
anonymous
  • anonymous
Which would make it .4364 ..... which is still wrong...
anonymous
  • anonymous
I feel like I'm making a silly mistake.
anonymous
  • anonymous
hmm, I just don't know why it isn't working out. that is definitely what you are supposed to do when you have a bernoulli problem and are using the normal curve to approximate it. I really don't know what to tell ya. did you try this one without the cont. correction? it would be 50%. If that doesn't work then I really have no idea:(
anonymous
  • anonymous
actually, more than 36 without cont.correction would be 37, which would give 1/3 and then 1-that would be around .367...I have no clue at this point though. wish you luck
anonymous
  • anonymous
:( Okay well thank you for helping me!

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