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.

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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
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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
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.
oops not 36.5-36 that's for another question. Ignore the bottom part.

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I guess I should say, you should then use 35
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.
35-36=-1, then divide by 3 is -1/3 39-36=3 then divide by 3 is 1
those are the z-values you want to use...Is that still giving you the wrong answer?
should be around 47% or so yes?
I know I got those answers, but when I subtracted the area of those...it was still wrong. .8413-.3821 = .4592
Still not right....
what is the answer? maybe they want you to use a continuity correction
I don't know the answer, they won't give it to me. I have to keep on trying until I get it right.
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
Yes I have a TI-83. How would I do that? 35C0 + 36C.....those?
48C35*p^35*(1-p)^13+48C36*p^36*(1-p)^12+...
.5795 is what I got
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
.5795 is incorrect too... Thanks for trying to help me though!
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?
with continuity correction i got around .57. other than that, I got nothing...though this problem will bother me for a while!
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?
greater than means you have to take 1-minus the value from the z-score
the z-table gives you less than percentage, not greater than percentage
Which would make it .4364 ..... which is still wrong...
I feel like I'm making a silly mistake.
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:(
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
:( Okay well thank you for helping me!

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