MHB How to Solve a Problem Using Binomial Distribution and Normal Table?

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To solve the problem using binomial distribution and normal approximation, the assignment requires calculating P(X<65) for X ~ B(100, 0.7). The mean and variance can be approximated as N(70, 21). The cumulative standard normal function, F, is then used to find P(X<65) by calculating F((65-70)/sqrt(21)), which yields approximately 0.1379 without continuity correction. However, applying a continuity correction gives a more accurate result of about 0.163. This approach effectively utilizes the normal distribution table for larger binomial problems.
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I have an assignment
1z22b9f.png
which is a bit different,
I have to use Mathematics Handbook for Sience and Engineering to solve the problem,
I can look it up in tables. But the tables for binomial functions is only up to 20,
Normal Distribution to 3.4 and Poisson up to 24 in some cases.
So how do I do it? Approximation of some kind?
 
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isa said:
I have an assignment image.png - Speedy Share - upload your files here which is a bit different,
I have to use Mathematics Handbook for Sience and Engineering to solve the problem,
I can look it up in tables. But the tables for binomial functions is only up to 20,
Normal Distribution to 3.4 and Poisson up to 24 in some cases.
So how do I do it? Approximation of some kind?

If you really want help try posting your question in a form that does not require helpers to jump through multiple hoops just to see it.

Your question is:

P(X<65) where X ~B(100,0.7)

CB
 
CaptainBlack said:
Your question is:

P(X<65) where X ~B(100,0.7)

CB

X is approximately ~N(70,21) (since the mean is Np and the variance is Np(1-p)), so P(X<65) ~= F((65-70)/sqrt(21)) where F denotes the cumulative standard normal, which is about 13.8%.

CB
 
CaptainBlack said:
X is approximately ~N(70,21) (since the mean is Np and the variance is Np(1-p)), so P(X<65) ~= F((65-70)/sqrt(21)) where F denotes the cumulative standard normal, which is about 13.8%.

CB

If I understan you corectly F((65-70)/sqrt(21)) = F(1.09108..)? And then look it up in the Normal table witch is 0.8621 or 0.8643 I'm not quite sure.
And then take 1-0.8621 and you get 0.1379?
Is that how you do it?

You do not have to include F((0-70)/sqrt(21)) ?
Since it is < 65? so like F((65-70)/sqrt(21)) -F((0-70)/sqrt(21))?
 
isa said:
If I understan you corectly F((65-70)/sqrt(21)) = F(1.09108..)? And then look it up in the Normal table witch is 0.8621 or 0.8643 I'm not quite sure.
And then take 1-0.8621 and you get 0.1379?
Is that how you do it?

You do not have to include F((0-70)/sqrt(21)) ?
Since it is < 65? so like F((65-70)/sqrt(21)) -F((0-70)/sqrt(21))?

More or less, but a continuity correction may be appropriate:

The continuity corrections would give F( (65.5-70)/sqrt(21) ) ~= 0.163

CB
 
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