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

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Discussion Overview

The discussion revolves around solving a problem using the binomial distribution and normal approximation, specifically focusing on how to handle calculations when limited to certain tables in a Mathematics Handbook for Science and Engineering. The problem involves calculating probabilities for a binomial random variable.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Homework-related

Main Points Raised

  • One participant expresses confusion about how to use limited tables for binomial, normal, and Poisson distributions to solve a specific problem.
  • Another participant reformulates the question to clarify the probability to be calculated: P(X<65) where X ~B(100,0.7).
  • A participant suggests that X can be approximated by a normal distribution with mean Np and variance Np(1-p), leading to the calculation P(X<65) using the cumulative standard normal function.
  • Further discussion includes the calculation of F((65-70)/sqrt(21)) and the interpretation of results from the normal table, with some uncertainty about the exact values.
  • Another participant introduces the idea of continuity correction, suggesting that using F((65.5-70)/sqrt(21)) may yield a different probability estimate.

Areas of Agreement / Disagreement

Participants appear to agree on the general approach of using normal approximation for the binomial distribution, but there is uncertainty regarding the exact calculations and the necessity of continuity corrections. No consensus is reached on the final probability value or the method of calculation.

Contextual Notes

Participants express uncertainty about the values obtained from the normal table and the application of continuity corrections, indicating that these aspects may require further clarification or refinement.

isa2
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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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