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

In summary: More or less, but a continuity correction may be appropriate:The continuity corrections would give F( (65.5-70)/sqrt(21) ) ~= 0.163CBIn summary, to solve the given problem of P(X<65) where X ~B(100,0.7), you can approximate X to be approximately ~N(70,21) and then use the cumulative standard normal function (F) to find the probability. A continuity correction can also be applied for a more accurate result. So, the final calculation would be F((65.5-70)/sqrt(21)) which is approximately 0.163.
  • #1
isa2
2
0
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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  • #2
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
 
  • #3
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
 
  • #4
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))?
 
  • #5
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
 

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

What is the binomial distribution?

The binomial distribution is a probability distribution that describes the probability of obtaining a certain number of successes in a fixed number of independent trials with only two possible outcomes (success or failure) and a constant probability of success for each trial.

What are the parameters of the binomial distribution?

The parameters of the binomial distribution are the number of trials (n) and the probability of success in each trial (p). These parameters determine the shape and characteristics of the distribution.

How is the binomial distribution different from the normal distribution?

The binomial distribution is different from the normal distribution in that it is discrete, meaning that the possible outcomes are countable whole numbers, while the normal distribution is continuous. Additionally, the binomial distribution is skewed to the right if the probability of success is less than 0.5, while the normal distribution is symmetrical.

What is the formula for calculating the probability of a specific number of successes in the binomial distribution?

The formula for calculating the probability of obtaining x successes in a binomial distribution with n trials and a probability of success p is: P(x) = (nCx)(px)(1-p)n-x, where nCx represents the number of combinations of n objects taken x at a time.

How is the binomial distribution used in real life?

The binomial distribution is commonly used in real life to model various phenomena, such as the number of heads in a series of coin flips, the number of defective products in a batch, or the number of successful drug trials out of a total number of trials. It is also used in statistical inference and hypothesis testing to determine the likelihood of obtaining certain results.

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