Question 1193397: Fifty percent of midsize manufacturers planned to have management representative visit Canada and Mexico to take advantage of opportunities created by the North American Free Trade Agreements. An export-import group in Toronto , Canada , has invited 20 US midsize manufacturers to participate in the conference to explore trade opportunities. What is the probability that 12 or more of these manufacturers will send representatives?
Found 2 solutions by ikleyn, math_tutor2020:
Answer by ikleyn(52781)   (Show Source):
You can put this solution on YOUR website! .
Fifty percent of midsize manufacturers planned to have management representative visit Canada and Mexico
to take advantage of opportunities created by the North American Free Trade Agreements.
An export-import group in Toronto , Canada , has invited 20 US midsize manufacturers to participate
in the conference to explore trade opportunities.
What is the probability that 12 or more of these manufacturers will send representatives?
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It is a binomial distribution probability problem.
The number of trials is n=20; the individual probability of success is 0.5;
the number of success is k>=12.
Use thhis online free of charge calculator
https://stattrek.com/online-calculator/binomial.aspx
The ANSWER is P(n=20; k>=12; p=0.5) = 0.2517 (rounded).
Solved.
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If you want to see many similar (or different) solved problems, look into the lessons
- Simple and simplest probability problems on Binomial distribution
- Typical binomial distribution probability problems
in this site.
After reading these lessons, you will be able to solve such problems on your own.
Answer by math_tutor2020(3817)  (Show Source):
You can put this solution on YOUR website!
Answer: 0.251722 (approximate)
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Explanation:
x = number of successes = number of companies that send a representative
n = sample size = 20
p = probability a company sends a representative = 50% = 0.5
B(x) = binomial probability
B(x) = (n C x)*(p)^(x)*(1-p)^(n-x)
B(x) = (20 C x)*(0.5)^(x)*(1-0.5)^(20-x)
B(x) = (20 C x)*(0.5)^(x)*(0.5)^(20-x)
The n C x refers to the nCr combination formula.
Plug in x = 12
B(x) = (20 C x)*(0.5)^(x)*(0.5)^(20-x)
B(12) = (20 C 12)*(0.5)^(12)*(0.5)^(20-12)
B(12) = (125,970)*(0.5)^(12)*(0.5)^(8)
B(12) = 0.120134 approximately
This is the approximate probability exactly twelve companies send a representative.
Repeat for x = 13
B(x) = (20 C x)*(0.5)^(x)*(0.5)^(20-x)
B(13) = (20 C 13)*(0.5)^(13)*(0.5)^(20-13)
B(13) = (77,520)*(0.5)^(13)*(0.5)^(7)
B(13) = 0.073929 approximately
This is the approximate probability exactly thirteen companies send a representative.
These steps are repeated for x = 14, x = 15, all the way up to x = 20
I'll let you do those steps, but you should get these approximate results:
B(14) = 0.036964
B(15) = 0.014786
B(16) = 0.004621
B(17) = 0.001087
B(18) = 0.000181
B(19) = 0.000019
B(20) = 0.000001
All of which are approximate to 6 decimal places.
The last step is to add B(12), B(13), all the way up to B(20)
You should get 0.251722 when doing so.
There's about a 25.1722% chance of twelve or more companies sending a representative.
Here's one free calculator to check your work
https://calculator-online.net/binomial-distribution-calculator/
there are tons of other free options out there as well.
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