Normal an Poisson approximation to binomial distribution
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In each month, the proportion of "Prize" bonds that win a prize is 1 in 11000. There is a large number of prizes and all bonds are equally likely to win each prize. Show that, for a given month, the probability that a bondholder with 5000 bonds wins at least one prize is 0.365.
For a given month find:
1) The probability taht in a group of 10 bondholders each holding 5000 bonds, four or more win at least one prize,
2) The probability in a group of 100 bondholders each holding 5000 bonds, 40 or more win at least one prize.
Find the expected number of prizes for a bondholder holding 550 bonds for 24 months.
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Solution Summary
In each month, the proportion of "Prize" bonds that win a prize is 1 in 11000. There is a large number of prizes and all bonds are equally likely to win each prize. Show that, for a given month, the probability that a bondholder with 5000 bonds wins at least one prize is 0.365.
For a given month find:
1) The probability taht in a group of 10 bondholders each holding 5000 bonds, four or more win at least one prize,
2) The probability in a group of 100 bondholders each holding 5000 bonds, 40 or more win at least one prize.
Find the expected number of prizes for a bondholder holding 550 bonds for 24 months.
Solution Preview
Let X be the number of prizes won.
X ~ Bin(5000, 1/11000)
Since n is large, p is very small,
such that np = 5/11
We use Poisson approximation to binomial distribution.
X ~ P(5/11) approx
P( X >= 1) = 1 - P(X=0)
= 1 - e^-5/11
= 0.365 ...
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