Homework Soursethe probability that a child is handicapped is 103 find the
the probability that a child is handicapped is 10^-3 find the probability p1 that in a school of 1800 children, one is handicapped, and the probability p2 that more than two children are handicapped: (a) exactly; (b) using approximation.
Solution
A) EXACTLY, using binomial distribution:
For p1:
Note that the probability of x successes out of n trials is
P(n, x) = nCx p^x (1 - p)^(n - x)
where
n = number of trials = 1800
p = the probability of a success = 0.001
x = the number of successes = 1
Thus, the probability is
P ( 1 ) = 0.297567724 [ANSWER, p1]
....
For p2:
Note that P(more than x) = 1 - P(at most x).
Using a cumulative binomial distribution table or technology, matching
n = number of trials = 1800
p = the probability of a success = 0.001
x = our critical value of successes = 2
Then the cumulative probability of P(at most x) from a table/technology is
P(at most 2 ) = 0.730647909
Thus, the probability of at least 3 successes is
P(more than 2 ) = 0.269352091 [ANSWER, p2]
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b)
Using Poisson approximation,
u (mean) = n p = 1800*0.001 = 1.8.
For p1:
Note that the probability of x successes out of n trials is
P(x) = u^x e^(-u) / x!
where
u = the mean number of successes = 1.8
x = the number of successes = 1
Thus, the probability is
P ( 1 ) = 0.297537999 [ANSWER, p1]
.....
For p2:
Note that P(more than x) = 1 - P(at most x).
Using a cumulative poisson distribution table or technology, matching
u = the mean number of successes = 1.8
x = our critical value of successes = 2
Then the cumulative probability of P(at most x) from a table/technology is
P(at most 2 ) = 0.730621086
Thus, the probability of at least 3 successes is
P(more than 2 ) = 0.269378914 [ANSWER, p2]
