Question

In: Statistics and Probability

In 2003, it was reported that 40% of all car accidents in Connecticut were alcohol related....

In 2003, it was reported that 40% of all car accidents in Connecticut were alcohol related. Suppose we take a random sample of 100 car accidents in Connecticut and let X be the number that are alcohol related.

Find the probability that at least 40 were alcohol related and at least 40 were not alcohol related.

Expert Solution

np=100*0.40=40>10

n(1-p)=100(1-0.40)=60>10

We can use here normal approximation to binomial.

the probability that at least 40 were alcohol related

................by using continuity correction.

=P[Z>-0.10]

=1-0.4602...............................by using normal probability table.

=0.5398

Therefore,the probability that at least 40 were alcohol related is 0.5398

b) the probability that at least 40 were not alcohol related

p=1-0.40=0.60

=P[Z>-4.18]

=0.99999

the probability that at least 40 were not alcohol related is approximately 1

orchestra answered 2 years ago

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