Posted by **AMber** on Saturday, August 4, 2012 at 10:03am.

If 95% of all people between 21 and 50 have had at least one adult beverage in their life, find these probabilities for a sample of 10 people in that age group:

a) Exactly 10 have had an adult beverage

b) At least 6 have had an adult beverage

c) At most 8 have had an adult beverage

d) At most 1 has had an adult beverage.

- stats -
**MathMate**, Saturday, August 4, 2012 at 1:37pm
Here are some properties of the experiment:

1. Each trial is Bernoulli (i.e. yes/no)

2. Number of trials is fixed.

3. probability of success does not vary from trial to trial.

4. each trial is independent of the others (assumed random sample).

All these point to the binomial distribution with N=10, p=0.95, q=(1-0.95)=0.05, and success="have had at least one adult beverage".

P(X=i)=(N,i)p^i q^(N-i)

and (N,i) = number of ways to choose i objects from N distinct objects.

(N,i)=N!/(i!(N-i)!)

a). N=10, i=10, p=0.95, q=0.05

P(X=10)=(10,10)0.95^10 0.05^0

=1*0.95^10*1

b) At least 6 => X=6,7,8,9,10

So

P(X≥6)=∑P(X=i), i=6 to 10

c) At most 8 => X=1,2,3...8

=> X≠9 and X≠10

We can use the identity

P(X≤8)=∑P(X=i), i=1 to 8

=1-P(X≥9)

=1- [∑P(X=i), i=9 to 10]

which simplifies the calculations.

d) at most 1 => X=0 or X=1.

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