Suppose ten people are randomly selected from a population w

Suppose ten people are randomly selected from a population where it is known that 22 percent of the population are smokers.

For this example, define what a trial would be, what a success would be, and what a failure would be. Also, state the values of n, p and q for this example.

What is the probability that all ten of the people are smokers?

What is the probability that none of the ten are smokers?

What is the probability that at least three of them are smokers?

What is the probability that no more than two of them are smokers?

What would be the average or expected amount of smokers out of a sample of ten people from this population?

Solution

a)

A trial would be testing whether a person is a smoker or not.

A success is that person being a smoker.

A failure is that person being a non-smoker.

Here,

n = 10

p = 0.22

q = 1-0.22 = 0.78

************************

b)

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 =    10      
p = the probability of a success =    0.22      
x = the number of successes =    10      
          
Thus, the probability is          
          
P (    10   ) =    2.65599*10^-7 [ANSWER]

*********************

c)

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 =    10      
p = the probability of a success =    0.22      
x = the number of successes =    0      
          
Thus, the probability is          
          
P (    0   ) =    0.083357758

***************************

d)

Note that P(at least x) = 1 - P(at most x - 1).          
          
Using a cumulative binomial distribution table or technology, matching          
          
n = number of trials =    10      
p = the probability of a success =    0.22      
x = our critical value of successes =    3      
          
Then the cumulative probability of P(at most x - 1) from a table/technology is          
          
P(at most   2   ) =    0.616880294
          
Thus, the probability of at least   3   successes is  
          
P(at least   3   ) =    0.383119706 [ANSWER]

***********************

e)

Using a cumulative binomial distribution table or technology, matching          
          
n = number of trials =    10      
p = the probability of a success =    0.22      
x = the maximum number of successes =    2      
          
Then the cumulative probability is          
          
P(at most   2   ) =    0.616880294 [ANSWER]

*************************

f)

E(x) = n p = 10*0.22 = 2.2 [ANSWER]