A federal study reported that 75 of the US workforce has a d

A federal study reported that 7.5% of the U.S. workforce has a drug problem. A drug enforcement official for the State of Indiana wished to investigate this statement. In her sample of 20 employed workers: a-1. How many would you expect to have a drug problem? (Round your answer to 1 decimal place.) Mean a-2. What is the standard deviation? (Round your answer to 4 decimal places.) Standard deviation b. What is the likelihood that none of the workers sampled has a drug problem? (Round your answer to 4 decimal places.) Likelihood c. What is the likelihood at least one has a drug problem? (Round your answer to 4 decimal places.) Likelihood

Solution

a)

Mean = n p = 20*0.075 = 1.5 [answer]

b)

Standard deviation = sqrt [n p (1 - p)] = sqrt(20*0.075*(1-0.075)) = 1.177921899 [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 =    20      
p = the probability of a success =    0.075      
x = the number of successes =    0      
          
Thus, the probability is          
          
P (    0   ) =    0.210297764 [answer]

d)

P(at least one) = 1 - P(0) = 0.789702236 [answer]