A federal report finds that lie detector tests given to trut

A federal report finds that lie detector tests given to truthful persons have probability about 0.2 of suggesting that the person is deceptive.

(a)A company asks 12 job applicants about theft from previous employers, using lie detector to assess their truthfulness. Suppose that all 12 answer truthfully. What is the probability that the lie detector says all 12 are truthful? What is the probability that lie detector says at least 1 is deceptive?

(b)What is the mean number among 12 truthful persons who will be classified as deceptive? What is the standard deviation of this number?

(c)What is the probability that the number classified as deceptive is less than the mean?

(d)If the company asks 200 employees to take the lie detector test, what is the probability that at most 10 will be classifies as deceptive?

Solution

Binomial Distribution

PMF of B.D is = f ( k ) = ( n k ) p^k * ( 1- p) ^ n-k
Where   
k = number of successes in trials
n = is the number of independent trials
p = probability of success on each trial

a.1) ALL ARE TRUE
P( X = 12 ) = ( 12 12 ) * ( 0.8^12) * ( 1 - 0.8 )^0
= 0.0687
a.2)
ATLEAST 1 DECEPTIVE
P( X < 1) = P(X=0) +   
= ( 12 0 ) * 0.2^0 * ( 1- 0.2 ) ^12 +
= 0.0687
P( X > = 1 ) = 1 - P( X < 1) = 0.9313
b)
Mean ( np ) = 2.4
Standard Deviation ( npq )= 12*0.2*0.8 = 1.3856
c)
Normal Distribution = Z= X- u / sd                   
P(X < 2.4) = (2.4-2.4)/1.3856
= 0/1.3856= 0
= P ( Z <0) From Standard NOrmal Table
= 0.5                  
d)
Mean ( np ) = 200* 0.20 = 40
Standard Deviation ( npq )= 200*0.2*0.8 = 5.6569
Normal Distribution = Z= X- u / sd                   

P(X < = 10) = (10-40)/5.6569
= -30/5.6569= -5.3033
= P ( Z <-5.3033) From Standard NOrmal Table
= 0