Question

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 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?a. What is the mean number among 12 truthful persons who will be classified as deceptive? What is the standard deviation of this number?b. What is the probability that the number classified as deceptive is less than the mean?c. 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?

Answer 1

a)
`P(X=12)=(12C12)(0.2)^(12) (1-0.2)^(12-12)=4.096x10^(-9)`

b)
`P(X \geq 1) =1-P(X<1)= 1-P(X=0)=1-0.0687=0.9313`

c)
`E(X) = np = 12*0.2= 2.4`

d)
`Sd(X) = √(np(1-p))=√(12*0.2*(1-0.2))=1.386`

e)
`P(X<2.4) =P(X\leq2) =P(X=0) +P(X=1)+P(X=2)= 0.558`

f)
`P(X\leq 10) =1.1x10^(-9)`

Explanation:

Previous concepts

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

Solution to the problem

Let X the random variable of interest, on this case we now that:

`X \sim Binom(n=12, p=0.2)`

The probability mass function for the Binomial distribution is given as:

`P(X)=(nCx)(p)^x (1-p)^(n-x)`

Where (nCx) means combinatory and it's given by this formula:

`nCx=(n!)/((n-x)! x!)`

Part a

For this case we want to find this probability:

`P(X=12)=(12C12)(0.2)^(12) (1-0.2)^(12-12)=4.096x10^(-9)`

Part b

`P(X \geq 1) =1-P(X<1)= 1-P(X=0)`

`P(X=0)=(12C0)(0.2)^(0) (1-0.2)^(12-0)=0.0687`

`P(X \geq 1) =1-P(X<1)= 1-P(X=0)=1-0.0687=0.9313`

Part c

The expected value is given by:

`E(X) = np = 12*0.2= 2.4`

Part d

The standard deviation is given by:

`Sd(X) = √(np(1-p))=√(12*0.2*(1-0.2))=1.386`

Part e

If we want the probability that the number classified as deceptive would be lower than the mean we want:

`P(X<2.4) =P(X\leq2) =P(X=0) +P(X=1)+P(X=2)`

`P(X=0)=(12C0)(0.2)^(0) (1-0.2)^(12-0)=0.0687`

`P(X=1)=(12C1)(0.2)^(1) (1-0.2)^(12-1)=0.2062`

`P(X=2)=(12C2)(0.2)^(2) (1-0.2)^(12-2)=0.2835`

`P(X<2.4) =P(X\leq2) =P(X=0) +P(X=1)+P(X=2)= 0.558`

Part f

For this case our random variable would be:

`X \sim Binom(n=200, p=0.2)`

And we want this probability:

`P(X\leq 10) = P(X=0)+P(X=1)+ .......+P(X=10)`

And we can use the following excel code to find the answer:

"=BINOM.DIST(10;200;0.2;TRUE)"

And we got:
`P(X\leq 10) =1.1x10^(-9)`