Question

Lie detectors Refer to Exercise 82. Let Y = the number of people who the lie detector says are telling the truth.

Answer 1

Answer:

a) $P(Y\geq 10) = PX \leq 2) = 0.558$

b) $E(X) = \mu_X = np = 12*0.2 = 2.4$

$\sigma_X = \sqrt{np(1-p)}=\sqrt{12*0.2*(1-0.2)}=1.386$

$E(Y) = \mu_Y = np = 12*0.8 = 9.6$

$\sigma_Y = \sqrt{np(1-p)}=\sqrt{12*0.8*(1-0.8)}=1.386$

For this case the expected value of people lying is 2.4 and the complement is 9.6 and that makes sense since we have a total of 12 poeple.

And the deviation for both variables are the same.

Step-by-step explanation:

Assuming this previous info : "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 thefts from previous employers, using  a lie detector to assess their truthfulness. Suppose that all 12 answer truthfully. Let X = the number of people who the lie  detector says are being deceptive"

For this case the distribution of X is binomial $X \sim N(n=12, p=0.2)And we define the new random variable Y="the number of people who the lie detector says are telling the truth" so as we can see y is the oppose of the random variable X, and the distribution for Y would be given by:[tex] Y \sim Bin (n=12,p=1-0.2=0.8)$

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".

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:

$nCx=\frac{n!}{(n-x)! x!}$

Part a

For this case we want to find this probability:

$P(Y \geq 10) = P(Y=10)+P(Y=11) +P(Y=12)$

And if we find the individual probabilites we got:

$P(Y=10)=(12C10)(0.8)^{10} (1-0.8)^{12-10}=0.283$

$P(Y=11)=(12C11)(0.8)^{11} (1-0.8)^{12-11}=0.206$

$P(Y=12)=(12C12)(0.8)^{12} (1-0.8)^{12-12}=0.069$

And if we replace we got:

$P(Y \geq 10) =0.283+0.206+0.069=0.558$

And for this case if we find $P(X\leq 2)=P(X=0) +P(X=1)+P(X=2)$  for the individual probabilites we got:

$P(X=0)=(12C0)(0.2)^0 (1-0.2)^{12-0}=0.069$

$P(X=1)=(12C1)(0.2)^1 (1-0.2)^{12-1}=0.206$

$P(X=2)=(12C2)(0.2)^2 (1-0.2)^{12-2}=0.283$

$P(X\leq 2)=0.283+0.206+0.069=0.558$

So as we can see we have $P(Y\geq 10) = P(X \leq 2) = 0.558$

Part b

Random variable X

For this case the expected value is given by:

$E(X) = \mu_X = np = 12*0.2 = 2.4$

And the deviation is given by:

$\sigma_X = \sqrt{np(1-p)}=\sqrt{12*0.2*(1-0.2)}=1.386$

Random variable Y

For this case the expected value is given by:

$E(Y) = \mu_Y = np = 12*0.8 = 9.6$

And the deviation is given by:

$\sigma_Y = \sqrt{np(1-p)}=\sqrt{12*0.8*(1-0.8)}=1.386$

For this case the expected value of people lying is 2.4 and the complement is 9.6 and that makes sense since we have a total of 12 poeple.

And the deviation for both variables are the same.