Answer:
b) 0.9313
c) mean = 2.4
standard deviation= 1.3856
d) 0.5583
Explanation:
Given:
p = 0.2
n = 12
a) X= number of applicants classified as deceptive.
Probability mass function of X will be:
![P(X=x) = \left(\begin{array}{c}12\\x\end{array}\left) (0.2)^x(1-0.2)^1^2^-^x,x=0, 1, 2, .....,12](https://img.qammunity.org/2021/formulas/mathematics/college/s6iz9yt6peia7pvasmmeca0f7dgdi4c0mx.png)
b) Probability that the polygraph says at least 1 is deceptive:
![P(X≥1) = 1 -P(X=0) = 1 -\left(\begin{array}{c}12\\0\end{array}\left) (0.2)^0(1-0.2)^1^2^-^0](https://img.qammunity.org/2021/formulas/mathematics/college/fjtt1ykt5yh0ol80nb75hygj14y7iib3q4.png)
= 1 - 0.0687
= 0.9313
c) The mean number among 12 truthful persons who will be classified as deceptive:
E(X) = n•p
= 12 * 0.2
= 24
Standard deviation:
![s.d = √(12*0.2*(1-0.2))](https://img.qammunity.org/2021/formulas/mathematics/college/lu3yx008px1s2bqi0s7h5skqquto7sb1u7.png)
= 1.3856
d) Probability that the number classified deceptive is less than the mean:
![P(X<2.4) = P(X≤2) = E^2_x_=_0 \left(\begin{array}{c}12\\0\end{array}\left) (0.2)^0(1-0.2)^1^2^-^0](https://img.qammunity.org/2021/formulas/mathematics/college/5uobawqtybq4nqwt9l4g7cjcupcwc1f9ff.png)
= 0.5583