# Team Assignment 6 An optical inspection system is Team Assignment 6 An optical inspection system is used to distinguish among different part types. The probability of a correct classification is 0.98. Suppose that three parts are inspected and that the classifications are independent. Let the random variable X denote the number of parts correctly classified. a) Determine the probability mass function of X. b) Find the mean, E(X). c) Find the standard deviation of X. 1

Solution Let Ri be the event that part i classified correctly. Let Wi be the event that part i classified incorrectly (i.e., the classification is wrong). 2 Solution: Part a P( X 0) P(W1W2W3 ) (0.02)(0.02)(0.02) (0.02)3 . P( X 1) P( R1W2W3 ) P(W1 R2W3 )

P(W1W2 R3 ) (0.98)(0.02)(0.02) (0.02)(0.98)(0.02) (0.02)(0.02)(0.98) (3)(0.98)(0.02) 2 . P( X 2) P( R1 R2W3 ) P( R1W2 R3 ) P(W1 R2 R3 ) (0.98)(0.98)(0.02) (0.98)(0.02)(0.98) (0.02)(0.98)(0.98) (3)(0.98) 2 (0.02). P( X 3) P( R1 R2 R3 ) (0.98)(0.98)(0.98) (0.98)3 .

3 Solution: Part a f (0) P ( X 0) (0.02)3 f (1) P ( X 1) (3)(0.98)(0.02) 2 f (2) P ( X 2) (3)(0.98) 2 (0.02) f (3) P ( X 3) (0.98)3 4 Solution: Part b 3 xf ( x)

x 0 (0)( f (0)) (1)( f (1)) (2)( f (2)) (3)( f (3)) (1)((3)(0.98)(0.2) 2 ) (2)((3)(0.98) 2 (0.2)) (3)(0.98)3 2.94. 5 Solution: Part c 3 V ( X ) x 2 f ( x) 2 x 0

(02 )( f (0)) (12 )( f (1)) (22 )( f (2)) (32 )( f (3)) 2.942 (1)((3)(0.98)(0.2) 2 ) (4)((3)(0.98) 2 (0.2)) (9)(0.98)3 8.6436 0.0588 V (X ) 0.0588 0.2425 6