Question

In the manufacture of electroluminescent lamps, several different layers of ink are deposited onto a plastic substrate. The thickness of these layers is critical if specifications regarding the final color and intensity of light are to be met. Let X and Y denote the thickness of two different layers of ink. It is known that X is normally distributed with a mean of 0.1 millimeter and a standard deviation of 0.00031 millimeter, and Y is normally distributed with a mean of 0.23 millimeter and a standard deviation of 0.00017 millimeter. The value of ρ for these variables is equal to 0. Specifications call for a lamp to have a thickness of the ink corresponding to X in the range of 0.099535 to 0.100465 millimeter and Y in the range of 0.22966 to 0.23034 millimeter. What is the probability that a randomly selected lamp will conform to specifications? (Hint: when two variables are independent of each other P(A and B) ).

Answer 1

X is normally distributed with a mean of 0.1 millimeter and a standard deviation of 0.00031 millimeter

Y normally distributed with a mean of 0.23 millimeter and a standard deviation of 0.00017 millimeter

Specification for X :in the range of 0.099535 to 0.100465 millimeter

Specification of Y : in the range of 0.22966 to 0.23034 millimeter

A : Event of X conform to specifications

P(A) : Probability that X confirm to specifications : P(0.099535 X 0.100465)

B : Event of Y conform to specifications

P(B) = Probability that Y confirm to specifications : P(0.22966 Y 0.23034)

Probability that a randomly selected lamp will conform to specifications

= Probability that (X conforms to specifications and Y conform to specifications)

= P(A and B)

Given the correlation of these variables is equal to 0 i.e Event A and Event B are independent

i.e

P(A and B) = P(A) P(B)

P(A) : Probability that X confirm to specifications : P(0.099535 X 0.100465)=P(X 0.100465) -P(X 0.099535)

Z-score for 0.099535 = (0.099535-0.1)/0.00031 = -0.000465/0.00031 = -1.5

Z-score for 0.100465 = (0.100465-0.1)/0.00031 = 0.000465/0.00031 = 1.5

From standard normal tables, P(Z -1.5) =0.0668 ; P(Z1.5)=0.9332

P(X 0.100465) -P(X 0.099535) = P(Z1.5) - P(Z -1.5) = 0.9332-0.0668=0.8664

P(A) : Probability that X confirm to specifications : P(0.099535 X 0.100465) = P(X 0.100465) -P(X 0.099535)=0.8664

P(B) : Probability that Y confirm to specifications :P(0.22966 Y 0.23034) = P(Y 0.23034) - P(Y 0.22966)

Z-score for 0.22966 = (0.22966-0.23)/0.00017 = -0.00034/0.00017 = -2

Z-score for 0.23034 = (0.23034-0.23)/0.00017 = 0.00034/0.00017 = 2

From standard normal tables, P(Z-2) = 0.0228 P(Z2) = 0.9772

P(Y 0.23034) - P(Y 0.22966) = P(Z2) - P(Z-2) = 0.9772-0.0228=0.9544

P(B) : Probability that Y confirm to specifications :P(0.22966 Y 0.23034) = P(Y 0.23034) - P(Y 0.22966) = 0.9544

P(A and B) = P(A) P(B) = 0.8664 x 0.9544 = 0.82689216

Probability that a randomly selected lamp will conform to specifications = 0.82689216