Question

Semiconductor wafers at a fabrication

plant are classified according to their

diameter into

1 inch and 3 inch wafers. Let X1 (resp. X2) denote the number of 1

inch (resp. 3 inch) wafers produced

in a day. Assume that X1 (resp. X2) has a

m

ean and standard deviations of

μ

1

= 1100,

σ

1

= 100

(res

p.

σ

2

= 900,

μ

2

= 200

)

.

(i) Assuming that the production of wafers types are independent processes,

find the mean and

standard deviation of the total number of wafers produced

in a day. (Justify your answer.)

(ii) Further, assume each of previous variables is distributed as a Gaussian.

What is the probability

that at least 1900 wafers are produced in a day?

(iii) How should the standard deviation of the 1 inch wafer production be

lowered so that at least

1500 w

afers are produced in a day with 99%

probability?

Answer 1

(i) Assuming that the production of wafers types are independent processes, find the mean and standard deviation of the total number of wafers produced in a day.

u= M1 + 2 = 1100 + 200 = 1300

and, since X1 and X2 are independent, Y has a standard deviation

0= Vo+oj = V1002 + 9002 = 905.54

(ii) Further, assume each of previous variables is distributed as a Gaussian. What is the probability that at least 1900 wafers are produced in a day?

P(Y > 1900) = 1- P(Y <1900) = 1-0 1900 – 1300 905.54 -) = 1- (0.663)

  

= (-0.7) = 0.2419

(iii) Keeping σ1 as an unknown, the standard deviation of Y is a function,

0= Vož +810000

Using this, we set up the equation:

200 P(Y > 1500) = 1 - P(Y <1500) = 1 - 0 = 0.99

2001 = () - 251 - 2 - ਲ - 7.52

Then, we have that,

0 = 77.822 – 900 = -803944