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 mean and standard deviations of μ1= 1100, σ1= 100(resp. σ2= 900, μ2= 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 wafers are produced in a day with 99% probability?

Answer 1

X₁ denote number of wafers produces whose diameter is 1 inch.

M1 = 1100,01 = 100

X2 denote number of wafers produces whose diameter is 3 inch.

U2 = 900,02 = 200

1)

If X₁ and X₂ are independently distributed variables then,

E(X1 + X2) = E(X1) + E(X2) = 1100 + 900 = 2000

→ MX1 + X2) = 2000

Var (X1 + X2) = Var (X1) + Var (X2) = 1002 + 2002 = 500000

o(X1 + X2) = 50000 =0= 223.6068

Mean of number of total wafers produced is 2000 and standard deviation is 223.6068

2)

If X₁ and X2 follows Normal distribution (Gaussian distribution) then

X₁ + X₂ follows normal distribution with

mean = 41+ uz
and   
variance = oí + oź = 0 = V0 +ož

From part 1) we got,

u=2000 and 0 = 223.6068

Therefore,

(X1 + X2 ~ Nu = 2000,0 = 223.6068)

For simplicity purpose let us denote X₁ + X₂ by X.

We have to find probability that number of wafers produced in a day is at least 1900

i.e

P(> 1900)

P(2 > 1900) = P(:>

Plar > 1900) = P => 1900 - 2000 223.6068

—P(2> -0.45) = 1 - Plz<-0.45)

(z score probabilities obtained from left tailed z table)

PI > 1900) = 1 -0.3264 = 0.6736

Hence probability that at least 1900 wafers are produced in a day is 0.6736

3)

We want to set standard deviation such that,

P(>1500) = 0.99

We know that,

Plz > -2.33) = 0.99

Use conversion formula of 'z'

=== = ==

Plugging in values we get,

0 = 1500 - 2000 -=O= 214.5923 -2.33

We have to keep $\sigma_{2}$ same and adjust $\sigma_{1}$ such that we get

o = 214.5923

0= Voſ + o2 = 214.5923 = V oſ + 40000

= 46049.84435 = σ? + 40000

0 = 6049.84435 0= 6049.84435

0= 77.78074538 78

Hence standard deviation of the 1 inch wafer production should be lowered from 100 to 78

It should be lowered by approximately 22 units.