The thickness of photoresist applied to wafers in semiconductor manufacturing at a particular on the wafer is uniformly distributed between 0.2032 and 0.2147 micrometers. (a) Determine the proportion of wafers that exceeds 0.2125 micrometers of photoresist thickness. (Round your answer to 3 decimal places.) (b) Determine the mean of X (in micrometers) (Round your answer to 4 decimal places.) (c) Determine the variance of X (in micrometers2) (Round your answer to 8 decimal places.)

Respuesta :

Answer:

a) P=0.191

b) E(T)=0.2090

c) V(T)= 0.00001102  

Step-by-step explanation:

The thickness T is a random variable uniformly distributed between 0.2032 and 0.2147.

a) Determine the proportion of wafers that exceeds 0.2125 micrometers of photoresist thickness.

The proportion is equal to the probability of T>0.2125.

[tex]P(T>0.2125)=1-\frac{T-T_{min}}{T_{max}-T_{min}} =1-\frac{0.2125-0.2032}{0.2147-0.2032} =1-\frac{0.0093}{0.0115} =1-0.809=0.191[/tex]

b) Determine the mean of X (in micrometers)

The mean of a random variable with a uniform distribution is:

[tex]E(T)=0.5(T_{min}+T_{max})=0.5*(0.2032+0.2147)=0.5*0.4179=0.2090[/tex]

c) Determine the variance of X

The variance of a random variable with a uniform distribution is:

[tex]V(T)=\frac{(T_{max}-T_{min})^2}{12} =\frac{(0.2147-0.2032)^2}{12}=\frac{0.0115^2}{12}=\frac{0.00013225}{12} = 0.00001102[/tex]

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