The line width for semiconductor manufacturing is assumed to be normally distrib

Question

The line width for semiconductor manufacturing is assumed to be normally distributed with a mean of 1.0 micrometer and a standard deviation of 0.07 micrometer. Suppose a random sample of n = 30 was drawn. What is the probability that a line width is between 0.8 and 1.2 micrometer? The line width of 50% of samples is below what value? What is the mean of the sample mean xbar of the random sample with size n = 30? What is the standard deviation of the sample mean x bar of the random sample with size n = 30? What is the probability that the sample mean is greater than 1.05?

Explanation / Answer

Mean ( u ) =1
Standard Deviation ( sd )=0.07
Number ( n ) = 30
Normal Distribution = Z= X- u / (sd/Sqrt(n) ~ N(0,1)                  
a.
To find P(a <= Z <=b) = F(b) - F(a)
P(X < 0.8) = (0.8-1)/0.07/ Sqrt ( 30 )
= -0.2/0.0128
= -15.6492
= P ( Z <-15.6492) From Standard Normal Table
= 0
P(X < 1.2) = (1.2-1)/0.07/ Sqrt ( 30 )
= 0.2/0.0128 = 15.6492
= P ( Z <15.6492) From Standard Normal Table
= 1
P(0.8 < X < 1.2) = 1-0 = 1                  

b.
P ( Z < x ) = 0.5
Value of z to the cumulative probability of 0.5 from normal table is 0
P( x-u/s.d < x - 1/0.01278 ) = 0.5
That is, ( x - 1/0.01278 ) = 0
--> x = 0 * 0.01278 + 1 = 1                  
                  

c.
Mean will as same as population, which is 30

d.
when n=30 is drawn = 0.07/Sqrt(30) = 0.01278

e.
P(X > 1.05) = (1.05-1)/0.01278/ Sqrt ( 30 )
= 0.05/0.002= 21.4289
= P ( Z >21.4289) From Standard Normal Table
= 0      

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