A process that produces carbon nanotubes has yield rate of 9

A process that produces carbon nanotubes has yield rate of 90%, that is, 10% of the nanotubes produced have flaws and cannot be used. Among six randomly selected nanotubes, how likely is it that only one has flaws and cannot be used?

The probability mass function (pmf) for X = the number of major problems on a randomly selected machine of a certain type is

A particular normal random variable has an unknown mean and variance. However, a sample of 40 observations yields a sample mean of 28.12 and a sample standard deviation of 2.34. Compute a 95% confidence interval for the true population variance.

  x  

  0  

  1  

  2  

  3  

  4  

  p(x)  

  .08  

  .15  

  .45  

  .27  

  .05  

Solution

Q2.

Confidence Interval

CI = (n-1) S^2 / ^2 right < ^2 < (n-1) S^2 / ^2 left
Where,
S = Standard Deviation
^2 right = (1 - Confidence Level)/2
^2 left = 1 - ^2 right
n = Sample Size

Since aplha =0.05
^2 right = (1 - Confidence Level)/2 = (1 - 0.95)/2 = 0.05/2 = 0.025
^2 left = 1 - ^2 right = 1 - 0.025 = 0.975
the two critical values ^2 left, ^2 right at 39 df are 58.1201 , 23.654
S.D( S^2 )=2.34
Sample Size(n)=40
Confidence Interval = [ 39 * 5.4756/58.1201 < ^2 < 39 * 5.4756/23.654 ]
= [ 213.5484/58.1201 < ^2 < 213.5484/23.6543 ]
[ 3.674 , 9.026 ]