Question

Thickness measurements of a coating process are made to the nearest hundredth of a millimeter. The thickness measurements are uniformly distributed with values 0.14, 0.15, 0.16, 0.17, 0.18. Determine the mean and variance of the coating thickness for this process. Round your answers to four decimal places (e.g. 98.7654).

Answer 1

`\bar X = (0.14+0.15+0.16+0.17+0.18)/(5)= 0.16`

We can calculate the sample variance with the following formula:

`s^2 = (\sum_(i=1)^n (X_i -\bar X)^2)/(n-1)`

And replacing we got:

`s^2 = 0.00025`

And the deviation is given by this:

`s=√(0.00025)= 0.0158`

If we want to find the population deviation we just need to use this formula:

`\sigma^2 = (\sum_(i=1)^n (X_i -\bar X)^2)/(n)`

And replacing we got:

`\sigma^2 = 0.0002`

And the population deviation would be:

`\sigma = 0.0141`

Explanation:

For this case we have the following values:

0.14, 0.15, 0.16, 0.17, 0.18.

We can calculate the mean with the following formula:

`\bar X = (\sum_(i=1)^n X_i)/(n)`

And replacing we got:

`\bar X = (0.14+0.15+0.16+0.17+0.18)/(5)= 0.16`

We can calculate the sample variance with the following formula:

`s^2 = (\sum_(i=1)^n (X_i -\bar X)^2)/(n-1)`

And replacing we got:

`s^2 = 0.00025`

And the deviation is given by this:

`s=√(0.00025)= 0.0158`

If we want to find the population deviation we just need to use this formula:

`\sigma^2 = (\sum_(i=1)^n (X_i -\bar X)^2)/(n)`

And replacing we got:

`\sigma^2 = 0.0002`

And the population deviation would be:

`\sigma = 0.0141`