Question

Consider the following sample of observations on coating thickness for low-viscosity paint.

Answer 1

Answer:

a) $\bar X = \frac{\sum_{i=1}^n X_i}{n}$

And for this case if we use this formula we got:

$\bar x = 1.3538$

b) Since we have n =16 values for the sample the median can be calculated as the average between position 8th anf 9th and we got:

$Median = \frac{1.31+1.46}{2}= 1.385$

c) $P(X>a)=0.1$   (a)

$P(X$   (b)

Both conditions are equivalent on this case. We can use the z score again in order to find the value a.  

As we can see on the figure attached the z value that satisfy the condition with 0.9 of the area on the left and 0.1 of the area on the right it's z=1.28. On this case P(Z<1.28)=0.9 and P(z>1.28)=0.1

If we use condition (b) from previous we have this:

$P(X$  

$P(z$

But we know which value of z satisfy the previous equation so then we can do this:

$z=1.28$

And if we solve for a we got

$a=1.3538 +1.28*0.3505=1.8024$

So the value of height that separates the bottom 90% of data from the top 10% is 1.8024.  

d) $Median= \frac{x_{8} +x_{9}}{2}$

The variance for this estimator is given by:

$Var(\frac{x_{8} +x_{9}}{2}) = \frac{1}{4} Var(X_{8} +X_{9})$

We can assume the obervations independent so then we have:

$Var(\frac{x_{8} +x_{9}}{2}) = \frac{1}{4} (2\sigma^2) = \frac{\sigma^2}{2}$

And replacing we got:

$Var(\frac{x_{8} +x_{9}}{2})= \frac{0.3105^2}{2}= 0.0482$

And the standard error would be given by:

$Sd(\frac{x_{8} +x_{9}}{2})= \sqrt{0.0482}=0.2196$

Step-by-step explanation:

Data given:

0.86 0.88 0.88 1.07 1.09 1.17 1.29 1.31  1.46 1.49 1.59 1.62 1.65 1.71 1.76 1.83

Part a

We can calculate the mean with the following formula:

$\bar X = \frac{\sum_{i=1}^n X_i}{n}$

And for this case if we use this formula we got:

$\bar x = 1.3538$

Part b

For this case in order to calculate the median we need to put the data on increasing way like this:

0.86 0.88 0.88 1.07 1.09 1.17 1.29 1.31 1.46 1.49  1.59 1.62 1.65 1.71 1.76 1.83

Since we have n =16 values for the sample the median can be calculated as the average between position 8th anf 9th and we got:

$Median = \frac{1.31+1.46}{2}= 1.385$

Part c

For this case we can assume that the mean is $\mu = 1.3538$

And we can calculate the population deviation with the following formula:

$\sigma = \sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{N}}$

And if we replace we got:  $\sigma= 0.3105$

And assuming normal distribution we have this:

$X \sim N (\mu = 1.3538, \sigma= 0.3105)$

For this part we want to find a value a, such that we satisfy this condition:

$P(X>a)=0.1$   (a)

$P(X$   (b)

Both conditions are equivalent on this case. We can use the z score again in order to find the value a.  

As we can see on the figure attached the z value that satisfy the condition with 0.9 of the area on the left and 0.1 of the area on the right it's z=1.28. On this case P(Z<1.28)=0.9 and P(z>1.28)=0.1

If we use condition (b) from previous we have this:

$P(X$  

$P(z$

But we know which value of z satisfy the previous equation so then we can do this:

$z=1.28$

And if we solve for a we got

$a=1.3538 +1.28*0.3505=1.8024$

So the value of height that separates the bottom 90% of data from the top 10% is 1.8024.  

Part d

The median is defined as :

$Median= \frac{x_{8} +x_{9}}{2}$

The variance for this estimator is given by:

$Var(\frac{x_{8} +x_{9}}{2}) = \frac{1}{4} Var(X_{8} +X_{9})$

We can assume the obervations independent so then we have:

$Var(\frac{x_{8} +x_{9}}{2}) = \frac{1}{4} (2\sigma^2) = \frac{\sigma^2}{2}$

And replacing we got:

$Var(\frac{x_{8} +x_{9}}{2})= \frac{0.3105^2}{2}= 0.0482$

And the standard error would be given by:

$Sd(\frac{x_{8} +x_{9}}{2})= \sqrt{0.0482}=0.2196$