Step 1

The data of coating thickness for low-viscosity paint for $n = 16$ observations is given below:

X= 0.83, 0.88, 0.88, 1.04, 1.09, 1.12, 1.29, 1.31, 1.48, 1.49, 1.59, 1.62, 1.65, 1.71, 1.76, 1.83

 
Step 2

(a)

The sample mean is unbiased estimators of population mean, thus, sample mean can be used as point estimate of population mean.

The estimator sample mean is calculated as follows:

\[\begin{array}{c} \bar x = \frac{{\sum\limits_{i = 1}^{16} {{x_i}} }}{n}\\ = \frac{{0.83 + 0.88 + 0.88 + ... + 1.83}}{{16}}\\ = \frac{{21.57}}{{16}}\\ = 1.348 \end{array}\]

The point estimate of population mean is $\hat \mu = 1.348$.

 
Step 3

(b)

The position of the middle terms is:

\[\begin{array}{c} {1^{st}}\;{\rm{term}} = \frac{{16}}{2}\\ = {8^{th}}\;{\rm{term}}\\ {{\rm{2}}^{nd}}\;{\rm{term}} = \frac{{16}}{2} + 1\\ = {9^{th}}\;{\rm{term}} \end{array}\]

The point estimate of the median of the coating thickness distribution is calculated as follows:

\[\begin{array}{c} \tilde x = \frac{{{8^{th}} + {9^{th}}}}{2}\\ = \frac{{1.31 + 1.48}}{2}\\ = 1.395 \end{array}\]

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The data is taken from a normal distribution which is symmetric. Thus mean is equal to median. The estimator used can be sample median or sample mean. Here sample median is used and value of point estimate is 1.395.

 
Step 4

(c)

The sample standard deviation is the point estimate of the population standard deviation.

The sample standard deviation is calculated using the table below:

x	${\left( {x - \bar x} \right)^2}$
0.83	0.268
0.88	0.219
0.88	0.219
1.04	0.095
1.09	0.067
1.12	0.052
1.29	0.003
1.31	0.001
1.48	0.017
1.49	0.020
1.59	0.059
1.62	0.074
1.65	0.091
1.71	0.131
1.76	0.170
1.83	0.232
0.83	0.268
0.88	0.219
0.88	0.219
1.04	0.095

The sample variance is calculated as follows:

\[\begin{array}{c} {s^2} = \frac{1}{{n - 1}}\sum\limits_{i = 1}^{16} {{{\left( {{x_i} - \bar x} \right)}^2}} \\ = \frac{1}{{16 - 1}}\left( {1.719} \right)\\ = \frac{{1.719}}{{15}}\\ = 0.115 \end{array}\]

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The sample standard deviation that is the point estimate of population standard deviation is calculated as follows:

\[\begin{array}{c} s = \sqrt {0.115} \\ = 0.339 \end{array}\]

The point estimate of population standard deviation is $\hat \sigma = 0.339$.

For a normal distribution 100th percentile is given as $\mu + {z_p}\sigma $ where ${z_p}$ is the z-score.

The suitable value of ${z_p}$ is such that $\Phi \left( {{z_p}} \right) = 0.90$.

Using z-table the value of ${z_p}$ is equal to 1.28.

A point estimate of the value that separates largest 10% of the thickness from the remaining 90% is calculated as follows:

\[\begin{array}{c} \hat \mu + {z_p}\hat \sigma = \bar x + 1.28s\\ = 1.348 + 1.28\left( {0.339} \right)\\ = 1.782 \end{array}\]
 
Step 5

(d)

It is known that $X$ has normal distribution. Therefore, the probability of $X < 1.5$ is calculated as follows:

\[\begin{array}{c} P\left( {X < 1.5} \right) = P\left( {\frac{{X - \bar x}}{s} < \frac{{1.5 - 1.348}}{{0.339}}} \right)\\ = P\left( {Z < 0.45} \right)\\ = \Phi \left( {0.45} \right) \end{array}\]

Using z-table, the value of $\Phi \left( {0.45} \right)$ is equal to 0.6736.

Substituting the value;

\[P\left( {X < 1.5} \right) = 0.6736\]
 
Step 6

(e)

The estimate standard error of the estimator is calculated as follows:

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\[\begin{array}{c} {{\hat \sigma }_{\bar x}} = \frac{s}{{\sqrt n }}\\ = \frac{{0.339}}{{\sqrt {16} }}\\ = \frac{{0.339}}{4}\\ = 0.0847 \end{array}\]