Question

(a) Calculate a point estimate of the value that separates the largest 10% of all values in the thickness distribution from the remaining 90%. [Hint: Express what you are trying to estimate in terms of ? and ?.] (Round your answer to four decimal places.)
(b) Estimate P(X < 1.5), i.e., the proportion of all thickness values less than 1.5. [Hint: If you knew the values of ? and ?, you could calculate this probability. These values are not available, but they can be estimated.] (Round your answer to four decimal places.)
(c) What is the estimated standard error of the estimator that you used in part (b)? (Round your answer to four decimal places.)

Consider the following sample of observations on coating thickness for low-viscosity paint. 0.83 0.8 0.8 1.07 1.09 125 1.29 1.31 1.42 1.49 1.59 1.62 1.65 1.71 1.76 1.83 Assume that the distribution of coating thickness is normal (a normal probability plot strongly supports this assumption)

Answer 1

a) Mean

$\overline{x} =\frac{1}{n}\sum x = \frac{21.67}{16}= 1.3544$

Standard deviation

$s =\sqrt{\frac{1}{n-1}\sum \left (x - \overline{x} \right )^2}= \sqrt{\frac{1.6462}{15}} =0.3313$

For largest 10%, the critical value of Z is 1.28

The point estimate of the value that separates the largest 10% of all values in the thickness distribution from the remaining 90% is

$X = \mu &plus;Z\sigma = 1.3544&plus;1.28*0.3313 = 1.7784$

b)  

$P(X<1.5) =P(Z<\frac{1.5-1.3544}{0.3313})=P(Z<0.44)$

= 0.5+0.1700 = 0.6700

c)  the estimated standard error of the estimator

$\widehat{\sigma }_{\overline{x}} = \frac{s}{\sqrt{n}} = \frac{0.3313}{\sqrt{16}} =0.0828$