Its all part of one question. I need each part that’s unanswered Consider the following sample of observations on coating thickness for low-viscosity paint.0.831.420.88 0.88 1.04 1.09 1.17 1.29 1.311.491.59 1.62 1.65 1.71 1.76 1.83USE SALTAssume that the distribution of coating thickness is normal (a normal probability plot strongly supports this assumption).(a) Calculate a point estimate of the mean value of coating thickness. (Round your answer to four decimal places.)State which estimator you used.O pOsOs/xXOx(b) Calculate a point estimate of the median of the coating thickness distribution. (Round your answer to four decimalplaces.)State which estimator you used and which estimator you might have used instead. (Select all that apply.)Os✓XO pOs/X✓X(c) Calculate a point estimate of the value that separates the largest 10% of all values in the thickness distribution fromthe remaining 90%. [Hint: Express what you are trying to estimate in terms of μ and o.] (Round your answer to fourdecimal places.)OsOxState which estimator you used.Ox10th percentile90th percentile(d) Estimate P(X < 1.5), i.e., the proportion of all thickness values less than 1.5. [Hint: If you knew the values of μ and o,you could calculate this probability. These values are not available, but they can be estimated.] (Round your answer tofour decimal places.)(e) What is the estimated standard error of the estimator that you used in part (b)? (Round your answer to four decimalplaces.)

Answered: Consider the following sample of…

Calculus For The Life Sciences

2nd Edition

ISBN:9780321964038

Author:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.

Publisher:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.

Chapter13: Probability And Calculus

Section13.CR: Chapter 13 Review

Problem 59CR

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Question

Its all part of one question. I need each part that’s unanswered

Consider the following sample of observations on coating thickness for low-viscosity paint.
0.83
1.42
0.88 0.88 1.04 1.09 1.17 1.29 1.31
1.49
1.59 1.62 1.65 1.71 1.76 1.83
USE SALT
Assume that the distribution of coating thickness is normal (a normal probability plot strongly supports this assumption).
(a) Calculate a point estimate of the mean value of coating thickness. (Round your answer to four decimal places.)
State which estimator you used.
O p
Os
Os/x
X
Ox
(b) Calculate a point estimate of the median of the coating thickness distribution. (Round your answer to four decimal
places.)
State which estimator you used and which estimator you might have used instead. (Select all that apply.)
Os
✓X
O p
Os/X
✓X
(c) 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 μ and o.] (Round your answer to four
decimal places.)
Os
Ox
State which estimator you used.
Ox
10th percentile
90th percentile
(d) Estimate P(X < 1.5), i.e., the proportion of all thickness values less than 1.5. [Hint: If you knew the values of μ and o,
you could calculate this probability. These values are not available, but they can be estimated.] (Round your answer to
four decimal places.)
(e) What is the estimated standard error of the estimator that you used in part (b)? (Round your answer to four decimal
places.)

Expert Solution

Step 1

(a)

Calculated point estimate of the mean value of coating thickness is : . (Rounded of my answer to four decimal places.)

1.3475

(b)

The point estimate of the population median can be obtained from the sample median. In this case, since there are 16 data values, the sample median will be the average of the 8th and 9th data values in the ordered data set. The 8th value is 1.31 and the 9th value is 1.42, so the median will be calculated as follows:

Median = (1.31 + 1.42) / 2 = 1.365

Therefore, the point estimate of the population median is 1.365.

(c)

First we need to find the standard deviation of the data. Following table shows the calculations:

	X	(X-mean)^2
	0.83	0.267
	0.88	0.2185
	0.88	0.2185
	1.04	0.0945
	1.09	0.0663
	1.17	0.0315
	1.29	0.0030
	1.31	0.0014
	1.42	0.0052
	1.49	0.0203
	1.59	0.05880
	1.62	0.07425
	1.65	0.09150
	1.71	0.1314
	1.76	0.1701
	1.83	0.2328
Total	21.56	1.7555

$s a m p l e s i z e n = 16 s a m p l e s u m : \sum_{}^{} x = 21.56 s a m p l e s u m : \frac{\sum_{}^{}}{n} = 21.56 / 16 = 1.3475 s a m p l e s \tan d a r d d e v i a t i o n : s = \sqrt{\frac{\sum_{}^{} {(x - x^{-})}^{2}}{n - 1}} = \sqrt{\frac{1.7555}{16 - 1}} = 0.3421$

Now we need z-score that has 0.10 area to its left. The z-score -1.28 has 0.10 area to its left. Since data is distributed normally so the required value is

$m - z_{c} s = 1.3475 - 1.28 . 0.3421 = 0.909612$