Consider the fallowing sample of observations an coating thickness for low-visco

Question

Consider the fallowing sample of observations an coating thickness for low-viscosity paint. .81 0.88 0.8 1.03 1.09 21 29 1.31 1.36 1.491.59 1.62 1.65 .71 1.76 1.83 Assume that the distnibution af coating thickness is normal (a normal probability plat strangly supports this essumption). (a) Calculate e point estimate of the mean value of coating thickness. (Round your answer to four decimal places. 093625 State which estimater you used ?sfx (b) Calculate a pcint estimate of the median of the coating thickness distrbution. (Round your answer to four decimal places.) State which estimator you used and which estimator you might have used instead. (Select all that apply.) have c ca ulate a point estimate of the value that separates the largest 10% of all values in the thi ess et ution o the remaining 90% Hint Express hat you are t in to estimate in terms and Ro n your answer to our decima places. State which estimater you used. 90th percentile 10th percentile d stir te ??1.6 i e the proportion a all thickness values less than 1.6 Hint: f you knew the values o ? and o au could calcula a this prabability. These values are not available, but they can ba st mated Round your an we to four de mal p ac s (e) What is the estimated standard emor of the estimator that you used in part(b)? (Round your answer to four decimal places.)

Explanation / Answer

Solution:- Given that sample 0.81,0.88,0.88,1.03,1.09,1.21,1.29,1.31,1.36,1.49,1.59,1.62,1.65,1.71,1.76,1.83

(a) Mean = 1.3444

=> option D.

(b) Median = 1.335.

=> option B.

(c) X = mu + Z * sigma
= 1.3444 + (1.28*0.3363) = 1.7749

=> option E.

(d)P(X < 1.6)

= > P(X < 1.6) = P(Z < (1.6-1.3444)/0.3363)
= P(Z < 0.76)
= 0.7764

(e) standard error = 0.3363/sqrt(16) = 0.0841

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