Question

Consider the following sample of observations on coating thickness for low-viscosity paint. 1.09 0.85 0.88 0.88 1.04 1.14 1.29 1.31 1.39 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). (a) Calculate a point estimate of the mean value of coating thickness. (Round your answer to four decimal places.) State which estimator you used. f ix s x os/ (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.) s/x P s (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 u and o.] (Round your answer to four decimal places.) State which estimator you used. x 90th percentile S 10th 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 u 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.)

Answer 1

from given sample Calculated, X = EX = 21052 = 1.3450 16 n. s Elxi-22 1.6742 15 = 0.3341 n-1 a) point estimate for mean is X. x = 1.3450 we used estimator X b) point estimate for median. for normal distribution we know that, mean = median = mode ů = x = 1.3450 median = 1.3450 i median = Here iep [ < x- we used estimator X. instead we might have used ã (sample median). c) point estimate of value that seperates the largest lot from remaining goy. is - M 0.90 @ i= 1.3550 & os 0.3341 X-1.3450 = 0.90 0.3341 from Standard normal table, 2c 1.2816] 0.90 بنا Here û = .. P[z 2 < P[zc1. X- 1.3450 21.2816 0.3341 x = (1.2816 X 0.3341) + 1.3450 [x = 1.7732 Here we used estimator goth percentile,

d) Pexclus) by a central limit theorem, X-4 <1.5. 1:3450 0.334) Ep[z< 0,46 ) 0,6772 ispcxc15) = 0.6772' e) estimated emor E' slun standard 0.334) 16. 0.083525

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