A production manager is interested in the mean weight of items turned out by a p

Question

A production manager is interested in the mean weight of items turned out by a particular process. He feels that the weight of items from the process is normally distributed with mean mu and that mu is assumed to be continuous and the prior distribution for mu is a normal distribution with mean 110 and variance 0.4. From past experience, he is willing to assume that the process variance is sigma^2 = 4. He randomly selects five items from the process and weighs them, with the following results: 108, 109, 107.4, 109.6, and 112. Find the production manager's posterior distribution and compute the means and the variances of the prior and posterior distributions.

Explanation / Answer

Solution:-

Mean variance for 108, 109, 107.4, 109.6, 112

added up all of the numbers:

107.4 + 108 + 109 + 109.6 + 112 = 546

Mean = 546/5

= 109.2

546 x 546 = 298116

298116 / 5 = 59623.2

squared them individually this time, and added them all up:

(107.4 x 107.4) + (108 x 108) + (109 x 109) + (109.6 x 109.6) + (112 x 112) = 59635.92

59635.92 - 59623.2 = 12.72

12.72 / 4 = 3.18

Then.

The production manager's posterior and the mean = (110+109.2)/2= 109.6

Variance = 3.18-0.40

=2.78

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