Question

The fish were caught in a particular the level of so e mercury in seafood allowed by FDA is 1ppm (1 part of mercury in a million parts seafood). A sample of n=5 region, sample mean level of me. hercury is found to be x = 995 ppm, and sumple standard deviation is found to be a s.ol. Let r be the true amount of mercury in all such fish, and assume that the distribution of levels of all such fish is normal. meu the mercury

Answer 1

mer solotion : 9,995 Given, sample mean level of encong sample Bize - 5rasap Sample standard deviation, = troue pe mean amount , conficlence level, c- 99% - 0.99 level of significance --05L-Cloob in all such fitnes, Here, meneany Now, the 99% prediction interval for the level of given by x + Zao II + 1 / We know, In of 5719 = 560.01) 0.022 it from danmal 2.576 get Tabe. So, The given by : Now, . z a = 20.005 [you right end point of the interval *+ 29.-[L* * +{2:576 * J"} - {0.0546 * 1.09543 0,995 +. $2.576 *.0.022.x uit 0.495 = 0.995 + 0.06199 1.0569 1.06. (Ang) Thanks you :) х