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Let x_(1),x_(2),dots,x_(100) denote the actual net weights of 100 randomly selected 50-lb bags of feetilizer.a. If expected weight of each bag is mu=50 and the variance is 6^(2)=1P(49.9 <= bar(x) <= 50.1)=P((49.9-mu)/(6//sqrtn) <= (( bar(x))-4)/(6//sqrtn) <= (50.1-4)/(6//sqrtn))Since aceording to Central limit Theoren (CLT), as ' n ' becomes sutficiently laeye say n >= 30, sampling distribution of sample mean tends to normal distribution.{:[:.P(49.9 <= bar(x) <= 50.1)=P((49.9-50)/(1//sqrt100) <= z <= (50. ... See the full answer