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I need help to get started with this question.

A grain mill manufactures 100 pound bags of flour for sale in restaurant supply warehouses. Historically the weights of bags of flour manufactured at the mill were normally distributed with a mean of 100 pounds and a standard deviation of 15 pounds.

A. what is the probability that the weight of a randomly selected bag of flour falls between 94 and 106 pounds?

B. if samples of 36 bags are taken, what is the standard error of the mean?

c. what is the probability that a sample of 36 bags of four has a mean weight between 94 and 106 pounds?

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  • $\begingroup$ Determine the z-score of 94 and or 106. z(94) = (94-100)/15 = -6/15 = -2/5 = -0.4 z(106) = (106-100)/15 = 6/15 = +0.4 P(94<= x <=106) = P(-0.4<= z <=0.4) = 0.3108 =================================================== ...the Probability that the sample of 36 bags which will be n=36 has a mean weight between 94 and 106 pounds. Is that correct sir for the first part? $\endgroup$ – Hani Abdullah Apr 21 '14 at 20:20
  • $\begingroup$ Your answer to A) is correct. For C), if $\bar{X}$ is the average of a sample of $36$, then $\bar{X}$ is normal mean $100$ standard deviation $15/\sqrt{36}=2.5$. Now use same procedure, we want the probability that a standard normal falls between $-6/2.5$ and $6/2.5$. $\endgroup$ – André Nicolas Apr 21 '14 at 20:27
  • $\begingroup$ The std for the distribution of sample means is 15/sqrt(36) = 15/6 = 5/2 =================== z(94) = (94-100)/(5/2) = -6/(5/2) = -12/5 = -2.8 z(106) = (106-100)/(5/2) = +2.8 ---- P(96<- x-bar <=106) = P(-2.8<= z <=2.8) = 0.9949. Is that correct sir? $\endgroup$ – Hani Abdullah Apr 21 '14 at 20:28

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