A Manufacturing process produces bags of cookies. The distribution of the weight

Question

A Manufacturing process produces bags of cookies. The distribution of the weights of the contents of these bags is approximately Normal with a mean of 15.0oz and standard deviation 1.0oz.

A.) Suppose we take many samples of size n=16, from this pop weigh each bag in the sample and then compute the mean weight for each sample of 16. The Central Limit Theorem tells us that the shape of the distribution of the means of those samples as the number of samples taken increase will look _____, with a mean of ______ and standard Dev of ______.

B.) We Draw one sample of N=16 and find the mean of the weights of the bags in that sample is 14.6 oz. A 98% confidence interval to estimate the population mean "u" would be?

C.) Suppose we draw random samples of size n from the initial pop want to construct a confidence interval the same level as in B BUT reduce the margin of error to a value of no more than 3. What sample size is needed to achieve this level of accuracy?

D.) If sample sizes of N=100 are selected randomly from the original population, the probability that the sample mean will be between 14.868 and 15.132 oz is about??

E.) Suppose we want to take samples of size n from the initial population. If we want the distribution of the sample means from those samples to have a standard deviation of .03, what is the sample size n that we must use to achieve this goal??

Thank you in advance!

Explanation / Answer

a)  Suppose we take many samples of size n=16, from this pop weigh each bag in the sample and then compute the mean weight for each sample of 16. The Central Limit Theorem tells us that the shape of the distribution of the means of those samples as the number of samples taken increase will look _Normal distribution, with a mean of 15 and standard Dev of 1 / sqrt(16)

b ) n = 16 , mean = 14.6 , std.deviation = 1

z value at 98% confidence interval = 2.33

CI = mean + / - z * ( s/sqrt(n))

= 14.6 + / -2.33 * ( 1/sqrt(16))

= (14.02 , 15.18)

c) ME = 3 , n = 16 , mean = 15 , s = 1

ME = t * (s / sqrt(n))

= 2.33 * ( 1/sqrt(16))

= 0.5825

d) n =100 , mean =15 , s = 1

p( 14.868 < x < 15.132)

z = ( x - mean) / ( s / sqrt(n))

= p ( (14.868 - 15 ) / ( 1 / sqrt(100) < z < (15.132 - 15 ) / ( 1 / sqrt(100))

= p ( -1.32 < z < 1.32)

Now, we need to find p ( -1.32 < z < 1.32)

p( 14.868 < x < 15.132) = p ( -1.32 < z < 1.32) = 0.8132

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