Question 1101850: The local bakery bakes more than a thousand 1-pound loaves of bread daily, and the weights of these loaves varies. The mean weight is 2 lb. and 1 oz., or 936 grams. Assume the standard deviation of the weights is 25 grams and a sample of 34 loaves is to be randomly selected.
(a) This sample of 34 has a mean value of x, which belongs to a sampling distribution. Find the shape of this sampling distribution.
skewed right
(b) Find the mean of this sampling distribution. (Give your answer correct to nearest whole number.)
grams
(c) Find the standard error of this sampling distribution. (Give your answer correct to two decimal places.)
(d) What is the probability that this sample mean will be between 928 and 944? (Give your answer correct to four decimal places.)
(e) What is the probability that the sample mean will have a value less than 932? (Give your answer correct to four decimal places.)
(f) What is the probability that the sample mean will be within 4 grams of the mean? (Give your answer correct to four decimal places.)
Found 2 solutions by stanbon, Boreal:
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! The local bakery bakes more than a thousand 1-pound loaves of bread daily, and the weights of these loaves varies.
The mean weight is 2 lb. and 1 oz., or 936 grams. Assume the standard deviation of the weights is 25 grams and a sample of 34 loaves is to be randomly selected.
(a) This sample of 34 has a mean value of x, which belongs to a sampling distribution. Find the shape of this sampling distribution.
skewed right
(b) Find the mean of this sampling distribution. (Give your answer correct to nearest whole number.)
mean of the sample means = 936
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(c) Find the standard error of this sampling distribution. (Give your answer correct to two decimal places.)
std = 25/sqrt(34) = 4.29
(d) What is the probability that this sample mean will be between 928 and 944? (Give your answer correct to four decimal places.)
z(928) = (928-936)/4.29 = -1.8648
z(944) = (944-936)/4.29 = +1.8648
P(928 < x < 944) = P(-1.8648 < z < 1.8648) = normalcdf(-1.8648,1.8648) = 0.9378
(e) What is the probability that the sample mean will have a value less than 932? (Give your answer correct to four decimal places.)
Find the z-value of 932.
Find the probability that z is below that z-value
(f) What is the probability that the sample mean will be within 4 grams of the mean? (Give your answer correct to four decimal places.)
Lower:: (932-936)/4.29 = -4/4.29 = -0.9324
Upper:: (940-936)/4.29 = 4/4.29 = 0.9324
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P(-0.9324 < z < 0.9324) = normalcdf(-0.9324,0.9324) = 0.6489
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Cheers,
Stan H.
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Answer by Boreal(15235)  (Show Source):
You can put this solution on YOUR website! This is a t-distribution with mean 936 gm and sd= 25/sqrt(34)=4.29(which is the standard error of the mean)
t=(928-936)/4.29=-1.86
t=(944-936)/4.29=+1.86
probability is 0.9284
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less than 932 is t< (932-936)/4.29<-0.93
probability is 0.1795
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within 4 grams is between -0.93 and +0.93, and that is 1-2(0.1795)=0.6410
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