the Weight of crackers in a box is stated to be 16 oz. the amount that the packa
ID: 3437862 • Letter: T
Question
the Weight of crackers in a box is stated to be 16 oz. the amount that the packaging machine puts in the boxes is believed to have a normal model with a mean 16.15 oz. and standard deviation of 0.3 oz. A) What is the probability that the mean weight of a 10-box case of crackers is below 16 oz?B) What is the range of the weights for the middle 95% of boxes of crackers?
C) What is the range of the mean weights for the middle 95% of 10-box cases of crackers? the Weight of crackers in a box is stated to be 16 oz. the amount that the packaging machine puts in the boxes is believed to have a normal model with a mean 16.15 oz. and standard deviation of 0.3 oz. A) What is the probability that the mean weight of a 10-box case of crackers is below 16 oz?
B) What is the range of the weights for the middle 95% of boxes of crackers?
C) What is the range of the mean weights for the middle 95% of 10-box cases of crackers? the Weight of crackers in a box is stated to be 16 oz. the amount that the packaging machine puts in the boxes is believed to have a normal model with a mean 16.15 oz. and standard deviation of 0.3 oz. A) What is the probability that the mean weight of a 10-box case of crackers is below 16 oz?
B) What is the range of the weights for the middle 95% of boxes of crackers?
C) What is the range of the mean weights for the middle 95% of 10-box cases of crackers?
Explanation / Answer
the Weight of crackers in a box is stated to be 16 oz. the amount that the packaging machine puts in the boxes is believed to have a normal model with a mean 16.15 oz. and standard deviation of 0.3 oz.
A) What is the probability that the mean weight of a 10-box case of crackers is below 16 oz?
Standard error = sd/sqrt(n) = 0.3/SQRT(10) =0.0949
Z value for 16, z=(16-16.15)/0.0949 = -1.58
P( x < 16) =P( z < -1.58)
= 0.0571
B) What is the range of the weights for the middle 95% of boxes of crackers?
Z value for middle 95% = -1.96 and 1.96
The limits are( mean ± z*sd)
Lower weight =16.15-1.96*0.3 = 15.562
upper weight =16.15+1.96*0.3 = 16.738
The required range =(15.562, 16.738)
C) What is the range of the mean weights for the middle 95% of 10-box cases of crackers?
Z value for middle 95% = -1.96 and 1.96
The limits are( mean ± z*se)
Lower weight =16.15-1.96*0.0949 = 15.964
upper weight =16.15+1.96*0.0949 = 16.336
The required range =(15.964, 16.336)
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