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?
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)