Crossett Trucking Company claims that the mean weight of its

Crossett Trucking Company claims that the mean weight of its delivery trucks when they are fully loaded is 6,000 pounds and the standard deviation is 150 pounds. Assume that the population follows the normal distribution. Forty trucks are randomly selected and weighed. Within what limits will 95% of the sample means occur? (Round your z-value to 2 decimal places and final answers to 1 decimal place.) Sample means

Solution

Note that              
              
Lower Bound = X - z(alpha/2) * s / sqrt(n)              
Upper Bound = X + z(alpha/2) * s / sqrt(n)              
              
where              
alpha/2 = (1 - confidence level)/2 =    0.025          
X = sample mean =    6000          
z(alpha/2) = critical z for the confidence interval =    1.96          
s = sample standard deviation =    150          
n = sample size =    40          
              
Thus,              
              
Lower bound =    5953.514518          
Upper bound =    6046.485482          
              
Thus, the confidence interval is              
              
(   5953.514518   ,   6046.485482   ) [ANSWER]