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 6,178 pounds and the standard deviation is 193 pounds. Assume that the population follows the normal distribution. Sixty three trucks are randomly selected and weighed. Within what limits will 95% of the sample means occur?

The limits are: __and__

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 =    6178          
z(alpha/2) = critical z for the confidence interval =    1.959963985          
s = sample standard deviation =    193          
n = sample size =    63          
              
Thus,              
              
Lower bound =    6130.342075          
Upper bound =    6225.657925          
              
Thus, the confidence interval is              
              
(   6130.342075   ,   6225.657925   ) [ANSWER]