Asurvey found that the american family generates an average of 17.2 pounds of glass garbage each year. assume the standard deviation of the distriution is 2.5 pounds. find the probability that the mean of a sample of 55 families will be between 17 and 18 pounds

Given:
μ = 17.2 lb, the population mean
σ = 2.5 lb, the population standard deviation

The sample size is n = 55.
σ/√n =2.5/√55 = 3.0339

For the random variable x = 14 lb, with sample size = 55,
z = (x - μ)/(σ√n) = (14 - 17.2)/3.0339 = -1.0547
From standard tables, P(x<14) = 0.1458

For the random variable x = 18 lb,
z = (18 - 17.2)/3.0339 = 0.2637
P(x<18) = 0.604

Therefore 
P(14 < x < 18) = 0.604 - 0.1458 = 0.4582

0.458 or  45.8%