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 310 pounds. Assume that the population follows the normal distribution. Fifty-five trucks are randomly selected and weighed.

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 310 pounds. Assume that the population follows the normal distribution. Fifty-five trucks are randomly selected and weighed. Within what limits will 95% of the sample mean occur

Subtract 1 from sample size to find degree of freedom(df). Here sample size is 55,so

df= 55-1= 54

To determine α, subtract confidence interval from 1 and then divide by 2. Here confidence interval is 95% or 0.95, so

α= (1-0.95)/2= 0.025

Use t-distribution table(see attachment) to find t-value for α=0.025 and df=54. So t=2.021(since df=54 is not listed in the table, I have used the table row corresponding to the next lowest value of df that is 40)

divide sample deviation, 310, by root of sample size that is 55. So,

[tex]\frac{310}{\sqrt{55} }[/tex]= 41.8

Now multiply the answers from last two steps 41.8 × 2.021= 84.5

lower limit= 6000-84.5=5915.5
upper limit= 6000+84.5=6084.5

95% of 55 trucks have weights between 5915.5 lbs and 6084.5 lbs