Question

Weight of boxes of a particular type of cargo follows some distribution with mean 60 pounds and standard deviation 10 pounds. (a) (6 points) If a random sample (i.i.d.) of 140 such boxes are taken, what is the probability that the sample average weight will be between 58 pounds and 61 pounds. (b) (6 points) A particular shipment consists of 100 such boxes. Assume the weights of the boxes in the shipment are i.i.d. Find the 95th percentile of the total shipment weight.

Answer 1

Solution :

Given that,

mean = = 60

standard deviation = = 10

a) n = 140

= = 60

= / n = 10 / 140 = 0.845

P(58 < < 61)  

= P[(58 - 60) / 0.845 < ( - ) / < (61 - 60) / 0.845)]

= P(-2.37 < Z < 1.18 )

= P(Z < 1.18) - P(Z < -2.37)

Using z table,  

= 0.8810 - 0.0089

= 0.8721

b) n = 100

= = 60

= / n = 10 / 100 = 1

Using standard normal table,

P(Z < z) = 95%

= P(Z < z ) = 0.95

= P(Z < 1.645) = 0.95  

z = 1.645

Using z-score formula  

= z * +

= 1.645 * 1 + 60

= 61.65