Question

Lorry carries heavy Cartons of Goods. The weights of these cartons are distributed about a mean of 100kg and with a standard deviation of 7kg. Find how many cartons the Lorry can carry so that the probability of the total load exceeding 4500kg is less than 0.05.

Answer 1

Given mean= 100;sd=7

Let the load L be the sum of the weights of n cartons. 

Then L ~ N(100n, 49n) approximately (provided n is “large”). 

We require P(L > 4500) < 0.05,

that is $P(Z>\frac{4500-100n}{7\sqrt{n}})$ <0.05

so we require $\frac{4500-100n}{7\sqrt{n}}$ >1.6449

This ⇒ √n < 6.651 ⇒ n < 44.2, so maximum number of cartons to be carried is 44.