Question

Suppose that a truck has a legal load carrying capacity of 10,000 lbs. We want to load product, with an average weight of 21 lbs. and a standard deviation of 4 lbs. onto the truck. If we have 473 of these units, how many can we load on the truck if we only want to have a 1% risk chance of “overloading” the truck?

	a.	473
	b.	472
	c.	471
	d.	470
	e.	463

Answer 1

Let N be the number of units to be loaded.

Average weight of total N units = 21N lbs

Variance of N units = N = 16N

Standard deviation of N units = = 4

Let T be the total weight of N units. Then by Central limit theorem, T ~ (21N , 16N)

P(T > 10,000) = 0.01

P[Z > (10000 - 21N) / (4 ) ] = 0.01

(10000 - 21N) / (4 ) = 2.326

10000 - 21N = 9.304

Squaring we get,

Using quadratic equation formula,

N = (420086.6 - 8529.449 ) /882 , (420086.6 + 8529.449 ) /882

N = 466.6181 ,  485.9592

We will take vales less than 473 units. So, N = 467 (Rounding to nearest value)

467 answer is not given. We will choose the value lower than 467 which is 463

e. 463