Question

The weight distribution of parcels sent in a certain manner is normal with mean value 15 lb and standard deviation 3.3 lb. The parcel service wishes to establish a weight value c beyond which there will be a surcharge. What value of c is such that 99% of all parcels are at least 1 lb under the surcharge weight?

Answer 1

Answer: 23.7 lb

Step-by-step explanation:

Mean m = 15 lb

Standard deviation d = 3.3 lb

To determine the surcharge weight, we need to know the highest weight of 99% of the parcels.

P(z<x) = 0.99 = ¢(Z)

Z = 2.33

Since Z = (x - m)/d

x = dZ + m

x = 3.3*2.33 + 15

x = 22.689 lb approximately

x = 22.7 lb

Therefore, the highest weight for 99% if the parcels is 22.7 lb.

That is, the surcharge weight = 22.7 + 1 = 23.7 lb