Question

The weight distribution of parcels sent in a certain manner is normal with mean of 12 lb and standard deviation of 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 under the surcharge weight?

Answer 1

Answer:

21.16

Step-by-step explanation:

Starting from the theory we have the following equation:

$fi*P(x$

Using the data supplied in the exercise, we have subtracting the mean and dividing by the standard deviation:

$P( z \leq \frac{c-1-12}{3.5}) =0.99/fi$

solving for "c", knowing that fi is a tabulating value:

$\frac{c-13}{3.5}=0.99/fi\\\frac{c-13}{3.5}=2.33\\c-13=2.33*3.5\\c = 8.155 +13\\c = 21.155$

therefore the value of c is equal to 21.16