Question

The weight distribution of parcels sent in a certain manner is normal with meanvalue 12 pounds and standard deviation 3.5 pounds. The parcel service wishes to establish aweight valuecbeyond which there will be a surcharge. What value ofcis such that 99% ofall parcels are under the surcharge weight?

Answer 1

Answer:

The parcel with weight less than 20.14 pounds are 99% of all parcels under the surcharge weight.

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 12 pounds

Standard Deviation, σ = 3.5 pounds

We are given that the distribution of weights is a bell shaped distribution that is a normal distribution.

Formula:

$z_{score} = \displaystyle\frac{x-\mu}{\sigma}$

We have to find the value of x such that the probability is 0.99

$P( X < x) = P( z < \displaystyle\frac{x - 12}{3.5})=0.99$  

Calculation the value from standard normal z table, we have,  

$\displaystyle\frac{x - 12}{3.5} = 2.326\\\\x = 20.141\approx 20.14$  

Thus, parcel with weight less than 20.14 pounds are 99% of all parcels under the surcharge weight.