2.CommerceThe 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

the value of c is  20.155 such that 99% of all parcels are under the surcharge weight.

Step-by-step explanation:

Given that :

The mean value = 12

The standard deviation = 3.5

Let Consider Q to be the weight of the parcel that is normally distributed .

Then;

Q Norm(12,3.5)

The objective is to determine thewight  value of c under which there is a surcharge

Also, let's not that 99% of all the parcels are below the surcharge

However ;

From the Percentiles table of Standard Normal Distribution;

At 99th percentile; the value for Z = 2.33

The formula for the Z-score is:

2.33 × 3.5 = X - 12

8.155 = X - 12

- X = - 12 - 8.155

- X = -20.155

 X = 20.155

the weight  value of c under which there is a surcharge = X + 1 (0) since all the  pounds are below the surcharge

c = 20.155 + 1(0)

c = 20.155

Thus ; the value of c is  20.155 such that 99% of all parcels are under the surcharge weight.

$30.32

step-by-step explanation:

65.00*.12=2.80

65.00-2.80=57.20

57.20*1.06=60.632

60.632/2=30.316