Answer:
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 [tex]\mu[/tex] = 12
The standard deviation [tex]\sigma[/tex] = 3.5
Let Consider Q to be the weight of the parcel that is normally distributed .
Then;
Q [tex]\sim[/tex] 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:
[tex]Z = \dfrac{X- \mu}{\sigma}[/tex]
[tex]2.33 = \dfrac{X - 12}{3.5}[/tex]
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.
