Question

The weight distribution of parcels sent in a certain manner is normal with mean value 17 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? (Round your answer to three decimal places.) c = lb

Answer 1

Answer: The value of c would be 26.514 lb.

Step-by-step explanation:

Since we have given that

Mean = 17 lb

Standard deviation = 3.3 lb

At 99%  level of significance, z = 2.58

So, it becomes,

$Z=\dfrac{X-\mu}{\sigma}\\\\2.58=\dfrac{X-17}{3.3}\\\\2.58\times 3.3=X-17\\\\8.514+17=X\\\\X=25.514$

So, the weight c would be

$c=25.514+1\\\\c=26.514$

Hence, the value of c would be 26.514 lb.