The weight distribution of parcels sent in a certain manner is normal with mean value 12 lb and standard deviation 3.5 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?

Solution 47E Step1 of 2: We have The weight distribution of parcels sent in a certain manner is normal with mean value 12 lb and standard deviation 3.5 lb. That is =12 and = 3.5 We need to establish a weight value c beyond which there will be a surcharge and we need to find What value of c is such that 99% of all parcels are at least 1 lb under the surcharge weight Step2 of 2: Consider, = 99% = 0.99 Z Z 0.99 We need to find the value of Z score and it is calculated by using standard normal table(area under normal curve). We have to see where 0.99 falls in standard normal table, it falls in row 2.3 under column 0.04. Hence, Z = 2.34. 0.99 Now, The Z statistics is given by Z = x Where, Z = z score x = random variable(c) = mean = standard deviation x Z = x12 2.34 = 3.5 2.34×3.5 = c - 12 8.19 = c - 12 8.19 + 12 = x 20.19 = x Therefore the value of c is given by c = x + 1 = 20.19 + 1 c = 21.19.