Question

8.The weight distribution of parcels sent in a certain manner is normal with mean value 12lb 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

Answer 1

The value is c = 21.1445.

Explanation:

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean
`\mu` and standard deviation
`\sigma`, the z-score of a measure X is given by:

`Z = (X - \mu)/(\sigma)`

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

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

This means that
`\mu = 12, \sigma = 3.5`

What value of c is such that 99% of all parcels are at least 1 lb under the surcharge weight?

This 1 added to the value of X for the 99th percentile, which is X when Z has a p-value of 0.99, so X when Z = 2.327.

`Z = (X - \mu)/(\sigma)`

`2.327 = (X - 12)/(3.5)`

`X - 12 = 2.327*3.5`

`X = 20.1445`

1 + 20.1445 = 21.1445

The value is c = 21.1445.