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

Answer:

The value is c = 21.1445.

Step-by-step 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:

$Z = \frac{X - \mu}{\sigma}$

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.

$Z = \frac{X - \mu}{\sigma}$

$2.327 = \frac{X - 12}{3.5}$

$X - 12 = 2.327*3.5$

$X = 20.1445$

1 + 20.1445 = 21.1445

The value is c = 21.1445.