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 [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
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 [tex]\mu = 12, \sigma = 3.5[/tex]
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.
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]2.327 = \frac{X - 12}{3.5}[/tex]
[tex]X - 12 = 2.327*3.5[/tex]
[tex]X = 20.1445[/tex]
1 + 20.1445 = 21.1445
The value is c = 21.1445.
