Question

3. Uniform A package delivery service breaks up its shipping charges into weight classes, where the package weights are uniformly distributed WITHIN each weight class. a. Suppose one of the shipping classes is 10 to 12 lbs. What proportion of packages in this class would weigh less than 11 lbs? (draw the distribution as I did in class) b. What proportion would weigh more than 11.5 lbs? c. What would the average weight for a package in this class be?

Answer 1

a) 50% of packages in this class would weigh less than 11 lbs.

b) 25% would weigh more than 11.5 lbs.

c) The average weight for a package in this class is 11 lbs.

Explanation:

An uniform probability is a case of probability in which each outcome is equally as likely.

For this situation, we have a lower limit of the distribution that we call a and an upper limit that we call b.

The probability that we find a value X lower than x is given by the following formula.

`P(X \leq x) = (x - a)/(b-a)`

For this problem, we have that:

a. Suppose one of the shipping classes is 10 to 12 lbs. What proportion of packages in this class would weigh less than 11 lbs?

Uniform distribution from 10 lbs to 12 lbs, this means that
`a = 10, b = 12`.

The answer is
`P(X \leq 11)`.

`P(X \leq 11) = (11 - 10)/(12-10) = 0.5`

50% of packages in this class would weigh less than 11 lbs.

b. What proportion would weigh more than 11.5 lbs?

Either a package weighs 11.5 lbs or less, or it weighs more than 11.5 lbs. The sum of the probabilities of these events is decimal 1. So:

`P(X \leq 11.5) + P(X > 11.5) = 1`

`P(X > 11.5) = 1 - P(X \leq 11.5)`

`P(X > 11.5) = 1 - (11.5-10)/(12-10)`

`P(X > 11.5) = 0.25`

25% would weigh more than 11.5 lbs.

c. What would the average weight for a package in this class be?

The mean M of the uniform distribution is:

`M = (a+b)/(2)`

So

`M = (10+12)/(2) = 11`

The average weight for a package in this class is 11 lbs.