Question

Two companies, A and B, make express delivery for small-item packages in a city. Company A charges a flat fee of RM70 per package regardless of weight. Company B charges RM20 plus RM3.50 per kilogram of item. The weights of small-item packages delivered in the city have a normal distribution with mean of 9 kilograms and a standard deviation of 6.1 kilograms.
Find
a) the probability that Company B would charge more than Company A to deliver a small-item package.
b) the mean and standard deviation of amount charged by Company B to deliver a small-item package.​

Answer 1

Using the normal probability distribution and the central limit theorem, it is found that:

a) There is a 0.1922 = 19.22% probability that Company B would charge more than Company A to deliver a small-item package.

b) The mean amount charged is of RM 51.5, with a standard deviation of 21.35.

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

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

• It 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, which is the percentile of X.

By the Central Limit Theorem:

• When a fixed constant k multiplies a variable, the mean is $k\mu$ and the standard deviation is $k\sigma$
• When two normal variables are subtracted, the mean is the subtraction of the means, while the standard deviation is the square root of the sum of the variances.

In this question, b is needed to solve a, so I am going to place the solution to item b first.

Item b:

Flat fee of RM20(not considered for the standard deviation), plus a variable fee of RM3.5, hence:

$\mu_B = 20 + 3.5(9) = 51.5$

$\sigma = 6.1(3.5) = 21.35$

The mean amount charged is of RM 51.5, with a standard deviation of 21.35.

Item a:

This probability is $P(B - A) > 0$. For the distribution, we have that:

$\mu_{B-A} = \mu_B - \mu_A = 51.5 - 70 = -18.5$

Since A has a constant fee, it's standard deviation is 0, hence:

$\sigma_{B-A} = \sqrt{\sigma_A^2 + \sigma_B^2} = \sqrt{21.35^2} = 21.35$

The probability is 1 subtracted by the p-value of Z when X = 0, so:

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

$Z = \frac{0 + 18.5}{21.35}$

$Z = 0.87$

$Z = 0.87$ has a p-value of 0.8078.

1 - 0.8078 = 0.1922.

0.1922 = 19.22% probability that Company B would charge more than Company A to deliver a small-item package.

A similar problem is given at brainly.com/question/25403659