Question

Coal is carried from a mine in West Virginia to a power plant in New York in hopper cars on a long train. The automatic hopper car loader is set to put 88 tons of coal into each car. The actual weights of coal loaded into each car are normally distributed, with mean µ = 88 tons and standard deviation σ = 0.5 ton.

Required:
a. What is the probability that one car chosen at random will have less than 49.5 tons of coal?
b. What is the probability that 35 cars chosen at random will have a mean load weight of less than 49.5 tons of coal?

Answer 1

a) 0% probability that one car chosen at random will have less than 49.5 tons of coal.

b) 0% probability that 35 cars chosen at random will have a mean load weight of less than 49.5 tons of coal.

Explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

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.

In this question:

`\mu = 88, \sigma = 0.5`

a. What is the probability that one car chosen at random will have less than 49.5 tons of coal?

This is the p-value of Z when X = 49.5, so:

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

`Z = (49.5 - 88)/(0.5)`

`Z = -77`

`Z = -77` has a p-value of 0.

0% probability that one car chosen at random will have less than 49.5 tons of coal.

b. What is the probability that 35 cars chosen at random will have a mean load weight of less than 49.5 tons of coal?

Now
`n = 35, s = (0.5)/(√(35))`

So

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

By the Central Limit Theorem

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

`Z = (49.5 - 88)/((0.5)/(√(35)))`

`Z = -455.5`

`Z = -455.5` has a p-value of 0.

0% probability that 35 cars chosen at random will have a mean load weight of less than 49.5 tons of coal