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 56 tons of coal into each car. The actual weights of coal loaded into each car are normally distributed, with mean µ = 56 tons and standard deviation Ï = 1.1 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

When the distribution is normal, we use the z-score formula.

In a set with mean
`\mu` and standard deviation
`\sigma`, the zscore 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 pvalue, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean
`\mu` and standard deviation
`\sigma`, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean
`\mu` and standard deviation
`s = (\sigma)/(√(n))`.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

In this question, we have that:

`\mu = 56, \sigma = 1.1`

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

This is the pvalue of Z when X = 49.5. So

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

`Z = (49.5 - 56)/(1.1)`

`Z = -5.9`

`Z = -5.9` has a pvalue of 0

So 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 we have
`n = 35`, applying the Central limit theorem
`s = (1.1)/(√(35)) = 0.1859`

This is the pvalue of Z when X = 49.5. So

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

By the Central Limit Theorem

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

`Z = (49.5 - 56)/(0.1859)`

`Z = -35`

`Z = -35` has a pvalue of 0

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