Question

Suppose price changes of a particular commodity are independent and identically distributed random variables with µ = 1.1 and variance σ2 = 0.16. You collect a sample of 36 such price changes. Use the Central Limit Theorem to compute the probability that the average price change of this sample is less than 1.0.

Answer 1

0.0668 = 6.68% probability that the average price change of this sample is less than 1.0.

Explanation:

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.

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.

Mean µ = 1.1 and variance σ2 = 0.16. You collect a sample of 36 such price changes.

So the standard deviation is:

`\sigma = √(\sigma^2) = √(0.16) = 0.4`

Sample of 36:

This means that
`s = (0.4)/(√(36)) = 0.0667`

Probability that the average price change of this sample is less than 1.0.

This is the pvalue of Z when X = 1. So

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

`Z = (1 - 1.1)/(0.0667)`

`Z = -1.5`

`Z = -1.5` has a pvalue of 0.0668

0.0668 = 6.68% probability that the average price change of this sample is less than 1.0.