Question

The gypsy moth is a serious threat to oak and aspen trees. A state agriculture department places traps throughout the state to detect the moths. When traps are checked periodically, the mean number of moths trapped is only 0.7, but some traps have several moths. The distribution of moth counts is discrete and strongly skewed, with standard deviation 0.8.

A) What are the mean and standard deviation of the average number of moths x in 60 traps?
B) Use the central limit theorem to find the probability (±0.01) that the average number of moths in 40 traps is greater than 0.4.

Answer 1

a) The mean is 0.7 and the standard deviation is 0.1033.

b) 99.11% probability that the average number of moths in 40 traps is greater than 0.4.

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 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.

When traps are checked periodically, the mean number of moths trapped is only 0.7, but some traps have several moths. The distribution of moth counts is discrete and strongly skewed, with standard deviation 0.8.

This means that
`\mu = 0.7, \sigma = 0.8`

A) What are the mean and standard deviation of the average number of moths x in 60 traps?

60 traps means that
`n = 60`

By the Central Limit Theorem

Mean
`\mu = 0.7`

Standard deviation
`s = (0.8)/(√(60)) = 0.1033`

The mean is 0.7 and the standard deviation is 0.1033.

B) Use the central limit theorem to find the probability (±0.01) that the average number of moths in 40 traps is greater than 0.4.

40 traps means that
`n = 40`

Mean
`\mu = 0.7`

Standard deviation
`s = (0.8)/(√(40)) = 0.1265`

This probability is 1 subtracted by the pvalue of Z when X = 0.4. So

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

By the Central Limit Theorem

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

`Z = (0.4 - 0.7)/(0.1265)`

`Z = -2.37`

`Z = -2.37` has a pvalue of 0.0089

1 - 0.0089 = 0.9911

99.11% probability that the average number of moths in 40 traps is greater than 0.4.