Question

ale mosquitos have pretty short lifespans. Males of a certain species have lifespans that are strongly skewed to the right with a mean of 888 days and a standard deviation of 666 days. A biologist collects a random sample of 363636 of these male mosquitos and observes them to calculate the sample mean lifespan. What is the probability that the mean lifespan from the sample of 363636 mosquitos \bar x x ˉ x, with, \bar, on top exceeds 101010 days

Answer 1

0.0228 = 2.28% probability that the mean lifespan from the sample exceeds 10 days.

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.

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 of 8, standard deviation of 6, sample of 36:

This means that
`\mu = 8, \sigma = 6, n = 36, s = (6)/(√(36)) = 1`

What is the probability that the mean lifespan from the sample exceeds 10 days?

This is 1 subtracted by the pvalue of Z when X = 10. So

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

By the Central Limit Theorem

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

`Z = (10 - 8)/(1)`

`Z = 2`

`Z = 2` has a pvalue of 0.9772

1 - 0.9772 = 0.0228

0.0228 = 2.28% probability that the mean lifespan from the sample exceeds 10 days.