Question

The lifespan (in days) of the common housefly is best modeled using a normal curve having mean 22 days and standard deviation 5. Suppose a sample of 25 common houseflies are selected at random. Would it be unusual for this sample mean to be less than 19 days?

Answer 1

Yes, it would be unusual.

Explanation:

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

Normal probability distribution

Problems of normally distributed samples are 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.

If
`Z \leq -2` or
`Z \geq 2`, the outcome X is considered unusual.

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 = 22, \sigma = 5, n = 25, s = (5)/(√(25)) = 1`

Would it be unusual for this sample mean to be less than 19 days?

We have to find Z when X = 19. So

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

By the Central Limit Theorem

`Z = (19 - 22)/(1)`

`Z = -3`

`Z = -3 \leq -2`, so yes, the sample mean being less than 19 days would be considered an unusual outcome.