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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?

User Hemal
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7 votes

Answer:

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.

User Silvans Solanki
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