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valina [46]
3 years ago
14

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?
Mathematics
1 answer:
Natasha_Volkova [10]3 years ago
7 0

Answer:

Yes, it would be unusual.

Step-by-step 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 = \frac{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 = \frac{\sigma}{\sqrt{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 = \frac{5}{\sqrt{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 = \frac{X - \mu}{\sigma}

By the Central Limit Theorem

Z = \frac{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.

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