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
and standard deviation
, the zscore of a measure X is given by:
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
or
, the outcome X is considered unusual.
Central Limit Theorem
The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean
and standard deviation
, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean
and standard deviation
.
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](https://tex.z-dn.net/?f=%5Cmu%20%3D%2022%2C%20%5Csigma%20%3D%205%2C%20n%20%3D%2025%2C%20s%20%3D%20%5Cfrac%7B5%7D%7B%5Csqrt%7B25%7D%7D%20%3D%201)
Would it be unusual for this sample mean to be less than 19 days?
We have to find Z when X = 19. So
By the Central Limit Theorem
![Z = -3](https://tex.z-dn.net/?f=Z%20%3D%20-3)
, so yes, the sample mean being less than 19 days would be considered an unusual outcome.