Respuesta :
Answer:
The probability that the mean daily precipitation will be of X inches or less for a random sample of n November days is the p-value of [tex]Z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}[/tex], in which [tex]\mu[/tex] is mean amount of inches of rain and [tex]\sigma[/tex] is the standard deviation.
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 normal distributions can be solved using the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
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 establishes that, for a normally distributed random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex].
For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.
In this question:
Mean [tex]\mu[/tex], standard deviation [tex]\sigma[/tex]
n days:
This means that [tex]s = \frac{\sigma}{\sqrt{n}}[/tex]
Applying the Central Limit Theorem to the z-score formula.
[tex]Z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}[/tex]
What is the probability that the mean daily precipitation will be of X inches or less for a random sample of November days?
The probability that the mean daily precipitation will be of X inches or less for a random sample of n November days is the p-value of [tex]Z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}[/tex], in which [tex]\mu[/tex] is mean amount of inches of rain and [tex]\sigma[/tex] is the standard deviation.
