Question

According to records, the amount of precipitation in a certain city on a November day has a mean of inches, with a standard deviation of inches. What is the probability that the mean daily precipitation will be inches or less for a random sample of November days (taken over many years)

Answer 1

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 $Z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}$, in which $\mu$ is mean amount of inches of rain and $\sigma$ 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 $\mu$ and standard deviation $\sigma$, the z-score 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 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 $\mu$ 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:

Mean $\mu$, standard deviation $\sigma$

n days:

This means that $s = \frac{\sigma}{\sqrt{n}}$

Applying the Central Limit Theorem to the z-score formula.

$Z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}$

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 $Z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}$, in which $\mu$ is mean amount of inches of rain and $\sigma$ is the standard deviation.