Problem 1 - A Modern Introduction to Probability and Statistics: Understanding Why and How

Question

Let \(X_{1}, X_{2}, \ldots, X_{144}\) be independent identically distributed random variables, each with expected value \(\mu=\mathrm{E}\left[X_{i}\right]=2\), and variance \(\sigma^{2}=\) \(\operatorname{Var}\left(X_{i}\right)=4\). Approximate \(\mathrm{P}\left(X_{1}+X_{2}+\cdots+X_{144}>144\right)\), using the central limit theorem.

Step-by-Step Solution

Answer

The probability \( \mathrm{P}(S_{144} > 144) \) is approximately 1.

Step 1: Understanding the Problem

We are tasked with finding the probability that the sum of 144 identically distributed random variables exceeds 144. We are given that the mean \( \mu \) of each variable is 2 and the variance \( \sigma^2 \) is 4.

Step 2: Define the Sum

Let \( S_{144} = X_1 + X_2 + \cdots + X_{144} \). The central limit theorem (CLT) states that if \( n \) is large, \( S_n \) is approximately normally distributed with mean \( n\mu \) and variance \( n\sigma^2 \).

Step 3: Calculate Mean and Variance of \( S_{144} \)

Using CLT, the mean of \( S_{144} \) is given by \( \mathrm{E}[S_{144}] = 144 \times 2 = 288 \). The variance is given by \( \operatorname{Var}(S_{144}) = 144 \times 4 = 576 \).

Step 4: Convert the Problem

We want to find \( \mathrm{P}(S_{144} > 144) \). Using the normal approximation, we convert it to standard normal distribution. With \( Z = \frac{S_{144} - 288}{\sqrt{576}} \), we find \( \mathrm{P}(Z > \frac{144 - 288}{24}) \).

Step 5: Calculate the Z-score

The Z-score for \( S_{144} = 144 \) is \( Z = \frac{144 - 288}{24} = -6 \). Therefore, \( \mathrm{P}(Z > -6) \) needs to be calculated.

Step 6: Probability from Z-table

The cumulative probability \( \mathrm{P}(Z < -6) \) is practically 0, so \( \mathrm{P}(Z > -6) = 1 - 0 = 1 \). Therefore, \( \mathrm{P}(S_{144} > 144) \approx 1 \).

Key Concepts

Normal Distribution

The normal distribution is a fundamental concept in statistics. It is characterized by its bell-shaped curve, noted for its symmetry around the mean. The spread of the distribution is determined by its standard deviation.
A few key features:

• In a normal distribution, the mean, median, and mode are all the same.
• The curve is completely described by two parameters: the mean (μ) and the standard deviation (σ).
• Approximately 68% of the data falls within one standard deviation of the mean, about 95% within two, and about 99.7% within three.

The normal distribution is so prevalent because many natural phenomena align with this pattern. In our exercise, the sum of variables outlined in the question turns into a normal distribution due to the Central Limit Theorem.
The Central Limit Theorem helps us understand that as we add more variables (like in the case with our 144 variables), the distribution of the total is approximately normal. This approximation allows us to use properties of the normal distribution to calculate probabilities.

Probability Calculation

Probability calculation in statistics refers to determining the likelihood of a particular outcome among all possible outcomes. In our exercise, we want to find the probability that the sum of 144 identical and independent random variables exceeds a certain threshold.
The process begins with understanding that the sum of these variables can be approximated using the normal distribution. Thanks to the Central Limit Theorem, even if the original variables are not normally distributed, their sum will tend to a normal distribution as the number of variables becomes large.
To calculate the probability, we do the following:

• Identify the new mean by multiplying the number of variables by their average value: here, it was 144 times 2 (the value of µ), giving a mean of 288.
• Determine the variance by multiplying the number of variables by their variance: 144 times 4, resulting in a variance of 576.
• Using these parameters (mean and variance), convert the original problem into a format suitable for using the normal distribution for probability calculation, allowing the use of the Z-score.

By converting our data into this format, the problem shifts into a standard, solvable statistical question, enabling the effective application of normal distribution principles.

Z-score

The Z-score is a statistical measurement that describes a value's relation to the mean of a group of values. It is measured in terms of standard deviations from the mean.
In simpler terms:

• A Z-score tells us how many standard deviations a certain data point is from the mean.
• A Z-score can be positive or negative, indicating whether the data point is above or below the mean.

For our exercise, once we have approximated our sum of 144 variables using a normal distribution, we have to convert the problem to a Z-score.
The formula used is:\[Z = \frac{X - \mu}{\sigma}\]where:

• \(X\) is the value of the data point (in our exercise, the threshold was 144).
• \(\mu\) is the mean (or average) of the distribution (was 288 in the exercise).
• \(\sigma\) is the standard deviation.

The Z-score tells us how far and in what direction the point is from the mean, measured in units of standard deviation. In the exercise, the calculation led us to a Z-score of -6, which indicates that our threshold is six standard deviations below the mean. This Z-score is a tool that allows us to find the probability related to our specific problem by consulting Z-tables, which show probabilities associated with different Z-scores.