Problem 2 - A Modern Introduction to Probability and Statistics: Understanding Why and How

Question

Let \(X_{1}, X_{2}, \ldots, X_{625}\) be independent identically distributed random variables, with probability density function \(f\) given by $$ f(x)= \begin{cases}3(1-x)^{2} & \text { for } 0 \leq x \leq 1 \\ 0 & \text { otherwise }\end{cases} $$ Use the central limit theorem to approximate \(\mathrm{P}\left(X_{1}+X_{2}+\cdots+X_{625}<170\right)\).

Step-by-Step Solution

Answer

The probability is approximately 0.8577.

Step 1: Find the Expected Value (Mean)

The first step is to calculate the expected value of a single random variable, denoted as \( E[X] \). The expected value \( E[X] \) can be found by integrating the density function, \( f(x) = 3(1-x)^2 \), over the interval \( [0, 1] \): \[E[X] = \int_0^1 x \cdot 3(1-x)^2 \, dx.\] Compute the integral: \[E[X] = \int_0^1 3x(1-2x+x^2) \, dx = \int_0^1 (3x - 6x^2 + 3x^3) \, dx.\] By evaluating it, we find \[E[X] = \left[ \frac{3x^2}{2} - \frac{6x^3}{3} + \frac{3x^4}{4} \right]_0^1 = \frac{3}{2} - 2 + \frac{3}{4} = \frac{1}{4}.\]

Step 2: Find the Variance

Next, calculate the variance \( \text{Var}(X) \) of a single random variable. First, find \( E[X^2] \): \[E[X^2] = \int_0^1 x^2 \cdot 3(1-x)^2 \, dx = \int_0^1 (3x^2 - 6x^3 + 3x^4) \, dx.\] Evaluate this integral: \[E[X^2] = \left[ \frac{3x^3}{3} - \frac{6x^4}{4} + \frac{3x^5}{5} \right]_0^1 = 1 - \frac{3}{2} + \frac{3}{5} = \frac{7}{20}.\] The variance is then \[\text{Var}(X) = E[X^2] - (E[X])^2 = \frac{7}{20} - \left(\frac{1}{4}\right)^2 = \frac{7}{20} - \frac{1}{16} = \frac{21}{80}.\]

Step 3: Apply the Central Limit Theorem

Since \( X_1, X_2, \ldots, X_{625} \) are iid, let \( S_n = X_1 + X_2 + \cdots + X_{625} \). The mean of \( S_n \) is \( nE[X] \) and variance \( n\text{Var}(X) \), where \( n = 625 \). \[E[S_{625}] = 625 \times \frac{1}{4} = 156.25,\] \[\text{Var}(S_{625}) = 625 \times \frac{21}{80} = \frac{13125}{80} = 164.0625.\] By the central limit theorem, \( S_{625} \) is approximately \( N(156.25, 164.0625) \).

Step 4: Standardize the Variable

To find \( \mathrm{P}(S_{625} < 170) \), standardize: \[Z = \frac{S_{625} - 156.25}{\sqrt{164.0625}}.\] Substitute and solve for \( Z \): \[Z = \frac{170 - 156.25}{\sqrt{164.0625}} \approx \frac{13.75}{12.806} \approx 1.0738.\]

Step 5: Use the Standard Normal Distribution

Using the standard normal distribution table (or a calculator), find \( \mathrm{P}(Z < 1.0738) \). This gives us the probability that the sum is less than 170. \( \mathrm{P}(Z < 1.0738) \approx 0.8577. \) This value indicates the probability that \( S_{625} < 170 \).

Key Concepts

Expected Value

The expected value, often referred to as the mean, is a fundamental concept in probability that describes the average outcome of a random variable over numerous trials. Imagine you throw a dice many times; the expected value will give you an idea of what value you'll get on average. To calculate this for a continuous random variable, like in our exercise, we use integration.

In the given problem, we found the expected value of a single random variable, denoted by \( E[X] \), using the probability density function (PDF). We integrated \( x \cdot f(x) \) over the interval \([0, 1]\), where \( f(x) = 3(1-x)^2 \). The computation showed us that the expected value \( E[X] = \frac{1}{4} \).

This result tells us that if we were to observe many values of this random variable, the average would be around 0.25. Expected value is critical in applications ranging from risk management in finance to decision-making in everyday life.

Variance

Variance measures the variability or spread of a set of data. In simple terms, it tells us how much the data points differ from the expected value (mean). High variance means data points are spread out widely, while low variance means they are clustered close to the mean. In equations, variance is often denoted as \( \text{Var}(X) \).

In our exercise, we calculated the variance by first finding \( E[X^2] \), the expected value of the squared random variable, which requires great attention to detail due to additional integration. Using the PDF \( f(x) = 3(1-x)^2 \), we found \( E[X^2] = \frac{7}{20} \). Then, the variance was assigned to be \( \text{Var}(X) = E[X^2] - (E[X])^2 \). This resulted in \( \text{Var}(X) = \frac{21}{80} \).

Variance provides insight into the reliability of the mean: the smaller the variance, the more dependable the mean as a descriptor of typical individual observations.

Probability Density Function

A probability density function (PDF) describes the likelihood of a random variable to take a specific value. For continuous random variables, the PDF represents the relative likelihood of the variable across its range. It differs from probability mass functions which are used for discrete variables. A helpful way to think of PDFs is as the equation for a probability curve over continuous data.

In the central limit theorem (CLT) exercise, the probability density function is given by \( f(x) = 3(1-x)^2 \). This function defines the distribution of each \( X_i \), showing that values around zero are more likely, as the height of the curve (probability) is higher closer to zero.

Understanding the PDF is crucial as it forms the foundation for calculating such key statistics as expected value and variance. It dictates the behavior of data under different scenarios and is highly instrumental when applying techniques like the Central Limit Theorem.

Standard Normal Distribution

The standard normal distribution is a special normal distribution with a mean of 0 and a standard deviation of 1. It is a symmetric, bell-shaped curve, often used because it simplifies statistical calculations. In practice, any normal distribution can be converted into a standard normal distribution via a process called standardization.

In our problem, after finding the cumulative sum \( S_{625} \), we needed to standardize it using the formula \( Z = \frac{S_n - n \cdot E[X]}{\sqrt{n \cdot \text{Var}(X)}} \). This step allowed us to transform \( S_{625} \) into a standard normal variable \( Z \), letting us use standard normal distribution tables or calculators. The value \( Z \approx 1.0738 \) facilitated finding the probability that \( S_{625} < 170 \) in a standard context.

This process of using the standard normal distribution highlights its importance in providing a universal scale and method to assess probabilities for any normal variable, a pivotal aspect of statistical tests and inferential statistics.