Problem 5

Question

Let $X$ be a $\operatorname{Bin}(n, p)$ distributed random variable. Show that the random variable $$ \frac{X-n p}{\sqrt{n p(1-p)}} $$ has a distribution that is approximately standard normal.

Step-by-Step Solution

Verified

Answer

By the Central Limit Theorem, the standardized binomial random variable becomes approximately standard normal as $n$ increases.

1Step 1: Understand the Problem

We are given a binomially distributed random variable $X$ with parameters $n$ (number of trials) and $p$ (probability of success). Our goal is to show that the standardized version of $X$, $\frac{X-n p}{\sqrt{n p(1-p)}}$, approximately follows a standard normal distribution as $n$ becomes large.

2Step 2: Recall the Central Limit Theorem

The Central Limit Theorem states that for a large enough $n$, the distribution of the sum (or average) of $n$ independent random variables becomes approximately normal, regardless of the original distribution. For a binomial random variable $X$, as $n$ becomes large, $X$ can be approximated by a normal distribution with mean $np$ and variance $np(1-p)$.

3Step 3: Standardize the Random Variable

Standardization is done by subtracting the mean from the random variable and then dividing by the standard deviation. Since $X$ has mean $np$ and variance $np(1-p)$, its standard deviation is $\sqrt{np(1-p)}$. Thus, the standardized form is \[ Z = \frac{X - np}{\sqrt{np(1-p)}} \].

4Step 4: Apply the Central Limit Theorem

With the standardization complete, according to the Central Limit Theorem, as $n$ approaches infinity, the standardized version $Z$ of $X$ will closely follow a standard normal distribution. This implies that $Z$ has a mean of 0 and a variance of 1. Hence, $Z$ approximates a $\mathcal{N}(0, 1)$ distribution.

Key Concepts

Standard Normal DistributionBinomial DistributionRandom Variable Standardization

Standard Normal Distribution

The standard normal distribution is a special type of normal distribution. It is distinguished by two key features: a mean ($\mu$) of 0, and a standard deviation ($\sigma$) of 1. In practice, it helps simplify calculations and comparisons between different statistical data sets.
Whenever data is transformed into a standard normal distribution, it is referred to as a Z-score. This allows one to easily determine how far a particular value deviates from the average in terms of standard deviations.
This concept is crucial in statistical analysis for tasks like hypothesis testing and creating confidence intervals. Once data is presented in the form of a standard normal distribution, probabilities and percentiles can be looked up in standard normal distribution tables, making analysis more convenient.

Binomial Distribution

The binomial distribution is a discrete probability distribution. It describes the number of successes in a fixed number of trials. Each trial is independent and has the same probability of success ($p$). Binomial random variables are denoted as $\operatorname{Bin}(n, p)$, where $n$ is the number of trials.
Some key features of a binomial distribution include:

The distribution is completely defined by two parameters: $n$ (trials) and $p$ (success probability).
The mean of a binomial distribution is given by $np$.
The variance is $np(1-p)$.

As $n$ increases, under certain conditions, a binomial distribution can be approximated by a normal distribution, thanks to the Central Limit Theorem. This makes it easier to work with in terms of probability and statistical analysis.

Random Variable Standardization

Standardization of a random variable is crucial for comparison across different scales and to apply certain statistical methods. The process involves converting an original random variable into a standardized form, usually focusing on achieving a normal distribution with specific characteristics.
To standardize, subtract the mean original value from the random variable. Then divide it by the standard deviation. For a binomial random variable $X$ with mean $np$ and standard deviation $\sqrt{np(1-p)}$, the standardized form becomes $\frac{X - np}{\sqrt{np(1-p)}}$.
This transformation is significant because it adjusts for both the location and the scale of the data set, allowing for the application of the Central Limit Theorem. As a result, for large $n$, the outcome is approximately standard normal, which has a mean of 0 and a variance of 1. This makes statistical analysis streamlined and facilitates the application of statistical tests.

Problem 4

Problem 6

Other exercises in this chapter

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Problem 6

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Problem 8

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