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Chapter 5: Problem 13

Let \(X\) be \(\chi^{2}(50) .\) Approximate \(\operatorname{Pr}(40

Short Answer

Expert verified
The approximate value of \(\operatorname{Pr}(40<X<60)\) using the Central Limit Theorem is 0.6827.

Step by step solution

01

Identify the Mean and Standard Deviation

The mean of a Chi-square distribution is its degrees of freedom (\(v\)), which here is 50, so \(\mu = 50\). The standard deviation (\(\sigma\)) is \(\sqrt{2v}\), so for this problem, \(\sigma = \sqrt{2*50} = 10\).
02

Standardize the Random Variable

To use the Central Limit Theorem we need to transform the random variable \(X\) to a standard normal variable \(Z\). This is done with the formula \(Z = \frac{X - \mu}{\sigma}\). Thus, our limits 40 and 60 will be transformed as: \(Z_1 = \frac{40 - 50}{10} = -1\) and \(Z_2 = \frac{60 - 50}{10} = 1\).
03

Apply Central Limit Theorem

With the Central Limit Theorem, the given probability \(\operatorname{Pr}(40<X<60)\) can be approximated as \(\operatorname{Pr}(-1<Z<1)\). And by the properties of standard normal distribution, this is approximately 0.6827.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Central Limit Theorem
The Central Limit Theorem is a key concept in statistics. It allows us to make predictions about sample means even when the original distribution isn't normal. Here's how it works:
  • When you take many large enough samples from any distribution, the means of these samples will form a normal distribution.
  • This happens regardless of the shape of the original distribution.
In our problem, we're looking at a Chi-square distribution with 50 degrees of freedom. Although Chi-square distributions are not symmetrical, the Central Limit Theorem tells us that if our degrees of freedom are large enough, we can approximate it with a normal distribution. This is crucial because it simplifies our problem. Instead of working directly with a Chi-square distribution, we can convert our variables using the standard normal distribution methods to find the probabilities we need.
standard normal distribution
The standard normal distribution is a special normal distribution with a mean of 0 and a standard deviation of 1. It serves as a reference point for many statistical tests and calculations. Here's how it helps:
  • Standardizing any normal distribution allows us to compare it to the standard normal distribution.
  • This is done using the Z-score formula: \(Z = \frac{X - \mu}{\sigma}\).
For our exercise, we took the limits of our Chi-square distribution (40 and 60) and transformed them into Z-scores. By transforming these values into Z-scores, we move from a specific range of a Chi-square distribution into the universal standard scale. This allows us to easily apply our knowledge of the standard normal distribution to solve the problem.
degrees of freedom
Degrees of freedom is a term that describes the number of independent values that can vary in a statistical calculation. In the context of the Chi-square distribution:
  • Degrees of freedom (\(v\)) directly influence the shape of the Chi-square distribution.
  • The mean of the Chi-square distribution is equal to its degrees of freedom.
  • With more degrees of freedom, the distribution becomes more symmetrical and normal-looking.
In our problem, the Chi-square distribution has 50 degrees of freedom. This is significant when using the Central Limit Theorem, as it indicates our distribution is sufficiently close to normal, enabling us to make accurate approximations for probabilities. Degrees of freedom, therefore, provide a vital link between our original distribution and the methods used for approximation, guiding how we handle transformations and analyzing the data.

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Most popular questions from this chapter

Let \(Y_{n}\) denote the \(n\) th order statistic of a random sample from a distribution of the continuous type that has distribution function \(F(x)\) and p.d.f. \(f(x)=F^{\prime}(x) .\) Find the limiting distribution of \(Z_{n}=n\left[1-F\left(Y_{n}\right)\right]\).

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