91Ó°ÊÓ

Let \(X_{1}, X_{2}, \ldots, X_{n}\) be independent \(\chi^{2}\) -distributed random variables, each with 1 df. Define \(Y\) as $$ Y=\sum_{i=1}^{n} X_{i} $$ It follows from Exercise 6.59 that \(Y\) has a \(x^{2}\) distribution with \(n\) df. a. Use the preceding representation of \(Y\) as the sum of the \(X\) 's to show that \(Z=(Y-n) / \sqrt{2 n}\) has an asymptotic standard normal distribution. b. A machine in a heavy-equipment factory produces steel rods of length \(Y\), where \(Y\) is a normally distributed random variable with mean 6 inches and variance. 2 . The cost \(C\) of repairing a rod that is not exactly 6 inches in length is proportional to the square of the error and is given, in dollars, by \(C=4(Y-\mu)^{2} .\) If 50 rods with independent lengths are produced in a given day, approximate the probability that the total cost for repairs for that day exceeds \(\$ 48\).

Step by step solution

01

Analyze the Standardization of Y

We start by analyzing part (a) of the problem. Given that \(Y = \sum_{i=1}^{n} X_i\) follows a \(\chi^2\) distribution with \(n\) degrees of freedom, we know that the mean \(\mu\) of a \(\chi^2(n)\) distribution is \(n\) and the variance \(\sigma^2\) is \(2n\). To standardize \(Y\), we use:\[Z = \frac{Y - n}{\sqrt{2n}}\]This takes \(Y\) and transforms it into a form that, according to the Central Limit Theorem, approaches a standard normal distribution as \(n\) becomes large.

02

Apply Central Limit Theorem

By the Central Limit Theorem (CLT), for large \(n\), the sum of \(n\) independent and identically distributed random variables (in this case, \(X_i\)) will tend towards a normal distribution. Since \(Y\) is the sum of \(\chi^2(1)\) variates, the standardized version, \(Z = \frac{Y-n}{\sqrt{2n}}\), will thus follow an asymptotic standard normal distribution as \(n \to \infty\). Therefore, \(Z\) becomes a \(N(0,1)\) distribution when \(n\) is large enough.

03

Defining Cost C and Identifying Distribution for Part b

For part b, \(C = 4(Y - \mu)^2\) represents the cost model related to the deviation of \(Y\) from its mean \(\mu = 6\). Since \(Y\) arrives from a \(N(6, 2)\) distribution for each rod and \(Y - \mu\) would be \(N(0,2)\) under normal approximation for each rod, we need to consider the cost for 50 rods. Let \(C_{total} = \sum_{i=1}^{50} 4(Y_i - 6)^2\), where each \(Y_i\) is independent.

04

Compute Expected Cost and Variance

For a single rod, \(Y_i - 6\) follows \(N(0, 2)\,\), whose mean is \(0\) and variance is \(2\). The expected cost per rod is:\[E[4(Y_i - 6)^2] = 4 \cdot (2) = 8\]Thus, for 50 rods, the expected total cost is:\[E[C_{total}] = 50 \cdot 8 = 400\]And the variance of the cost for 50 rods follows from variance properties:\[Var(C_{total}) = 50 \cdot Var(4(Y_i - 6)^2) = 50 \cdot 8 = 400\]This allows us to treat \(C_{total}\) as a normal distribution \(N(400, 400)\).

05

Find Probability of Cost Exceeding $48

To find the probability that the cost exceeds $48, standardize \(C_{total}\):\[Z' = \frac{C_{total} - 400}{\sqrt{400}} = \frac{C_{total} - 400}{20}\]We need to find \(P(C_{total} > 48)\), which after standardizing becomes:\[P(Z' > \frac{48 - 400}{20}) = P(Z' > -17.6)\]Given the negative and extreme standard score, we expect this probability to be very close to 1.

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Key Concepts

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

Normal Distribution

The normal distribution, often referred to as the bell curve, is a continuous probability distribution that is symmetrical around its mean. This means that the likelihood of outcomes is evenly spread around the average, which is the highest point on the graph. Some key aspects of a normal distribution include:

• The mean, median, and mode of the distribution all coincide at the same point, which is the center of the graph.
• It is defined by two parameters: the mean (\( \mu \)) and the variance (\( \sigma^2 \)). These determine the location and scale of the distribution.
• Approximately 68% of the data fall within one standard deviation of the mean, 95% within two, and 99.7% within three.

In the context of the exercise, when the lengths of steel rods produced by a factory follow a normal distribution, it implies that most rods have lengths close to the mean, with fewer rods being shorter or longer. This helps manufacturers predict rod length outcomes and estimate costs associated with deviations from the standard specifications.

Chi-Square Distribution

The chi-square distribution is a continuous distribution often used in statistics to assess variance in a set of data. It comes into play primarily in tests of independence and goodness-of-fit tests. Here are a few highlights:

• It is derived from the sum of squared standard normal variables.
• The shape of the chi-square distribution depends on the degrees of freedom, which is usually related to the number of independent observations available.
• With more degrees of freedom, the distribution shifts to the right, becoming more normal.

In the given exercise, the random variables are chi-square distributed with 1 degree of freedom. The variable \( Y \) represents the sum of these variables, giving \( Y \) a chi-square distribution with \( n \) degrees of freedom. This makes \( Y \) an appropriate candidate for applying the Central Limit Theorem when determining its asymptotic behavior.

Standardization

Standardization is the process of transforming a random variable into a standard normal distribution. This means that the transformed data has a mean of 0 and a standard deviation of 1, making it easier to compare different data sets. The transformation is given by the formula:
\[ Z = \frac{X - \mu}{\sigma} \]
Where:

• \( X \) is the original data point.
• \( \mu \) is the mean of the data.
• \( \sigma \) is the standard deviation.

Applying this concept in the exercise, the variable \( Y \) was standardized to create \( Z \) as explained in step 1. This step aligns with the Central Limit Theorem by indicating that, as the sample size \( n \) grows, \( Z \) will conform to a standard normal distribution. Standardization is a crucial step in many statistical analyses, enabling easier interpretation and comparison between different data sets.

Variance

Variance is a fundamental measure used in statistics to quantify the amount of variation or dispersion in a set of data points. Specifically, it assesses how much individual observations deviate from the mean. The formula for variance is given by:
\[ \sigma^2 = \frac{1}{n}\sum_{i=1}^{n} (X_i - \mu)^2 \]
Where:

• \( \sigma^2 \) is the variance.
• \( X_i \) is each data point.
• \( \mu \) is the mean of the data set.
• \( n \) is the number of observations.

Variance is important in this exercise as it helps calculate the standard error required for the normal approximation of the cost function. In practical applications, understanding variance allows businesses to predict and manage variability in processes, such as the lengths of rods produced in a factory. Greater variance implies less predictability, which can increase costs if not managed carefully.