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Similar to Exercise \(4.8 .2\) but now approximate \(\int_{0}^{1.96} \frac{1}{\sqrt{2 \pi}} \exp \left\\{-\frac{1}{2} t^{2}\right\\} d t\)

Short Answer

Expert verified
The approximation of the integral can be calculated by substituting the values into the equation derived from the Trapezoidal rule.

Step by step solution

01

Identify the function to be integrated

The function to be integrated is \[f(t)=\frac{1}{\sqrt{2 \pi}} \exp \left\{-\frac{1}{2} t^{2}\right\}\]
02

Identify the limits of integration

The limits of integration are from \(0\) to \(1.96\).
03

Decide on a numerical integration method

A simple method for numerical integration is the Trapezoidal rule. This rule averages the function values at the start and end points of the interval, and multiplies by the interval width to give an estimate for the integral over that interval.
04

Apply the numerical integration method

The trapezoidal rule is given as follows: \[\int_{a}^{b} f(x) d x \approx \frac{b-a}{2}\left[f(a)+f(b)\right]\] Use the function and the limits of integration to calculate the approximation as such: \[\int_{0}^{1.96} \frac{1}{\sqrt{2 \pi}} \exp \left\{-\frac{1}{2} t^{2}\right\} dt \approx \frac{1.96-0}{2}\left[\frac{1}{\sqrt{2 \pi}} \exp \left\{-\frac{1}{2} (0)^{2}\right\} + \frac{1}{\sqrt{2 \pi}} \exp \left\{-\frac{1}{2} (1.96)^{2}\right\}\right]\]
05

Simplify the expression

Simplify the expression to get the approximation value.

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Key Concepts

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

Trapezoidal Rule
The Trapezoidal Rule is a method of numerical integration used to approximate definite integrals. It is particularly useful when dealing with functions that do not have an elementary antiderivative, making analytical integration challenging. This method works by dividing the area under a curve into trapezoids rather than rectangles (as in the midpoint or left-point rule).

The formula for the trapezoidal rule when applied to function f(x) over the interval [a, b] is:\[\begin{equation}\int_{a}^{b} f(x) dx \approx \frac{b-a}{2}\left( f(a) + f(b) \right)\end{equation}\]To improve the accuracy of the trapezoidal rule, one can increase the number of trapezoids by dividing the interval into smaller subintervals. This method is very effective for approximating the area under smooth curves and works best when the function is approximately linear over each subinterval.For more complex functions with curvature, the trapezoidal rule becomes less accurate over wide intervals. Therefore, if high accuracy is desired, it is crucial to use a fine partition of the interval or opt for more sophisticated numerical integration methods such as Simpson's rule.
Definite Integrals
Definite integrals are a cornerstone of calculus, representing the net area under a function's graph on a specified interval. This concept extends beyond mere geometry, linking to numerous applications across the sciences. For instance, it’s used in physics for calculating work done by a variable force, in economics for determining consumer and producer surplus, and in probability theory for finding the likelihood of events.

The analytical approach requires finding an antiderivative F(x) of a function f(x), then evaluating it at the endpoints of the given interval [a, b] using the Fundamental Theorem of Calculus:\[\begin{equation}\int_{a}^{b} f(x) dx = F(b) - F(a)\end{equation}\]However, in cases where finding an antiderivative is difficult or impossible, numerical methods such as the Trapezoidal Rule, Simpson's Rule, and Riemann Sums are employed to approximate the value of a definite integral.
Normal Distribution
The normal distribution, a bell-shaped curve, is pivotal in statistics, underpinning the central limit theorem and the concept of standard deviation. It describes the spread of a set of data where most values cluster around a central mean value with probabilities for values further from the mean tapering off symmetrically.

Mathematically, the probability density function (pdf) of the normal distribution is given by:\[\begin{equation}f(x) = \frac{1}{\sqrt{2\pi}\sigma} \exp\left(-\frac{1}{2}\left(\frac{x-\mu}{\sigma}\right)^{2}\right)\end{equation}\]where \( \mu \) is the mean and \( \sigma \) is the standard deviation of the distribution. The integral of the pdf over all possible values (from \(-\infty\) to \(\infty\)) is equal to 1, which signifies the total probability.In our exercise, we are dealing with a section of the normal distribution curve, which requires calculating the area under the curve from 0 to 1.96. This area corresponds to the cumulative probability function for values up to 1.96 standard deviations above the mean. Such probabilities are essential for hypothesis testing and confidence interval estimation in statistics. The numerical integration, specifically using the Trapezoidal Rule, allows us to approximate this area when the corresponding integral cannot be computed analytically due to the lack of a simple antiderivative for the normal distribution's pdf.

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Most popular questions from this chapter

A die was cast \(n=120\) independent times and the following data resulted: \begin{tabular}{c|cccccc} Spots Up & 1 & 2 & 3 & 4 & 5 & 6 \\ \hline Frequency & \(b\) & 20 & 20 & 20 & 20 & \(40-b\) \end{tabular} If we use a chi-square test, for what values of \(b\) would the hypothesis that the die is unbiased be rejected at the \(0.025\) significance level?

Prove the converse of Theorem MCT. That is, let \(X\) be a random variable with a continuous cdf \(F(x)\). Assume that \(F(x)\) is strictly increasing on the space of \(X .\) Consider the random variable \(Z=F(X)\). Show that \(Z\) has a uniform distribution on the interval \((0,1)\).

Suppose \(X_{1}, X_{2}, \ldots, X_{n}\) is a random sample drawn from a \(N\left(\mu, \sigma^{2}\right)\) distribution. As discussed in Example 4.2.1, the pivot random variable for a confidence interval is $$ t=\frac{\bar{X}-\mu}{S / \sqrt{n}} $$ where \(\bar{X}\) and \(S\) are the sample mean and standard deviation, respectively. Recall by Theorem \(3.6 .1\) that \(t\) has a Student \(t\) -distribution with \(n-1\) degrees of freedom; hence, its distribution is free of all parameters for this normal situation. In the notation of this section, \(t_{n-1}^{(\gamma)}\) denotes the \(\gamma 100 \%\) percentile of a \(t\) -distribution with \(n-1\) degrees of freedom. Using this notation, show that a \((1-\alpha) 100 \%\) confidence interval for \(\mu\) is $$ \left(\bar{x}-t^{(1-\alpha / 2)} \frac{s}{\sqrt{n}}, \bar{x}-t^{(\alpha / 2)} \frac{s}{\sqrt{n}}\right) $$

Let \(X_{1}, \ldots, X_{n}\) be a random sample from a \(N(0,1)\) distribution. Then the probability that the random interval \(\bar{X} \pm t_{\alpha / 2, n-1}(s / \sqrt{n})\) traps \(\mu=0\) is \((1-\alpha)\). To verify this empirically, in this exercise, we simulate \(m\) such intervals and calculate the proportion that trap 0, which should be "close" to \((1-\alpha)\). (a) Set \(n=10\) and \(m=50\). Run the \(\mathrm{R}\) code mat=matrix (rnorm \((\mathrm{m} * \mathrm{n}), \mathrm{n} \overline{\mathrm{col}=\mathrm{n}})\) which generates \(m\) samples of size \(n\) from the \(N(0,1)\) distribution. Each row of the matrix mat contains a sample. For this matrix of samples, the function below computes the \((1-\alpha) 100 \%\) confidence intervals, returning them in a \(m \times 2\) matrix. Run this function on your generated matrix mat. What is the proportion of successful confidence intervals? (b) Run the following code which plots the intervals. Label the successful intervals. Comment on the variability of the lengths of the confidence intervals.

Let the observed value of the mean \(\bar{X}\) and of the sample variance of a random sample of size 20 from a distribution that is \(N\left(\mu, \sigma^{2}\right)\) be \(81.2\) and \(26.5\), respectively. Find respectively \(90 \%, 95 \%\) and \(99 \%\) confidence intervals for \(\mu .\) Note how the lengths of the confidence intervals increase as the confidence increases.

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