/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 1 Let \(\bar{X}\) denote the mean ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 5: Problem 1

Let \(\bar{X}\) denote the mean of a random sample of size 100 from a distribution that is \(\chi^{2}(50)\). Compute an approximate value of \(P(49<\bar{X}<51)\).

Short Answer

Expert verified
The probability \(P(49<\bar{X}<51)\) is approximately 0.5222.

Step by step solution

01

Understanding the Chi-Square distribution

Recall that if a random variable X follows a chi-square distribution with k degrees of freedom, it's mean and variance are given by \(E(X)= k\) and \(Var(X)= 2k\) respectively. Here the mean will be 50 and variance will be 100.
02

Application of the Central Limit Theorem

According to the central limit theorem, if the sample size is sufficiently large, the sample mean follows approximately a normal distribution. The mean of this distribution is the same as the population mean (\( \mu \) = 50 for the chi-square distribution), and the standard deviation (SD) for this distribution is \( \sigma/ \sqrt{n}\). Here n=100 and \( \sigma = \sqrt {2n} \) giving us the SD of \( \sqrt {2}\). Now we need to standardize the values such that they follow a standard normal distribution (mean=0, SD=1). The normalized values will be \( Z = \frac{X - \mu}{\sigma/ \sqrt{n}} \).
03

Normalize the boundaries

To find the probability we should normalize the boundaries of the interval. The normalized boundary for 49 will be \(Z_1 = \frac{49-50}{\sqrt {2}} = -0.71\) and for 51 will be \(Z_2 = \frac{51-50}{\sqrt {2}} = 0.71\).
04

Find the probability

Now we need to compute the probability that \( P(-0.71 < Z < 0.71)\). The Z values fall within the range of normal standard distribution. We should use Z-table or any statistical tool to find this probability.
05

Final Computation

From the Z-table, we get \( P(Z < 0.71) = 0.7611\) and \( P(Z < -0.71) = 1 - 0.7611 = 0.2389\). We subtract the lower boundary value from the upper boundary value to obtain \( P(49<\bar{X}<51)\) = \(P(Z < 0.71) - P(Z < -0.71)\) = \(0.7611 - 0.2389 = 0.5222\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Suppose \(\mathbf{X}_{n}\) has a \(N_{p}\left(\boldsymbol{\mu}_{n}, \boldsymbol{\Sigma}_{n}\right)\) distribution. Show that $$\mathbf{X}_{n} \stackrel{D}{\rightarrow} N_{p}(\boldsymbol{\mu}, \mathbf{\Sigma}) \text { iff } \boldsymbol{\mu}_{n} \rightarrow \boldsymbol{\mu} \text { and } \mathbf{\Sigma}_{n} \rightarrow \mathbf{\Sigma}$$

Let \(X_{1}, X_{2}, \ldots, X_{n}\) be a random sample from a Poisson distribution with mean \(\mu .\) Thus, \(Y=\sum_{i=1}^{n} X_{i}\) has a Poisson distribution with mean \(n \mu .\) Moreover, \(\bar{X}=Y / n\) is approximately \(N(\mu, \mu / n)\) for large \(n .\) Show that \(u(Y / n)=\sqrt{Y / n}\) is a function of \(Y / n\) whose variance is essentially free of \(\mu\).

Let \(Y\) be \(b\left(400, \frac{1}{5}\right)\). Compute an approximate value of \(P(0.25

Let \(p=0.95\) be the probability that a man, in a certain age group, lives at least 5 years. (a) If we are to observe 60 such men and if we assume independence, find the probability that at least 56 of them live 5 or more years. (b) Find an approximation to the result of part (a) by using the Poisson distribution. Hint: Redefine \(p\) to be \(0.05\) and \(1-p=0.95\).

Let \(X_{1}, X_{2}, \ldots, X_{n}\) be a random sample of size \(n\) from a distribution that is \(N\left(\mu, \sigma^{2}\right)\), where \(\sigma^{2}>0 .\) Show that the sum \(Z_{n}=\sum_{1}^{n} X_{i}\) does not have a limiting distribution.
See all solutions

Recommended explanations on Math Textbooks

Applied Mathematics

Read Explanation

Geometry

Read Explanation

Probability and Statistics

Read Explanation

Decision Maths

Read Explanation

Calculus

Read Explanation

Logic and Functions

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.