/*! 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 26 Compute \(P\left(Y_{3}<\xi_{0... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Compute \(P\left(Y_{3}<\xi_{0.5}

Short Answer

Expert verified
The probability \(P\left(Y_{3}<\xi_{0.5}<Y_{7}\right) = \frac{5}{9}\

Step by step solution

01

Assumption Validation

First, it is important to validate the condition provided that \(Y_{1}<\cdots<Y_{9}\) are the order statistics of a random sample of size 9 from a continuous distribution. This means they are ranked in increasing order from smallest to largest.
02

Understanding the Median

The median of a distribution lies at the point where half the observations are greater and half are smaller. It is often denoted with \(\xi_{0.5}\). With nine observations in numerical order, the median value falls at \(Y_5\). This means, \(Y_{3}< \xi_{0.5}< Y_{7}\) is the range within which the median value lies.
03

Computing the Probability

Since the distribution is continuous, each order statistic is equally likely to be the median. Therefore, there are 5 equally likely places for the median to be: \(Y_3\), \(Y_4\), \(Y_5\), \(Y_6\), or \(Y_7\). Thus, the probability requested is simply the sum of the probabilities for each of these cases. Since these are equally likely, each event has a probability of \(1/{9}\), for the 9 positions the median value can take in the ordered sequence. So, the total probability is \(P\left(Y_{3}<\xi_{0.5}<Y_{7}\right) = 5 \times \frac{1}{9}\)

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 91Ó°ÊÓ!

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

Let \(z^{*}\) be drawn at random from the discrete distribution which has mass \(n^{-1}\) at each point \(z_{i}=x_{i}-\bar{x}+\mu_{0}\), where \(\left(x_{1}, x_{2}, \ldots, x_{n}\right)\) is the realization of a random sample. Determine \(E\left(z^{*}\right)\) and \(V\left(z^{*}\right)\).

Twenty motors were put on test under a high-temperature setting. The lifetimes in hours of the motors under these conditions are given below. Suppose we assume that the lifetime of a motor under these conditions, \(X\), has a \(\Gamma(1, \theta)\) distribution. \(\begin{array}{cccccccccc}1 & 4 & 5 & 21 & 22 & 28 & 40 & 42 & 51 & 53 \\ 58 & 67 & 95 & 124 & 124 & 160 & 202 & 260 & 303 & 363\end{array}\) (a) Obtain a frequency distribution and a histogram or a stem-leaf plot of the data. Use the intervals \([0,50),[50,100), \ldots\) Based on this plot, do you think that the \(\Gamma(1, \theta)\) model is credible? (b) Obtain the maximum likelihood estimate of \(\theta\) and locate it on your plot. (c) Obtain the sample median of the data, which is an estimate of the median lifetime of a motor. What parameter is it estimating (i.e., determine the median of \(X\) )? (d) Based on the mle, what is another estimate of the median of \(X\) ?

In the baseball data set discussed in the last exercise, it was found that out of the 59 baseball players, 15 were left-handed. Is this odd, since the proportion of left-handed males in America is about \(11 \% ?\) Answer by using \((4.2 .7)\) to construct a \(95 \%\) approximate confidence interval for \(p\), the proportion of left-handed baseball players.

The data set on Scottish schoolchildren discussed in Example 4.1.5 included the eye colors of the children also. The frequencies of their eye colors are \(\begin{array}{lccl}\text { Blue } & \text { Light } & \text { Medium } & \text { Dark } \\ 2978 & 6697 & 7511 & 5175\end{array}\)

Assume that the weight of cereal in a "10-ounce box" is \(N\left(\mu, \sigma^{2}\right)\). To test \(H_{0}: \mu=10.1\) against \(H_{1}: \mu>10.1\), we take a random sample of size \(n=16\) and observe that \(\bar{x}=10.4\) and \(s=0.4\). (a) Do we accept or reject \(H_{0}\) at the \(5 \%\) significance level? (b) What is the approximate \(p\) -value of this test?

See all solutions

Recommended explanations on Math Textbooks

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.