/*! 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 57 a. Find a \(z_{0}\) such that \(... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 6: Problem 57

a. Find a \(z_{0}\) such that \(P\left(z>z_{0}\right)=.9750 .\) b. Find a \(z_{0}\) such that \(P\left(z>z_{0}\right)=.3594\).

Short Answer

Expert verified
Answer: The values of \(z_0\) are (a) \(z_0=-1.96\) and (b) \(z_0=0.35\).

Step by step solution

01

Find the probability \(P(z\leq z_0)\) for Part a

Given \(P(z>z_0)=0.9750\), we can determine the complementary probability for \(P(z\leq z_0)\) by using the fact that the total probability under the normal distribution curve is equal to 1. Therefore, we have: \(P(z\leq z_0)=1-P(z>z_0)=1-0.9750=0.0250\).
02

Look up the \(z\)-score for Part a

Now, we can use the standard normal \(z\)-score table to find the \(z_0\) corresponding to \(P(z\leq z_0)=0.0250\). Looking up this value in the table gives us a \(z\)-score of -1.96.
03

Result for Part a

The desired value of \(z_0\) for Part a is \(z_0=-1.96\) such that \(P(z>z_0)=0.9750\). **Part b:**
04

Find the probability \(P(z\leq z_0)\) for Part b

Given \(P(z>z_0)=0.3594\), we can determine the complementary probability for \(P(z\leq z_0)\) by using the fact that the total probability under the normal distribution curve is equal to 1. Therefore, we have: \(P(z\leq z_0)=1-P(z>z_0)=1-0.3594=0.6406\).
05

Look up the \(z\)-score for Part b

Now, we can use the standard normal \(z\)-score table to find the \(z_0\) corresponding to \(P(z\leq z_0)=0.6406\). Looking up this value in the table gives us a \(z\)-score of 0.35.
06

Result for Part b

The desired value of \(z_0\) for Part b is \(z_0=0.35\) such that \(P(z>z_0)=0.3594\).

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!

Key Concepts

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

Z-score
The Z-score is a way to describe a position within a standard normal distribution. Think of it as a score that tells you how far away a particular point is from the average (or mean) in terms of standard deviations. In simpler terms, it's a way to compare any data point with the "average."
  • If a Z-score is 0, that means the data point is exactly at the average.
  • A positive Z-score means it's above the average.
  • A negative Z-score indicates it's below the average.
To find a Z-score from a normal distribution, you typically use a Z-table, which tells you the probability or area under the curve to the left of that Z-score. This is crucial for determining where a specific value lies in the distribution and how probable certain outcomes are.
Probability
Probability is the measure of how likely an event is to occur. It's a number between 0 and 1, where 0 implies impossibility, and 1 indicates certainty. In the context of a normal distribution, probability helps us understand how likely it is for a random variable to fall within a particular range.
  • The probability of something is often expressed as a percentage or a decimal. For example, a probability of 0.3 can also be seen as a 30% chance.
  • In the exercise, probabilities were given to find corresponding Z-scores using Z-tables: "What is the chance that Z is greater than a certain value?".
Understanding probability is essential as it forms the basis for interpreting results from statistical analyses and insights from data.
Standard Normal Distribution
The standard normal distribution is a special type of normal distribution that has a mean of 0 and a standard deviation of 1. In this symmetrical bell curve, most data points are close to the mean.
  • The total area under the standard normal distribution curve equals 1. This area is linked to probability.
  • The Z-score tells us how many standard deviations a given point is from the mean of the distribution, as it aligns with the standard normal distribution.
When you find Z-scores in a standard normal distribution, you're actually pinpointing where the event falls within this standardized curve. You can compare scores from different data sets efficiently by converting them to Z-scores, since they all use the same standard normal distribution framework.
Complementary Probability
Complementary probability is a concept that refers to the probability of the "other" outcome of a mutually exclusive event. Simply put, if we know the probability of one event occurring, we can find the probability of it not occurring by subtracting from 1.
  • For example, if there is a 20% chance of rain, then there is an 80% chance it won't rain, since probabilities in this context must sum up to 1.
  • In the exercise, complementary probability was used to find the probability of Z being less than a certain value by subtracting the given probability from 1.
Understanding complementary probability is essential in scenario analysis because it allows us to quickly evaluate the likelihood of alternative outcomes and is a handy tool in solving problems related to typical Z-score tasks.

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

A normal random variable \(x\) has mean \(\mu=5\) and standard deviation \(\sigma=2\). Find the probabilities associated with the following intervals: a. \(1.27.5\) c. \(x \leq 0\)

Tall or Short? Do Americans tend to vote for the taller of the two major candidates in a presidential election? In 49 of our presidential elections for which the heights of all the major-party candidates are known, 26 of the winners were taller than their opponents. Assume that Americans are not biased by a candidate's height and that the winner is just as likely to be taller or shorter than his opponent. a. Is the observed number of taller winners in the U.S. presidential election unusual? Find the approximate probability of finding 26 or more of the 49 pairs in which the taller candidate wins. b. Based on your answer to part a, can you conclude that Americans might consider a candidate's height when casting their ballot?

Find the following probabilities for the standard normal random variable \(z\) : a. \(P(-1.96 \leq z \leq 1.96)\) b. \(P(z>1.96)\) c. \(P(z<-1.96)\)

Suppose that you must establish regulations concerning the maximum number of people who can occupy an elevator. A study indicates that if eight people occupy the elevator, the probability distribution of the total weight of the eight people is approximately normally distributed with a mean equal to 1200 pounds and a standard deviation of 99 pounds. What is the probability that the total weight of eight people exceeds 1300 pounds? 1500 pounds?

A publisher has discovered that the number of words contained in new manuscripts is normally distributed, with a mean equal to 20,000 words in excess of that specified in the author's contract and a standard deviation of 10,000 words. If the publisher wants to be almost certain (say, with a probability of .95 ) that the manuscript will have less than 100,000 words, what number of words should the publisher specify in the contract?
See all solutions

Recommended explanations on Math Textbooks

Theoretical and Mathematical Physics

Read Explanation

Pure Maths

Read Explanation

Discrete Mathematics

Read Explanation

Logic and Functions

Read Explanation

Decision Maths

Read Explanation

Probability and Statistics

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.