Critical Values. In Exercises 41–44, find the indicated critical value. Round results to two decimal places.

z0.02

Short Answer

Expert verified

The critical value for z0.02is 2.05.

Step by step solution

01

Given information

The critical value isz0.02.

02

Define the critical value 

The critical value zαat a particular value of denotes that is the area to the right of zαon the standard normal curve.

Thus,

Pz>zα=α

In this case of α=0.02, PZ>z0.02=0.02.

The notation of critical valuez0.02 is used to represent the z-score with an area of 0.02 to its right.

03

Find the critical value z0.02

Due to the one-to-one correspondence between the area and probability under the standard normal curve,

PZ>z0.02=0.021-PZ<z0.02=0.02PZ<z0.02=0.98

In the standard normal table, the value closest to 0.98 is 0.9798, and the corresponding row value 2.0 and the column value 0.05 correspond to the z-score of 2.05, which is the critical value z0.02.

The value of z0.02=2.05has an area of 0.02 to its right. So, z0.02=2.05corresponds to a cumulative left area of 0.98.

Thus, the critical value is z0.02=2.05.

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 Vaia!

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

Continuous Uniform Distribution. In Exercises 5–8, refer to the continuous uniform distribution depicted in Figure 6-2 and described in Example 1. Assume that a passenger is randomly selected, and find the probability that the waiting time is within the given range.

Between 2 minutes and 3 minutes

Standard normal distribution. In Exercise 17-36, assume that a randomly selected subject is given a bone density test. Those test scores are normally distributed with a mean of 0 and a standard deviation of 1. In each case, Draw a graph, then find the probability of the given bone density test score. If using technology instead of Table A-2, round answers to four decimal places.

Less than 2.56

SAT and ACT Tests Because they enable efficient procedures for evaluating answers, multiple choice questions are commonly used on standardized tests, such as the SAT or ACT.

Such questions typically have five choices, one of which is correct. Assume that you must make random guesses for two such questions. Assume that both questions have correct answers of “a.”

a. After listing the 25 different possible samples, find the proportion of correct answers in each sample, then construct a table that describes the sampling distribution of the sample proportions of correct responses.

b. Find the mean of the sampling distribution of the sample proportion.

c. Is the mean of the sampling distribution [from part (b)] equal to the population proportion of correct responses? Does the mean of the sampling distribution of proportions always equal the population proportion?

In Exercises 5–8, find the area of the shaded region. The graphs depict IQ scores of adults, and those scores are normally distributed with a mean of 100 and a standard deviation of 15 (as on the Wechsler IQ test).

Standard normal distribution, assume that a randomly selected subject is given a bone density test. Those test scores are normally distributed with a mean of 0 and a standard deviation of 1. In each case, Draw a graph, then find the probability of the given bone density test score. If using technology instead of Table A-2, round answers to four decimal places.

Greater than -3.75

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.

Sign-up for free