/*! 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 17 Find the following probabilities... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 6: Problem 17

Find the following probabilities for the standard normal distribution. a. \(P(z<-2.34)\) b. \(P(.67 \leq z \leq 2.59)\) c. \(P(-2.07 \leq z \leq-.93)\) d. \(P(z<1.78)\)

Short Answer

Expert verified
The answers will vary as they are dependent on the z-table or calculator function being used to find the probabilities for the given z-values. Refer to a standard normal distribution table (z-table) or use a calculator with z-table function to get the exact values.

Step by step solution

01

Understanding the z-value

Z-values, also known as standard score, indicate by how many standard deviations an element is from the mean. In a standard normal distribution, mean is 0 and standard deviation is 1. We are given the z-values, we just have to find the corresponding probabilities using the z-table (also known as standard normal distribution table). A z-table provides the probability that a standard normal random variable Z is less than z.
02

Find probability for \(P(z

By referring to the z-table or using a calculator, find the probability for z = -2.34. This represents the probability that the z-score is less than -2.34.
03

Find probability for \(P(.67 \leq z \leq 2.59)\)

We find the probability of two z-scores, z=.67 and z=2.59. This represents the probability that the z-score is between .67 and 2.59. We start by finding the probabilities for each z-score using the z-table, then subtract the two to get the probability between those scores.
04

Find probability for \(P(-2.07 \leq z \leq-.93)\)

The problem is asking for the probability that the z-score is between -2.07 and -.93. We find the probabilities for each z-score using the z-table, then subtract the two to get the probability between those scores.
05

Find probability for \(P(z

We need to find the probability that the z-score is less than 1.78. Using the z-table or calculator, we find this probability.

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

According to the records of an electric company serving the Boston area, the mean electricity consumption during winter for all households is 1650 kilowatt-hours per month. Assume that the monthly electric consumptions during winter by all households in this area have a normal distribution with a mean of 1650 kilowatt-hours and a standard deviation of 320 kilowatt-hours. The company sent a notice to Bill Johnson informing him that about \(90 \%\) of the households use less electricity per month than he does. What is Bill Johnson's monthly electricity consumption?

Find the value of \(z\) so that the area under the standard normal curve a. from 0 to \(z\) is \(.4772\) and \(z\) is positive b. between 0 and \(z\) is (approximately) \(.4785\) and \(z\) is negative c. in the left tail is (approximately). 3565 d. in the right tail is (approximately). 1530

Let \(x\) be a continuous random variable that is normally distributed with a mean of 65 and a standard deviation of \(15 .\) Find the probability that \(x\) assumes a value a. less \(\operatorname{than} 45\) b. greater than 79 c. greater than 54 d. less than 70

Hurbert Corporation makes font cartridges for laser printers that it sells to Alpha Electronics Inc. The cartridges are shipped to Alpha Electronics in large volumes. The quality control department at Alpha Electronics randomly selects 100 cartridges from each shipment and inspects them for being good or defective. If this sample contains 7 or more defective cartridges, the entire shipment is rejected. Hurbert Corporation promises that of all the cartridges, only \(5 \%\) are defective. a. Find the probability that a given shipment of cartridges received by Alpha Electronics will be accepted. b. Find the probability that a given shipment of cartridges received by Alpha Electronics will not be accepted.

An office supply company conducted a survey before marketing a new paper shredder designed for home use. In the survey, \(80 \%\) of the people who tried the shredder were satisfied with it. Because of this high satisfaction rate, the company decided to market the new shredder. Assume that \(80 \%\) of all people are satisfied with this shredder. During a certain month, 100 customers bought this shredder. Find the probability that of these 100 customers, the number who are satisfied is a. exactly 75 b. 73 or fewer c. 74 to 85
See all solutions

Recommended explanations on Math Textbooks

Calculus

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Geometry

Read Explanation

Statistics

Read Explanation

Discrete Mathematics

Read Explanation

Pure Maths

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.