/*! 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 19 Sketch the areas under the stand... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 6: Problem 19

Sketch the areas under the standard normal curve over the indicated intervals, and find the specified areas. To the right of \(z=1.52\)

Short Answer

Expert verified
The area to the right of \( z = 1.52 \) is approximately 0.0643.

Step by step solution

01

Understand the Standard Normal Curve

The standard normal curve, also known as the bell curve, is a graph of the normal distribution with a mean of 0 and a standard deviation of 1. Any point on this curve can be described using a z-score, which measures how many standard deviations an element is from the mean.
02

Identify the Relevant Area

The problem asks for the area to the right of \( z = 1.52 \). On the standard normal distribution, this is the area under the curve that extends indefinitely to the right starting at \( z = 1.52 \).
03

Use the Standard Normal Distribution Table

To find the area to the right of \( z = 1.52 \), first find the area to the left of \( z = 1.52 \) using a standard normal distribution table. The table gives the cumulative probability from the far left of the curve to the z-score.
04

Calculate the Right-Sided Area

Once you have the cumulative area to the left from the standard normal table (let's denote it as \( P(Z < 1.52) \)), subtract this value from 1 to find the area to the right. \[ P(Z > 1.52) = 1 - P(Z < 1.52) \]
05

Determine the Area Using Values

Referencing the standard normal distribution table, the cumulative probability for \( Z < 1.52 \) is approximately 0.9357. Therefore, the area to the right is:\[ P(Z > 1.52) = 1 - 0.9357 = 0.0643 \]
06

Sketch the Area

On a graph of the standard normal distribution, draw a vertical line at \( z = 1.52 \). Then, shade the region to the right of this line to represent the area we calculated, which is approximately 0.0643.

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.

Understanding Z-scores
Z-scores are a number that tells us how far away a point, or data, is from the mean in a standard normal distribution. The mean here is always zero, and for any normal curve, the standard deviation is one.
Simply put, a z-score tells you how unusual or typical a data point is within the context of the rest of the data.
  • If a data point, represented as a z-score, is zero, it means it is exactly at the mean.
  • A negative z-score indicates a data point below the mean.
  • A positive z-score suggests a data point above the mean.
In the exercise, the z-score of 1.52 tells us that this point is 1.52 standard deviations greater than the mean.
The Shape of the Normal Curve
The normal curve, often called the bell curve due to its shape, is a graph that represents the normal distribution. It is a symmetrical curve where the center, or highest point, corresponds to the mean of the data.
  • For a standard normal distribution, the mean is zero.
  • The width of the curve is determined by the standard deviation, which is one in the case of the standard normal distribution.
  • The tails of the curve extend indefinitely to the left and the right, approaching but never quite reaching the horizontal axis.
In our exercise, the area under this curve to the right of a z-score represents the probability of obtaining a value greater than that z-score.
Exploring Cumulative Probability
Cumulative probability is the likelihood that a variable is less than or equal to a certain value in a distribution.
  • This probability is represented by the area under the normal curve up to a given point.
  • In calculations, cumulative probability helps find out the likelihood of a random variable falling within a certain range.
For this exercise, we were interested in the area to the right of a z-score (\(z = 1.52\)), but to find this, we first looked at the cumulative probability to the left, which was approximately 0.9357.
Using Probability Tables
Probability tables, especially the standard normal distribution table, are essential tools in statistics. They provide the cumulative probabilities for different z-scores in a normal distribution.
  • These tables help us find areas under the curve quickly without doing complex math.
  • They list z-scores and their corresponding cumulative probabilities from the mean to that z-score.
In solving our problem, we used the table to find the cumulative probability for \(Z < 1.52\), which was approximately 0.9357. This value helped us determine that the probability of a random variable being greater than \(z = 1.52\) was 0.0643 by subtracting from 1.

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

Thickness measurements of ancient prehistoric Native American pot shards discovered in a Hopi village are approximately normally distributed, with a mean of \(5.1\) millimeters \((\mathrm{mm})\) and a standard deviation of \(0.9 \mathrm{~mm}\) (Source: Homol'ovi II: Archaeology of an Ancestral Hopi Village, Arizona, edited by E. C. Adams and K. A. Hays, University of Arizona Press). For a randomly found shard, what is the probability that the thickness is (a) less than \(3.0 \mathrm{~mm}\) ? (b) more than \(7.0 \mathrm{~mm}\) ? (c) between \(3.0 \mathrm{~mm}\) and \(7.0 \mathrm{~mm}\) ?

Coal is carried from a mine in West Virginia to a power plant in New York in hopper cars on a long train. The automatic hopper car loader is set to put 75 tons of coal into each car. The actual weights of coal loaded into each car are normally distributed, with mean \(\mu=75\) tons and standard deviation \(\sigma=0.8\) ton. (a) What is the probability that one car chosen at random will have less than \(74.5\) tons of coal? (b) What is the probability that 20 cars chosen at random will have a mean load weight \(\bar{x}\) of less than \(74.5\) tons of coal? (c) Suppose the weight of coal in one car was less than \(74.5\) tons. Would that fact make you suspect that the loader had slipped out of adjustment? Suppose the weight of coal in 20 cars selected at random had an average \(\bar{x}\) of less than \(74.5\) tons. Would that fact make you suspect that the loader had slipped out of adjustment? Why?

Binomial probability distributions depend on the number of trials \(n\) of a binomial experiment and the probability of success \(p\) on each trial. Under what conditions is it appropriate to use a normal approximation to the binomial?

The Denver Post stated that \(80 \%\) of all new products introduced in grocery stores fail (are taken off the market) within 2 years. If a grocery store chain introduces 66 new products, what is the probability that within 2 years (a) 47 or more fail? (b) 58 or fewer fail? (c) 15 or more succeed? (d) fewer than 10 succeed?

Assume that \(x\) has a normal distribution with the specified mean and standard deviation. Find the indicated probabilities. $$ P(50 \leq x \leq 70) ; \mu=40 ; \sigma=15 $$
See all solutions

Recommended explanations on Math Textbooks

Applied Mathematics

Read Explanation

Logic and Functions

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Probability and Statistics

Read Explanation

Decision Maths

Read Explanation

Geometry

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.