/*! 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 41 Determine the following standard... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Determine the following standard normal (z) curve areas: a. The area under the \(z\) curve to the left of 1.75 b. The area under the \(z\) curve to the left of -0.68 c. The area under the \(z\) curve to the right of 1.20 d. The area under the \(z\) curve to the right of -2.82 e. The area under the \(z\) curve between -2.22 and 0.53 f. The area under the \(z\) curve between -1 and 1 g. The area under the \(z\) curve between -4 and 4

Short Answer

Expert verified
The areas under the z curve for the given scenarios are: a. 0.9599 b. 0.2483 c. 0.1151 d. 0.9976 e. 0.6891 f. 0.6826 g. 1.0000 (approximately)

Step by step solution

01

Identify the given z-scores and the needed areas under the z curve

The given z-scores and the required areas under the curve are as follows: a. Left of 1.75 b. Left of -0.68 c. Right of 1.20 d. Right of -2.82 e. Between -2.22 and 0.53 f. Between -1 and 1 g. Between -4 and 4
02

Use standard normal table or calculator to find probabilities associated with z-scores

Using a standard normal table or a calculator with built-in z-table function, find the cumulative probabilities (areas to the left) up to the given z-scores. a. P(Z \(\le\) 1.75) = 0.9599 b. P(Z \(\le\) -0.68) = 0.2483 c. P(Z \(\le\) 1.20) = 0.8849 d. P(Z \(\le\) -2.82) = 0.0024 e1. P(Z \(\le\) -2.22) = 0.0132 e2. P(Z \(\le\) 0.53) = 0.7023 f1. P(Z \(\le\) -1) = 0.1587 f2. P(Z \(\le\) 1) = 0.8413 g1. P(Z \(\le\) -4) = 0.0000 (approximately) g2. P(Z \(\le\) 4) = 1.0000 (approximately)
03

Calculate the desired areas under the z curve

Use values from Step 2 to find the areas under the z curve for each scenario. a. Area left of 1.75 = 0.9599 b. Area left of -0.68 = 0.2483 c. Area right of 1.20 = 1 - 0.8849 = 0.1151 d. Area right of -2.82 = 1 - 0.0024 = 0.9976 e. Area between -2.22 and 0.53 = 0.7023 - 0.0132 = 0.6891 f. Area between -1 and 1 = 0.8413 - 0.1587 = 0.6826 g. Area between -4 and 4 = 1.0000 - 0.0000 = 1.0000 (approximately)
04

State the final results

The areas under the z curve for the given scenarios are: a. 0.9599 b. 0.2483 c. 0.1151 d. 0.9976 e. 0.6891 f. 0.6826 g. 1.0000 (approximately)

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!

Key Concepts

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

Z-Scores
Understanding the concept of 'z-scores' is fundamental when working with the standard normal curve. Z-scores, also known as standard scores, are a way to describe the position of a raw score within a distribution. A z-score signifies how many standard deviations away a point is from the mean of the distribution. For example, a z-score of 1.75 means the raw score is 1.75 standard deviations above the mean. Conversely, a z-score of -0.68 indicates the score is 0.68 standard deviations below the mean.

To find the corresponding probability for a given z-score, you usually need to look it up in a z-table or use a statistical software or calculator. This probability tells us the portion of the data that falls to the left of this z-score under the standard normal curve. Essentially, z-scores allow us to translate individual scores into a standardized form where they can be easily compared and probabilities can be assigned.
Cumulative Probabilities
Cumulative probabilities are closely tied with the concept of z-scores and play a critical role in statistics. They represent the probability for a variable to take on a value less than or equal to a specific point in a distribution. To put it simply, if you have a z-score, the cumulative probability tells you the percentage of data points that lie to the left of that z-score on a standard normal curve.

For example, if we have a z-score of 1.20, looking it up in the z-table or using a calculator would tell us the cumulative probability associated with that z-score, which is the area under the curve to the left of that z-score. If we need to find the area to the right, we subtract the cumulative probability from 1, as the total area under the normal distribution curve adds up to 1. Therefore, understanding and finding cumulative probabilities is crucial for interpreting z-scores and evaluating the relative standing of data points within a normal distribution.
Normal Distribution
The normal distribution is a bell-shaped curve that is symmetric about the mean and characterized by its mean (μ) and standard deviation (σ). It's a continuous probability distribution that describes many natural phenomena and is a cornerstone in the field of statistics. One of the exceptional aspects of the normal distribution is that it is fully described by the mean and standard deviation.

The standard normal distribution is a special case of the normal distribution with a mean of 0 and a standard deviation of 1. Any normal distribution can be transformed into the standard normal distribution using the z-score formula. The areas under the standard normal curve correspond to probabilities, and the total area under the curve is always equal to 1. We use this feature to find probabilities for ranges of values within the distribution, for example, the probability of a random variable falling between two values (e.g., between -1 and 1), which is a very common exercise in statistics education.

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 \(x\) denote the IQ of an individual selected at random from a certain population. The value of \(x\) must be a whole number. Suppose that the distribution of \(x\) can be approximated by a normal distribution with mean value 100 and standard deviation 15. Approximate the following probabilities. a. \(P(x=100)\) b. \(P(x \leq 110)\) c. \(P(x<110)\) (Hint: \(x<110\) is the same as \(x \leq 109 .\) ) d. \(P(75 \leq x \leq 125)\)

Airlines sometimes overbook flights. Suppose that for a plane with 100 seats, an airline takes 110 reservations. Define the random variable \(x\) as \(x=\) the number of people who actually show up for a sold-out flight on this plane From past experience, the probability distribution of \(x\) is given in the following table: $$ \begin{array}{|cc|} \hline \boldsymbol{x} & \boldsymbol{p}(\boldsymbol{x}) \\ \hline 95 & 0.05 \\ 96 & 0.10 \\ 97 & 0.12 \\ 98 & 0.14 \\ 99 & 0.24 \\ 100 & 0.17 \\ 101 & 0.06 \\ 102 & 0.04 \\ 103 & 0.03 \\ 104 & 0.02 \\ 105 & 0.01 \\ 106 & 0.005 \\ 107 & 0.005 \\ 108 & 0.005 \\ 109 & 0.0037 \\ 110 & 0.0013 \\ \hline \end{array} $$ a. What is the probability that the airline can accommodate everyone who shows up for the flight? b. What is the probability that not all passengers can be accommodated? c. If you are trying to get a seat on such a flight and you are number 1 on the standby list, what is the probability that you will be able to take the flight? What if you are number 3 ?

The paper referenced in Example 6.24 ("Estimating Waste Transfer Station Delays Using GPS," Waste Management [2008]: 1742-1750) describing processing times for garbage trucks also provided information on processing times at a second facility. At this second facility, the mean total processing time was 9.9 minutes and the standard deviation of the processing times was 6.2 minutes. Explain why a normal distribution with mean 9.9 and standard deviation 6.2 would not be an appropriate model for the probability distribution of the variable \(x=\) total processing time of a randomly selected truck entering this second facility.

Consider the variable \(x=\) time required for a college student to complete a standardized exam. Suppose that for the population of students at a particular university, the distribution of \(x\) is well approximated by a normal curve with mean 45 minutes and standard deviation 5 minutes. a. If 50 minutes is allowed for the exam, what proportion of students at this university would be unable to finish in the allotted time? b. How much time should be allowed for the exam if you wanted \(90 \%\) of the students taking the test to be able to finish in the allotted time? c. How much time is required for the fastest \(25 \%\) of all students to complete the exam?

A coin is flipped 25 times. Let \(x\) be the number of flips that result in heads (H). Consider the following rule for deciding whether or not the coin is fair: Judge the coin fair if \(8 \leq x \leq 17\). Judge the coin biased if either \(x \leq 7\) or \(x \geq 18\). a. What is the probability of judging the coin biased when it is actually fair? b. Suppose that a coin is not fair and that \(P(\mathrm{H})=0.9\) What is the probability that this coin would be judged fair? What is the probability of judging a coin fair if \(P(\mathrm{H})=0.1 ?\) c. What is the probability of judging a coin fair if \(P(\mathrm{H})=0.6 ?\) if \(P(\mathrm{H})=0.4 ?\) Why are these probabilities large compared to the probabilities in Part (b)? d. What happens to the "error probabilities" of Parts (a) and \((b)\) if the decision rule is changed so that the coin is judged fair if \(7 \leq x \leq 18\) and unfair otherwise? Is this a better rule than the one first proposed? Explain.

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.