/*! 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 1 Area under the curve, Part I. Wh... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 4: Problem 1

Area under the curve, Part I. What percent of a standard normal distribution \(N(\mu=0, \sigma=1)\) is found in each region? Be sure to draw a graph. (a) \(Z<-1.35\) (b) \(Z>1.48\) (c) \(-0.42\)

Short Answer

Expert verified
(a) 8.85%; (b) 6.94%; (c) 58.86%; (d) 4.56%.

Step by step solution

01

Understanding the Standard Normal Distribution

The standard normal distribution is a bell-shaped curve with a mean (渭) of 0 and a standard deviation (蟽) of 1. When working with standard normal distributions, each point on the curve is denoted by a Z-score, which measures the number of standard deviations a data point is from the mean.
02

Using the Z-table for Calculation

A Z-table provides the percentage (or probability) of scores below a given Z-score in a standard normal distribution. To find probabilities for specific Z-scores, you will reference a Z-table or use a calculator with statistical functions.
03

Solving for (a) Probability of Z < -1.35

Using the Z-table, find the proportion of data less than -1.35. According to the Z-table, P(Z < -1.35) 鈮� 0.0885, which means approximately 8.85% of the data fall below a Z-score of -1.35.
04

Solving for (b) Probability of Z > 1.48

Find the proportion of data greater than 1.48. First, locate P(Z < 1.48) using the Z-table, which is approximately 0.9306. To find P(Z > 1.48), subtract this value from 1: P(Z > 1.48) = 1 - 0.9306 鈮� 0.0694, or about 6.94%.
05

Solving for (c) Probability of -0.4 < Z < 1.5

To find this probability, calculate P(Z < 1.5) and P(Z < -0.4) from the Z-table. P(Z < 1.5) 鈮� 0.9332 and P(Z < -0.4) 鈮� 0.3446. The difference gives the probability: P(-0.4 < Z < 1.5) = 0.9332 - 0.3446 鈮� 0.5886, or about 58.86%.
06

Solving for (d) Probability of |Z| > 2

This represents the probability of Z < -2 or Z > 2. Using the Z-table, P(Z < -2) 鈮� 0.0228 and due to symmetry, P(Z > 2) 鈮� 0.0228 as well. Thus, P(|Z| > 2) 鈮� 0.0228 + 0.0228 = 0.0456, or about 4.56%.

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 calculation
A Z-score is a statistical measurement that describes a value's position relative to the mean of a group of values. It tells you how many standard deviations a particular score or data point is from the mean of the dataset. Z-scores are an essential part of standard normal distributions, making them crucial for statistical analysis.

Here's how Z-score calculation works:
  • The formula for calculating a Z-score is: \( Z = \frac{X - \mu}{\sigma} \)where \( X \) is the value, \( \mu \) is the mean, and \( \sigma \) is the standard deviation.
  • A positive Z-score indicates the value is above the mean.
  • A negative Z-score means the value is below the mean.
By understanding Z-scores, one can easily infer how extreme or typical a value is within the context of a dataset. This helps both in visualizing and comprehending the data's distribution.
Z-table usage
A Z-table, also known as the standard normal table, is a vital tool used in statistics to find probabilities related to the standard normal distribution. This table helps in understanding how data points are distributed around the mean.

Here's how you use the Z-table:
  • Find your Z-score. For example, if you want to find the probability of Z being less than -1.35, locate the Z-score of -1.35 in the table.
  • The table will provide a cumulative probability for that Z-score, which signifies the area under the curve to the left of the score.
  • For Z-scores that aren't in the table, you might need to interpolate or use a calculator.
By subtracting or adding cumulative probabilities, you can find probabilities for a range of values. Mastery of the Z-table allows for swift and accurate probability calculations.
Probability calculation
Probability calculation is key in determining the likelihood of different outcomes in a standard normal distribution. It involves understanding both cumulative probabilities and differences in probabilities over ranges of Z-scores.

Probability calculations in practice:
  • To find the probability of a score being less than a Z-score (e.g., \( Z < -1.35 \)), look up the Z-score in the Z-table directly.
  • For scores greater than a Z-score (e.g., \( Z > 1.48 \)), calculate 1 minus the probability of \( Z < 1.48 \).
  • To find probabilities between two Z-scores (e.g., \( -0.4 < Z < 1.5 \)), subtract the probability of \( Z < -0.4 \) from \( Z < 1.5 \).
Understanding these calculations can significantly enhance data analysis skills and aid in interpreting various statistical results.
Symmetrical properties of normal distribution
The symmetrical properties of a normal distribution are crucial for understanding data behavior under the bell curve. A standard normal distribution is perfectly symmetrical about its mean, which is zero.

Here鈥檚 why symmetry matters:
  • It implies that for any positive Z-score, there is a corresponding negative Z-score equidistant from the mean.
  • This symmetry makes it easier to calculate probabilities, such as \( Z < -2 \) and \( Z > 2 \), because both areas are identical around zero.
  • The mirrored nature of the curve ensures that probabilities of being above or below a certain Z-score involve simple subtractions or additions.
These properties ensure that working with normal distributions is intuitive and predictable, automatically providing insights into the behavior of the data at different points of distribution.

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

Spray paint, Part 1. Suppose the area that can be painted using a single can of spray paint is slightly variable and follows a nearly normal distribution with a mean of 25 square feet and a standard deviation of 3 square feet. Suppose also that you buy three cans of spray paint. (a) How much area would you expect to cover with these three cans of spray paint? (b) What is the standard deviation of the area you expect to cover with these three eans of spray paint? (c) The area you wanted to cover is 80 square feet. What is the probability that you will be able to cover this entire area with these three cans of spray paint?

With and without replacement. In the following situations assume that half of the specified population is male and the other half is female. (a) Suppose you're sampling from a room with 10 people. What is the probability of sampling two females in a row when sampling with replacement? What is the probability when sampling without replacement? (b) Now suppose you're sampling from a stadium with 10,000 people. What is the probability of sampling two females in a row when sampling with replacement? What is the probability when sampling without replacement? (c) We often treat individuals who are sampled from a large population as independent. Using your findings from parts (a) and (b), explain whether or not this assumption is reasonable.

Betting on dinner, Part I. Suppose a restaurant is running a promotion where prices of menu items are random following some underlying distribution. If you're lucky, you can get a basket of fries for \(\$ 3\), or if you're not so lucky you might end up having to pay \(\$ 10\) for the same menu item. The price of basket. of fries is drawn from a normal distribution with mean \(\$ 6\) and standard deviation of \(\$ 2 .\) The price of a fountain drink is drawn from a normal distribution with mean \(\$ 3\) and standard deviation of \(\$ 1 .\) What is the probability that you pay more than \(\$ 10\) for a dinner consisting of a basket of fries and a fountain drink?

Distribution of \(\mu\). Suppose the true population proportion were \(p=0.95 .\) The figure below shows what the distribution of a sample proportion looks like when the sample size is \(n=20, n=100,\) and \(n=500\). (a) What does each point (observation) in each of the samples represent? (b) Describe the distribution of the sample proportion, \(\hat{p}\). How does the distribution of the sample proportion change as \(n\) becomes larger?

Bernoulli, the standard deviation. Use the probability rules from Section 3.5 to derive the standard deviation of a Bernoulli random variable, i.e. a random variable \(X\) that takes value 1 with probability \(\mathrm{p}\) and value 0 with probability \(1-p\). That is, compute the square root of the variance of a generic Bernoulli random variable.
See all solutions

Recommended explanations on Math Textbooks

Decision Maths

Read Explanation

Applied Mathematics

Read Explanation

Mechanics Maths

Read Explanation

Calculus

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Logic and Functions

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.