/*! 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 23 Find the following probabilities... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 6: Problem 23

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 probabilities for the given z-scores are: \n(a) \(P(z<-2.34) = 0.0096\)\n (b) \(P(.67 \leq z \leq 2.59) = 0.7437\)\n (c) \(P(-2.07 \leq z \leq-.93) = 0.1561\)\n (d) \(P(z<1.78) = 0.9625\)

Step by step solution

01

Interpretation of z-scores

A standard normal distribution is a normal distribution with a mean of 0 and standard deviation of 1. A z-score measures how many standard deviations an element is from the mean. In the standard normal distribution, the z-score is equal to the data point under consideration because the mean is 0.
02

Use z-tables

To calculate \(P(z<-2.34)\), from the standard normal tables, an area of 0.9904 corresponds to a z-score of 2.34. Hence, since this is a two-tailed distribution, we subtract this from 1 to find the probability of z < -2.34. So, \(P(z<-2.34) = 1 - 0.9904 = 0.0096\). Similarly for the other data points, we find their respective probabilities using standard normal distribution tables and appropriate calculations.
03

Find Probabilities

For \(P(.67 \leq z \leq 2.59)\), we find the area under curve from z=.67 to z=2.59. Using z-table, we find these areas to be 0.2514 and 0.9951 respectively. So, \(P(.67 \leq z \leq 2.59) = 0.9951 - 0.2514 = 0.7437\). For \(P(-2.07 \leq z \leq-.93)\), we find the areas under curve from z=-2.07 to z=-.93 to be 0.0192 and 0.1753 respectively. So, \(P(-2.07 \leq z \leq-.93) = 0.1753 - 0.0192 = 0.1561\). Lastly, for \(P(z<1.78)\), from the standard normal tables, 0.9625 corresponds to z=1.78. Hence, \(P(z<1.78) = 0.9625\).

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 the Z-Score
When you're exploring a standard normal distribution, understanding the z-score is essential. Imagine the z-score as a translator between your data point and the standard normal world. It tells you how far away, and in what direction, a data point is from the mean, expressed in terms of standard deviations.
  • A positive z-score indicates the data point is above the mean.
  • A negative z-score suggests it's below the mean.
  • Z-score of zero means the data point is exactly at the mean.

Z-scores are crucial since they transform data into a standard format, allowing you to use the z-table for probability calculations. This transformation makes it easier to compare different data sets or check specific probabilities in a model.
How to Calculate Probabilities in the Standard Normal Distribution
Probabilities in the realm of normal distribution help determine the likelihood that a given data point falls within a certain range. To calculate these probabilities, start by determining the relevant z-scores for your data points.

Suppose you have a range such as 0.67 to 2.59. To find the probability that your data falls within this range, you perform the following steps:
  • Find the area under the curve (probability) associated with each z-score in the range using a z-table.
  • Subtract the smaller area from the larger area within your range.

This subtraction gives you the probability for that range. Performing this task requires referencing standard normal distribution tables, often referred to as z-tables, which map z-scores to probabilities.
The Role of the Z-Table in Finding Probabilities
The z-table is your go-to tool for unlocking the probabilities related to z-scores in a standard normal distribution. It provides the cumulative probability associated with a z-score from the mean up to your value of interest.

Here's how you efficiently use a z-table:
  • Find your z-score value in the table; it is typically divided into rows and columns representing tenths and hundredths of z-scores respectively.
  • Look up the corresponding cumulative probability next to the z-score.

For example, when looking for the probability of a z-score less than 1.78, locate 1.78 in the table, and you'll see 0.9625 as the cumulative probability. This cumulative value represents the probability of a randomly selected score falling below your z-score value. This step-by-step lookup in the z-table helps calculate probabilities efficiently, making it a fundamental component of statistical analysis.

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

One of the cars sold by Walt's car dealership is a very popular subcompact car called Rhino. The final sale price of the basic model of this car varies from customer to customer depending on the negotiating skills and persistence of the customer. Assume that these sale prices of this car are normally distributed with a mean of \(\$ 19,800\) and a standard deviation of \(\$ 350\). a. Dolores paid \(\$ 19,445\) for her Rhino. What percentage of Walt's customers paid less than Dolores for a Rhino? b. Cuthbert paid \(\$ 20,300\) for a Rhino. What percentage of Walt's customers paid more than Cuthbert for a Rhino?

Fast Auto Service provides oil and lube service for cars. It is known that the mean time taken for oil and lube service at this garage is 15 minutes per car and the standard deviation is \(2.4\) minutes. The management wants to promote the business by guaranteeing a maximum waiting time for its customers. If a customer's car is not serviced within that period, the customer will receive a \(50 \%\) discount on the charges. The company wants to limit this discount to at most \(5 \%\) of the customers. What should the maximum guaranteed waiting time be? Assume that the times taken for oil and lube service for all cars have a normal distribution.

Do the width and/or height of a normal distribution change when its standard deviation remains the same but its mean increases?

Tommy Wait, a minor league baseball pitcher, is notorious for taking an excessive amount of time between pitches. In fact, his times between pitches are normally distributed with a mean of 36 seconds and a standard deviation of \(2.5\) seconds. What percentage of his times between pitches are a. longer than 39 seconds? b. between 29 and 34 seconds?

quarter), Britons spend an a鈥� # According to a 2004 survey by the telecommunications division of British Gas (Source: http://www. literacytrust.org.uk/Database/texting.html#quarter), Britons spend an average of 225 minutes per day communicating electronically (on a fixed landline phone, on a mobile phone, by emailing, by texting, and so on). Assume that currently such times for all Britons are normally distributed with a mean of 225 minutes per day and a standard deviation of 62 minutes per day. What percentage of Britons communicate electronically for a. less than 60 minutes per day b. more than 360 minutes per day c. between 120 and 180 minutes per day d. between 240 and 300 minutes per day?
See all solutions

Recommended explanations on Math Textbooks

Geometry

Read Explanation

Statistics

Read Explanation

Discrete Mathematics

Read Explanation

Probability and Statistics

Read Explanation

Logic and Functions

Read Explanation

Calculus

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.