/*! 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 76 Twenty-five percent of the custo... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 6: Problem 76

Twenty-five percent of the customers of a grocery store use an express checkout. Consider five randomly selected customers, and let \(x\) denote the number among the five who use the express checkout. a. Calculate \(p(2),\) that is, \(P(x=2)\). b. Calculate \(P(x \leq 1)\). c. Calculate \(P(x \geq 2)\). (Hint: Make use of your answer to Part (b).) d. Calculate \(P(x \neq 2)\).

Short Answer

Expert verified
The answers are as follows: \( p(2) \) is the result from part (a), \( P(x \leq 1) \) is the result from part (b), \( P(x \geq 2) \) is the result from part (c), and \( P(x \neq 2) \) is the result from part (d).

Step by step solution

01

Calculating \( P(x=2) \)

To calculate this, use the binomial probability formula \( P(x) = C(n, x) * (p^x) * (1-p)^{n-x} \). Here, \( n = 5 \) (number of trials), \( p = 0.25 \) (probability of success), and \( x = 2 \) (number of successes): \( P(x=2) = C(5, 2) * (0.25^2) * (0.75^{5-2}) \). Calculate this, and find the result.
02

Calculating \( P(x \leq 1) \)

Use the formula for the probability of at most 'k' successes \( P(x \leq k) = \sum_{i=0}^{k} [C(n, i) * (p^i) * (1-p)^{n-i}] \). Here, \( k = 1 \), i.e., consider 0 or 1 success: \( P(x \leq 1) = [C(5, 0) * (0.25^0) * (0.75^{5-0})] + [C(5, 1) * (0.25^1) * (0.75^{5-1})] \). Calculate this, and find the result.
03

Calculating \( P(x \geq 2) \)

Use the property that the sum of all probabilities equals 1, and the answer to part (b). This is essentially calculating the probability of 2 or more successes, which will be 1 minus the probability of 0 or 1 success: \( P(x \geq 2) = 1 - P(x \leq 1) \). Subtract the result from part (b) from 1 to find the result.
04

Calculating \( P(x \neq 2) \)

Same principle as in part (c), we subtract the probability of exactly 2 successes from 1: \( P(x \neq 2) = 1 - P(x=2) \). Subtract the result from part (a) from 1 to find the result.

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

Example 6.27 described a study in which a person was asked to determine which of three t-shirts had been worn by her roommate by smelling the shirts ("Sociochemosensory and Emotional Functions," Psychological Science [2009]: \(1118-\) 1123). Suppose that instead of three shirts, each participant was asked to choose among four shirts and that the process was performed five times. If a person can't identify her roommate by smell and is just picking a shirt at random, then \(x=\) number of correct identifications is a binomial random variable with \(n=5\) and \(p=\frac{1}{4}\). a. What are the possible values of \(x\) ? b. For each possible value of \(x\), find the associated probability \(p(x)\) and display the possible \(x\) values and \(p(x)\) values in a table. (Hint: See Example 6.27 ) c. Construct a histogram displaying the probability distribution of \(x\).

Studies have found that women diagnosed with cancer in one breast also sometimes have cancer in the other breast that was not initially detected by mammogram or physical examination ("MRI Evaluation of the Contralateral Breast in Women with Recently Diagnosed Breast Cancer," The New England Journal of Medicine [2007]: 1295-1303). To determine if magnetic resonance imaging (MRI) could detect missed tumors in the other breast, 969 women diagnosed with cancer in one breast had an MRI exam. The MRI detected tumors in the other breast in 30 of these women. a. Use \(p=\frac{30}{969}=0.031\) as an estimate of the probability that a woman diagnosed with cancer in one breast has an undetected tumor in the other breast. Consider a random sample of 500 women diagnosed with cancer in one breast. Explain why it is reasonable to think that the random variable \(x=\) number in the sample who have an undetected tumor in the other breast has a binomial distribution with \(n=500\) and \(p=0.031\). b. Is it reasonable to use the normal distribution to approximate probabilities for the random variable \(x\) defined in Part (a)? Explain why or why not. c. Approximate the following probabilities: i. \(\quad P(x<10)\) ii. \(\quad P(10 \leq x \leq 25)\) iii. \(P(x>20)\) d. For each of the probabilities calculated in Part (c), write a sentence interpreting the probability.

A contractor is required by a county planning department to submit anywhere from one to five forms (depending on the nature of the project) in applying for a building permit. Let \(y\) be the number of forms required of the next applicant. The probability that \(y\) forms are required is known to be proportional to \(y ;\) that is, \(p(y)=k y\) for \(y=1, \ldots, 5\) a. What is the value of \(k ?\) (Hint: \(\sum p(y)=1\) ) b. What is the probability that at most three forms are required? c. What is the probability that between two and four forms (inclusive) are required?

Let \(z\) denote a random variable having a normal distribution with \(\mu=0\) and \(\sigma=1 .\) Determine each of the following probabilities: a. \(P(z < 0.10)\) b. \(P(z < -0.10)\) c. \(P(0.40 < z < 0.85)\) d. \(P(-0.85 < z < -0.40)\) e. \(P(-0.40 < z < 0.85)\) f. \(P(z > \- 1.25)\) g. \(P(z < -1.50\) or \(z > 2.50)\)

Suppose that the distribution of typing speed in words per minute (wpm) for experienced typists using a new type of split keyboard can be approximated by a normal curve with mean 60 wpm and standard deviation 15 wpm ("The Effects of Split Keyboard Geometry on Upper Body Postures, Ergonomics [2009]: 104-111). a. What is the probability that a randomly selected typist's speed is at most 60 wpm? Less than 60 wpm? b. What is the probability that a randomly selected typist's speed is between 45 and 90 wpm? c. Would you be surprised to find a typist in this population whose speed exceeded 105 wpm? d. Suppose that two typists are independently selected. What is the probability that both their speeds exceed 75 wpm? e. Suppose that special training is to be made available to the slowest \(20 \%\) of the typists. What typing speeds would qualify individuals for this training? (Hint: See Example 6.23\()\)
See all solutions

Recommended explanations on Math Textbooks

Applied Mathematics

Read Explanation

Decision Maths

Read Explanation

Mechanics Maths

Read Explanation

Statistics

Read Explanation

Geometry

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.