BIG IDEAS MATH Algebra 2: Common Core Student Edition 2015
BIG IDEAS MATH Algebra 2: Common Core Student Edition 2015
15th Edition
ISBN: 9781608408405
Author: HOUGHTON MIFFLIN HARCOURT
Publisher: Cengage Learning
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Chapter 10.6, Problem 13E

(a)

To determine

To graph:

The histogram of the binomial distribution for the survey

(a)

Expert Solution
Check Mark

Answer to Problem 13E

The Histograph of the binomial distribution for the survey are as shown:

  BIG IDEAS MATH Algebra 2: Common Core Student Edition 2015, Chapter 10.6, Problem 13E , additional homework tip  1

Histograph

Explanation of Solution

Given information:

  BIG IDEAS MATH Algebra 2: Common Core Student Edition 2015, Chapter 10.6, Problem 13E , additional homework tip  2

The probability distribution is given as:

  n=6p=0.27k=0,1,2,,6

Interpretation:

The binomial experiment, the probability of k successes in number of trial is:

  P(ksuccesses)=Cnkpk(1p)nk...(1)

Put the values n=6,p=0.27,k=0,1,2,,6 in equation (1):

  P(k=0)=C60(0.27)0(0.73)7P(k=0)=0.1513P(k=1)=C61(0.27)1(0.73)6P(k=1)=0.3358P(k=2)=C62(0.27)2(0.73)5P(k=2)=0.3105P(k=3)=C63(0.27)3(0.73)4P(k=3)=0.1531P(k=4)=C64(0.27)4(0.73)3P(k=4)=0.0425P(k=5)=C65(0.27)5(0.73)2P(k=5)=0.0063P(k=6)=C66(0.27)6(0.73)1P(k=6)=0.0005

The probability value of x is

  P(k=0)=0.1513,P(k=1)=0.3358,P(k=2)=0.3105,P(k=3)=0.1531,P(k=4)=0.0425,P(k=5)=0.0063,P(k=6)=0.0005

The Histograph of the binomial distribution for the survey are as shown:

  BIG IDEAS MATH Algebra 2: Common Core Student Edition 2015, Chapter 10.6, Problem 13E , additional homework tip  3

Histograph

(b)

To determine

To calculate:

The most possible outcomes of the survey

(b)

Expert Solution
Check Mark

Answer to Problem 13E

The most possible outcome of the survey is 2 .

Explanation of Solution

Given information:

  BIG IDEAS MATH Algebra 2: Common Core Student Edition 2015, Chapter 10.6, Problem 13E , additional homework tip  4

Calculation:

The possible outcome of this survey is the value of x for that P(x) is higher.

The probability distribution result shows the probability is great for 2 , nearly 0.34 .

The most possible outcome of the survey is 2 .

(c)

To determine

To calculate:

The probability at most 2 people have the class ring.

(c)

Expert Solution
Check Mark

Answer to Problem 13E

The probability of at most 2 people have the class ring is 0.7976 .

Explanation of Solution

Given information:

  BIG IDEAS MATH Algebra 2: Common Core Student Edition 2015, Chapter 10.6, Problem 13E , additional homework tip  5

Calculation:

The required probability distribution is:

  P(k2)=P(k=0)+P(k=1)+P(k=2)

Put the values P(k=0)=0.1513,P(k=1)=0.3358,P(k=2)=0.3105 in equation

  P(k2)=0.1513+0.3358+0.3105P(k2)=0.7976

The probability of at most 2 people have the class ring is 0.7976 .

Chapter 10 Solutions

BIG IDEAS MATH Algebra 2: Common Core Student Edition 2015

Ch. 10.1 - Prob. 11ECh. 10.1 - Prob. 12ECh. 10.1 - Prob. 13ECh. 10.1 - Prob. 14ECh. 10.1 - Prob. 15ECh. 10.1 - Prob. 16ECh. 10.1 - Prob. 17ECh. 10.1 - Prob. 18ECh. 10.1 - Prob. 19ECh. 10.1 - Prob. 20ECh. 10.1 - Prob. 21ECh. 10.1 - Prob. 22ECh. 10.1 - Prob. 23ECh. 10.1 - Prob. 24ECh. 10.1 - Prob. 25ECh. 10.1 - Prob. 26ECh. 10.1 - Prob. 27ECh. 10.1 - Prob. 28ECh. 10.1 - Prob. 29ECh. 10.1 - Prob. 30ECh. 10.1 - Prob. 31ECh. 10.1 - Prob. 32ECh. 10.1 - Prob. 33ECh. 10.1 - Prob. 34ECh. 10.2 - Prob. 1ECh. 10.2 - Prob. 2ECh. 10.2 - Prob. 3ECh. 10.2 - Prob. 4ECh. 10.2 - Prob. 5ECh. 10.2 - Prob. 6ECh. 10.2 - Prob. 7ECh. 10.2 - Prob. 8ECh. 10.2 - Prob. 9ECh. 10.2 - Prob. 10ECh. 10.2 - Prob. 11ECh. 10.2 - Prob. 12ECh. 10.2 - Prob. 13ECh. 10.2 - Prob. 14ECh. 10.2 - Prob. 15ECh. 10.2 - Prob. 16ECh. 10.2 - Prob. 17ECh. 10.2 - Prob. 18ECh. 10.2 - Prob. 19ECh. 10.2 - Prob. 20ECh. 10.2 - Prob. 21ECh. 10.2 - Prob. 22ECh. 10.2 - Prob. 23ECh. 10.2 - Prob. 24ECh. 10.2 - Prob. 25ECh. 10.2 - Prob. 26ECh. 10.2 - Prob. 27ECh. 10.2 - Prob. 28ECh. 10.2 - Prob. 29ECh. 10.2 - Prob. 30ECh. 10.2 - Prob. 31ECh. 10.2 - Prob. 32ECh. 10.2 - Prob. 33ECh. 10.3 - Prob. 1ECh. 10.3 - Prob. 2ECh. 10.3 - Prob. 3ECh. 10.3 - Prob. 4ECh. 10.3 - Prob. 5ECh. 10.3 - Prob. 6ECh. 10.3 - Prob. 7ECh. 10.3 - Prob. 8ECh. 10.3 - Prob. 9ECh. 10.3 - Prob. 10ECh. 10.3 - Prob. 11ECh. 10.3 - Prob. 12ECh. 10.3 - Prob. 13ECh. 10.3 - Prob. 14ECh. 10.3 - Prob. 15ECh. 10.3 - Prob. 16ECh. 10.3 - Prob. 17ECh. 10.3 - Prob. 18ECh. 10.3 - Prob. 19ECh. 10.3 - Prob. 20ECh. 10.3 - Prob. 21ECh. 10.3 - Prob. 22ECh. 10.3 - Prob. 23ECh. 10.3 - Prob. 24ECh. 10.3 - Prob. 25ECh. 10.3 - Prob. 26ECh. 10.3 - Prob. 27ECh. 10.3 - Prob. 28ECh. 10.3 - Prob. 29ECh. 10.3 - Prob. 1QCh. 10.3 - Prob. 2QCh. 10.3 - Prob. 3QCh. 10.3 - Prob. 4QCh. 10.3 - Prob. 5QCh. 10.3 - Prob. 6QCh. 10.3 - Prob. 7QCh. 10.3 - Prob. 8QCh. 10.3 - Prob. 9QCh. 10.3 - Prob. 10QCh. 10.3 - Prob. 11QCh. 10.4 - Prob. 1ECh. 10.4 - Prob. 2ECh. 10.4 - Prob. 3ECh. 10.4 - Prob. 4ECh. 10.4 - Prob. 5ECh. 10.4 - Prob. 6ECh. 10.4 - Prob. 7ECh. 10.4 - Prob. 8ECh. 10.4 - Prob. 9ECh. 10.4 - Prob. 10ECh. 10.4 - Prob. 11ECh. 10.4 - Prob. 12ECh. 10.4 - Prob. 13ECh. 10.4 - Prob. 14ECh. 10.4 - Prob. 15ECh. 10.4 - Prob. 16ECh. 10.4 - Prob. 17ECh. 10.4 - Prob. 18ECh. 10.4 - Prob. 19ECh. 10.4 - Prob. 20ECh. 10.4 - Prob. 21ECh. 10.4 - Prob. 22ECh. 10.4 - Prob. 23ECh. 10.4 - Prob. 24ECh. 10.4 - Prob. 25ECh. 10.4 - Prob. 26ECh. 10.5 - Prob. 1ECh. 10.5 - Prob. 2ECh. 10.5 - Prob. 3ECh. 10.5 - Prob. 4ECh. 10.5 - Prob. 5ECh. 10.5 - Prob. 6ECh. 10.5 - Prob. 7ECh. 10.5 - Prob. 8ECh. 10.5 - Prob. 9ECh. 10.5 - Prob. 10ECh. 10.5 - Prob. 11ECh. 10.5 - Prob. 12ECh. 10.5 - Prob. 13ECh. 10.5 - Prob. 14ECh. 10.5 - Prob. 15ECh. 10.5 - Prob. 16ECh. 10.5 - Prob. 17ECh. 10.5 - Prob. 18ECh. 10.5 - Prob. 19ECh. 10.5 - Prob. 20ECh. 10.5 - Prob. 21ECh. 10.5 - Prob. 22ECh. 10.5 - Prob. 23ECh. 10.5 - Prob. 24ECh. 10.5 - Prob. 25ECh. 10.5 - Prob. 26ECh. 10.5 - Prob. 27ECh. 10.5 - Prob. 28ECh. 10.5 - Prob. 29ECh. 10.5 - Prob. 30ECh. 10.5 - Prob. 31ECh. 10.5 - Prob. 32ECh. 10.5 - Prob. 33ECh. 10.5 - Prob. 34ECh. 10.5 - Prob. 35ECh. 10.5 - Prob. 36ECh. 10.5 - Prob. 37ECh. 10.5 - Prob. 38ECh. 10.5 - Prob. 39ECh. 10.5 - Prob. 40ECh. 10.5 - Prob. 41ECh. 10.5 - Prob. 42ECh. 10.5 - Prob. 43ECh. 10.5 - Prob. 44ECh. 10.5 - Prob. 45ECh. 10.5 - Prob. 46ECh. 10.5 - Prob. 47ECh. 10.5 - Prob. 48ECh. 10.5 - Prob. 49ECh. 10.5 - Prob. 50ECh. 10.5 - Prob. 51ECh. 10.5 - Prob. 52ECh. 10.5 - Prob. 53ECh. 10.5 - Prob. 54ECh. 10.5 - Prob. 55ECh. 10.5 - Prob. 56ECh. 10.5 - Prob. 57ECh. 10.5 - Prob. 58ECh. 10.5 - Prob. 59ECh. 10.5 - Prob. 60ECh. 10.5 - Prob. 61ECh. 10.5 - Prob. 62ECh. 10.5 - Prob. 63ECh. 10.5 - Prob. 64ECh. 10.5 - Prob. 65ECh. 10.5 - Prob. 66ECh. 10.5 - Prob. 67ECh. 10.5 - Prob. 68ECh. 10.5 - Prob. 69ECh. 10.5 - Prob. 70ECh. 10.5 - Prob. 71ECh. 10.5 - Prob. 72ECh. 10.5 - Prob. 73ECh. 10.5 - Prob. 74ECh. 10.5 - Prob. 75ECh. 10.5 - Prob. 76ECh. 10.5 - Prob. 77ECh. 10.5 - Prob. 78ECh. 10.5 - Prob. 79ECh. 10.5 - Prob. 80ECh. 10.5 - Prob. 81ECh. 10.6 - Prob. 1ECh. 10.6 - Prob. 2ECh. 10.6 - Prob. 3ECh. 10.6 - Prob. 4ECh. 10.6 - Prob. 5ECh. 10.6 - Prob. 6ECh. 10.6 - Prob. 7ECh. 10.6 - Prob. 8ECh. 10.6 - Prob. 9ECh. 10.6 - Prob. 10ECh. 10.6 - Prob. 11ECh. 10.6 - Prob. 12ECh. 10.6 - Prob. 13ECh. 10.6 - Prob. 14ECh. 10.6 - Prob. 15ECh. 10.6 - Prob. 16ECh. 10.6 - Prob. 17ECh. 10.6 - Prob. 18ECh. 10.6 - Prob. 19ECh. 10.6 - Prob. 20ECh. 10.6 - Prob. 21ECh. 10.6 - Prob. 22ECh. 10.6 - Prob. 23ECh. 10.6 - Prob. 24ECh. 10 - Prob. 1CRCh. 10 - Prob. 2CRCh. 10 - Prob. 3CRCh. 10 - Prob. 4CRCh. 10 - Prob. 5CRCh. 10 - Prob. 6CRCh. 10 - Prob. 7CRCh. 10 - Prob. 8CRCh. 10 - Prob. 9CRCh. 10 - Prob. 10CRCh. 10 - Prob. 11CRCh. 10 - Prob. 12CRCh. 10 - Prob. 13CRCh. 10 - Prob. 14CRCh. 10 - Prob. 15CRCh. 10 - Prob. 16CRCh. 10 - Prob. 17CRCh. 10 - Prob. 1CTCh. 10 - Prob. 2CTCh. 10 - Prob. 3CTCh. 10 - Prob. 4CTCh. 10 - Prob. 5CTCh. 10 - Prob. 6CTCh. 10 - Prob. 7CTCh. 10 - Prob. 8CTCh. 10 - Prob. 9CTCh. 10 - Prob. 10CTCh. 10 - Prob. 11CTCh. 10 - Prob. 12CTCh. 10 - Prob. 13CTCh. 10 - Prob. 14CTCh. 10 - Prob. 1CACh. 10 - Prob. 2CACh. 10 - Prob. 3CACh. 10 - Prob. 4CACh. 10 - Prob. 5CACh. 10 - Prob. 6CACh. 10 - Prob. 7CACh. 10 - Prob. 8CACh. 10 - Prob. 9CA
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