MATH 1324 Homework 12 Answers
Homework 12 for MATH 1324 Finite Math covers problems from Section 7.3. Multiple-choice answers for problem 8b and 10b are provided. Select the correct values to practice and test your understanding of the concepts in the section.
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MATH 1324 - Finite Math with Applications
Homework 12 (Untimed, 1 Attempt)
DUE November 25, 2022 11:59:00 PM!
QI. The choices for problem number 8 part b, from Section 7.3 in the book are given below.
* A. 6.9676
* B. 2.6396
* C. 4.5727
* D. 4.8776
* E. 8.0278
* F. None of the above.
* x___________I - 3 - 1 1 3 5
P(X = xj | 0.17 0.28 0.15 0.27 0.13
* Part B: Find the variance.
* E X = x 1 p.+ x2 p2 + x2 p2 = - 3 : 0 . 1 7 -11'0.28 1 0.15 + 3 '0.27' + 5' 0.131 = 0.82 = p
* Vor A’ = p. x x — | Z + p 2l x, - p 2+ p2
= 0.17(-3 - 0.82) 2 + 0.28(-l - 0.82) 2 + 0.15(1 - 0.82) + 0.27(3 - 0.82) 2 + 0.13(5 - 0.82) 2
= 6.9676
Q2. The choices for problem number 10 pan b, from Section 7.3 in die book are given belowT
.
* A. 312.4215
* B. 284.8275
* C. 248.9658
* D. 298.9746
* E. 315.9872
* F. None of the above.
* x | 0 15 30 45 60
P(A' = x) | 0.24 0.46 0.13 0.11 0.06
* Part B: Find die variance.
* E A" =x1 p + x2 p2 + x2 pt
= 0(0.24) + 15(0.46) + 30(0.13) + 45(0.11) + 60(0.06) = 19.35 = p
Vor A =p. x x —p) 2-pj.x 2 - p ' + p; x3 “ p ?
= 0.24(0 - 19.35) 2 + 0.46(15 - 19.35) 2 + 0.13(30 - 19.35) 2 + 0.11(45 - 19.35) 2 + 0.06(60 - 19.35) 2
= 284.8275
MATH 1324 - Finite Math with Applications
Homework 12 (Untimed, 1 Attempt)
DUE November 25, 2022 11:59:00 PM!
QI. The choices for problem number 8 part b, from Section 7.3 in the book are given below.
* A. 6.9676
* B. 2.6396
* C. 4.5727
* D. 4.8776
* E. 8.0278
* F. None of the above.
* x___________I - 3 - 1 1 3 5
P(X = xj | 0.17 0.28 0.15 0.27 0.13
* Part B: Find the variance.
* E X = x 1 p.+ x2 p2 + x2 p2 = - 3 : 0 . 1 7 -11'0.28 1 0.15 + 3 '0.27' + 5' 0.131 = 0.82 = p
* Vor A’ = p. x x — | Z + p 2l x, - p 2+ p2
= 0.17(-3 - 0.82) 2 + 0.28(-l - 0.82) 2 + 0.15(1 - 0.82) + 0.27(3 - 0.82) 2 + 0.13(5 - 0.82) 2
= 6.9676
Q2. The choices for problem number 10 pan b, from Section 7.3 in die book are given belowT
.
* A. 312.4215
* B. 284.8275
* C. 248.9658
* D. 298.9746
* E. 315.9872
* F. None of the above.
* x | 0 15 30 45 60
P(A' = x) | 0.24 0.46 0.13 0.11 0.06
* Part B: Find die variance.
* E A" =x1 p + x2 p2 + x2 pt
= 0(0.24) + 15(0.46) + 30(0.13) + 45(0.11) + 60(0.06) = 19.35 = p
Vor A =p. x x —p) 2-pj.x 2 - p ' + p; x3 “ p ?
= 0.24(0 - 19.35) 2 + 0.46(15 - 19.35) 2 + 0.13(30 - 19.35) 2 + 0.11(45 - 19.35) 2 + 0.06(60 - 19.35) 2
= 284.8275
2
Q3. The choices for problem number 12 part c, from Section 7 3 in the book are given below.
* A. 5.4866
' B. 7.4532
* C. 4.5727
' D. 5.0197
* E. 6.8147
* F. None of the above.
* x I -6 -1 ___0 8
P(X = x) | 0.21 0.25 0.26 0.28
* Part C: Find die standard deviation.
* Standard deviation = o = V Var (X J
* E(X) = -6(0.21) + (-1X0.25) + 0(0.26) + 8(0.28) = 0.73 = p
* Vai(X) = 0.21(-6 - 0.73) 2 - 0.25( 1 - 0.73) 2 + 0.26(0 - 0.73) 2 + 0.28(8 - 0.73) 3
= 25.1971
* o = <25.1971 « 5.0197
* 5.0197
Q4. The choices for problem number 20 part c, from Section 7.3 in the book are given below.
* A. 0.9897
* B. 0.7940
* C. 1.0135
* D. 0.9236
* E. 0.8688
* F. None of the above.
•
* The table below gives the number of days it takes for packages to be delivered from an online retailer
and the probability associated with each number of days.
* Number of Days | 3 4 5 6 7
Probability | 0.01 0.14 0.46 0.25 0.14
* Part C: Find the standard deviation of the random variable.
* Standard deviation = o = V Vor (x )
* E(X) = 3(0.01) + 4(0.14) + 5(0.46) + 6(0.25) + 7(0.14) = 5.37 = p
' Var(X) = 0.01(3 - 5.37) 2 + 0.14(4 - 5.37) 2 + 0.46(5 - 5.37) 2 + 0.25(6 - 5.37) 2 + 0.14(7 - 5.37) 2
= 0.8531
* cr = <0.8531 a 0.9236
* 0.9236
Q3. The choices for problem number 12 part c, from Section 7 3 in the book are given below.
* A. 5.4866
' B. 7.4532
* C. 4.5727
' D. 5.0197
* E. 6.8147
* F. None of the above.
* x I -6 -1 ___0 8
P(X = x) | 0.21 0.25 0.26 0.28
* Part C: Find die standard deviation.
* Standard deviation = o = V Var (X J
* E(X) = -6(0.21) + (-1X0.25) + 0(0.26) + 8(0.28) = 0.73 = p
* Vai(X) = 0.21(-6 - 0.73) 2 - 0.25( 1 - 0.73) 2 + 0.26(0 - 0.73) 2 + 0.28(8 - 0.73) 3
= 25.1971
* o = <25.1971 « 5.0197
* 5.0197
Q4. The choices for problem number 20 part c, from Section 7.3 in the book are given below.
* A. 0.9897
* B. 0.7940
* C. 1.0135
* D. 0.9236
* E. 0.8688
* F. None of the above.
•
* The table below gives the number of days it takes for packages to be delivered from an online retailer
and the probability associated with each number of days.
* Number of Days | 3 4 5 6 7
Probability | 0.01 0.14 0.46 0.25 0.14
* Part C: Find the standard deviation of the random variable.
* Standard deviation = o = V Vor (x )
* E(X) = 3(0.01) + 4(0.14) + 5(0.46) + 6(0.25) + 7(0.14) = 5.37 = p
' Var(X) = 0.01(3 - 5.37) 2 + 0.14(4 - 5.37) 2 + 0.46(5 - 5.37) 2 + 0.25(6 - 5.37) 2 + 0.14(7 - 5.37) 2
= 0.8531
* cr = <0.8531 a 0.9236
* 0.9236
3
Qz>. The choices for problem number 26 pan a, from Section 7.3 in the book are given below.
* A. 6
* B. 0.9722
* C. 0.8333
* D. 0.5556
* E. 0.4444
* F. None of the above.
A probability distribution has a mean of 12 and a standard deviation of . Use Chebyshev's inequality
to approximate the probability that an outcome of the experiment lies between:
p = 12, o = 7 Pan A: P(10 < x < 14)
solve for k
p - ko = small | p - ko = Big
12 “k 1 ; =10 | 12 + k j =14
~1 1
/ k = - 2 13 3
k = 6 | k = 6
1 1 1
I ’ T -1 = 1- — = 1 - RJ 0.9722
k 2 6s
0.9722
Q6. The choices for problem number 26 pan b, from Section 7.3 in the book are given below.
* A. 12
* B. 0.9167
* C. 0.4375
* D. 0.9931
* E. 0.5625
* F. None of the above.
* A probability distribution has a mean of 12 and a standard deviation of . Use Chebyshev's inequality
to approximate the probability that an outcome of the experiment lies between:
1
* p = 1 2 , o = 3 Pan B: P(8 < x < 16)
solve for k
* p - ko = small | p - ko = Big
12— k ’ i - S I 12 + k 1 1 = 1 6
3 1 3
| l k = 4
k = 12 | k = 12
=1 -lb = 1- -fh Rs 0.9931
144 -----
* 0.9931
Qz>. The choices for problem number 26 pan a, from Section 7.3 in the book are given below.
* A. 6
* B. 0.9722
* C. 0.8333
* D. 0.5556
* E. 0.4444
* F. None of the above.
A probability distribution has a mean of 12 and a standard deviation of . Use Chebyshev's inequality
to approximate the probability that an outcome of the experiment lies between:
p = 12, o = 7 Pan A: P(10 < x < 14)
solve for k
p - ko = small | p - ko = Big
12 “k 1 ; =10 | 12 + k j =14
~1 1
/ k = - 2 13 3
k = 6 | k = 6
1 1 1
I ’ T -1 = 1- — = 1 - RJ 0.9722
k 2 6s
0.9722
Q6. The choices for problem number 26 pan b, from Section 7.3 in the book are given below.
* A. 12
* B. 0.9167
* C. 0.4375
* D. 0.9931
* E. 0.5625
* F. None of the above.
* A probability distribution has a mean of 12 and a standard deviation of . Use Chebyshev's inequality
to approximate the probability that an outcome of the experiment lies between:
1
* p = 1 2 , o = 3 Pan B: P(8 < x < 16)
solve for k
* p - ko = small | p - ko = Big
12— k ’ i - S I 12 + k 1 1 = 1 6
3 1 3
| l k = 4
k = 12 | k = 12
=1 -lb = 1- -fh Rs 0.9931
144 -----
* 0.9931
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