Math in Society: Mathematics for liberal arts majors

1. Proposition
2. Not a proposition
3. Proposition
4. Proposition

1. Not a proposition
2. Proposition
3. Not a proposition
4. Proposition

1. I do not ride my bike to campus.
2. Portland is in Oregon.

1. You should not see this movie.
2. Lashonda is not wearing blue.

1. Exclusive
2. Inclusive

1. Exclusive
2. Inclusive

1. Inclusive
2. Exclusive

1. Inclusive
2. Inclusive

1. If it is sunny, then I will go swimming.
2. If it is Friday, then I will go see a movie.

1. If it is raining, then I carry an umbrella.
2. If it is the weekend, then I am hanging out with friends.

1. Elvis is not alive
2. Elvis is alive or Elvis is the King
3. Elvis is not alive and Elvis is the King
4. If Elvis is the King, then Elvis is not alive

1. I do not own an umbrella.
2. It rains in Oregon and I do not own an umbrella
3. If it rains in Oregon, then I own an umbrella
4. If I do not own an umbrella, then it rains in Oregon

1. B and A
2. If B, then A
3. If B, then A
4. If not B, then not A

1. A and not B
2. Not A
3. If A, then B
4. If not B, then not A

1. True
2. False

1. False
2. False

• Region A: Unemployed non-honor-students who identify as boys
• Region B: Unemployed honor-students who identify as boys
• Region C: Unemployed honor-students who identify as girls or nonbinary
• Region D: Employed non-honor-students who identify as boys
• Region E: Employed honor-students who identify as boys
• Region F: Employed honor-students who identify as girls or nonbinary
• Region G: Employed non-honor-students who identify as girls or nonbinary
• Region H: Unemployed non-honor-students who identify as girls or nonbinary

• Students enrolled for keyboard class: 24
• Students enrolled for keyboard class only: 8
• Students didn’t enroll at all: 6
• Students took all three classes: 4
• Students enrolled for guitar and drum: 6
• Students enrolled for guitar and drum only: 2

1. Thirty-six coffee drinkers like cream.
2. Fifty-five coffee drinkers like sugar.
3. Thirty-five coffee drinkers like sugar but not cream.
4. Sixteen coffee drinkers like cream but not sugar.
5. Twenty coffee drinkers like cream and sugar.
6. Seventy-one coffee drinkers like cream or sugar.
7. Twenty-nine coffee drinkers like neither cream nor sugar.

1. 
2. Nineteen said “It’s more filling!” but didn’t say “It tastes great!”
3. Forty-two said neither of those things.
4. 108 said “It’s more filling!” or said “It tastes great!”

1. The argument is inductive.
2. The argument is deductive.

1. The next term is 18.
2. The next term is 39.

1. Yes
2. Yes
3. No
4. No
5. Yes
6. Yes

1. I don’t take public transportation to get to class.
2. I didn’t go to a movie on Friday.
3. I want to go golfing today
4. I don’t love watching basketball.
5. Breylynn’s favorite color is not green.
6. Mirriam is not a theater major.

1. Inclusive
2. Inclusive
3. Exclusive
4. Exclusive
5. Exclusive

1. If you do homework, then it helps increase your grade in class.
2. If you ride public transportation, then it will help you save money.
3. If it is a squirrel, then it will bury their food.
4. If you eat to much candy, then you will get sick.
5. If you think you have the flu, then go too the doctor.

1. I will not buy an iPhone.
2. I do not learn how to use technology fast.
3. I will buy an iPhone or I learn how to use technology fast.
4. I will buy an iPhone and I don’t learn how to use technology fast.
5. If I learn to use new technology, then I will buy an iPhone.

1. \(27\) people only like cream in their coffee.
2. \(13+20=33\)
   \(33\) people put sugar in their coffee
3. \(27+13+20=60\)
   \(60\) people put sugar or cream in their coffee.
4. \(20+10=30\)
   \(30\) people don’t like cream in their coffee.

1. 
2. \(30-26=4\)
   \(4\) students are not taking either course.

1. 
2. \(7+22+28+1+3+2=63\text{.}\)
   63 people get thier news from Twitter or a website.
3. 53 people do not get their news from any of the three sources.
4. 7 people use only Twitter.
5. \(7+22+28=57\text{.}\)
   57 people use Twitter or a website, but not a newspaper.

1. =5780.23-5250 which gives $530.23
2. =530.23/5250 which gives approximately 0.100996, or approximately 10.1%
3. Exactly 200%
4. =5250*115.5% which gives exactly $6063.75

1. See the table at the bottom for part a and part b.
   After 1 year, the account holds $563.41.
2. See the table at the bottom for part a and part b.
   After 2 years, the account holds $634.87.
3. =A13/A1 which gives \(\approx 112.6825\%\) growth
4. =A25/A13 which gives \(\approx 112.6825\%\) growth (same)
5. Each starting value increases mathematically by a factor of \((1.01)^{12}\) each year, which is approximately 112.6825%. So yes, this pattern must continue indefinitely into future years.

1. =1000*103%*103% which gives $1060.90
2. =1000*(103%)^2 gives the same result of $1060.90, because raising 103% to the second power means the same as multiplying 103% by itself two times.
3. =1000*(103%)^15 which gives $1557.97 rounded to the nearest cent
4. Refer to the table at the bottom for part d and part e.
   Note the entry in cell B3 here is = B2*103% and the remaining cells are computed using the fill down feature.
   You will have to wait a minimum of 24 full years, in each case, in order for the balance to finally exceed twice the opening deposit amount.
5. Since \((103\%)^{23} \lt 2 \lt (103\%)^{24}\text{,}\) the minimum number of full years until the opening deposit doubles must be the same here, for any positive opening balance that we may choose for this account.

1. \begin{align*} A\amp=100+100(0.03)(0.5)\\ \amp=\$101.50 \end{align*}
2. \begin{align*} I\amp=101.50-100\\ \amp=\$1.50 \end{align*}

1. \begin{align*} I\amp=1000(0.025)(5)\\ \amp=\$125 \end{align*}
2. \begin{align*} A\amp=1000+1000(0.025)(5)\\ \amp=\$1{,}125 \end{align*}

1. \begin{align*} A\amp=20000+20000(0.05)(10)\\ \amp=\$30{,}000 \end{align*}
2. \begin{align*} A\amp=20000(1+0.05)^{10}\\ \amp=\$32{,}577.89 \end{align*}

=FV(0.085/12,12*20,0,4500)

=FV(0.07/52,52*20,0,1000)

1. \begin{align*} A\amp=3000\left(1+\frac{0.03}{4}\right)^{4*5}\\ \amp\approx \$3{,}483.55 \end{align*}
   The future value is $3,483.55.
   Or,

   =FV(0.03/4,4*5,0,3000)

2. \begin{align*} I\amp=3{,}483.53-3000\\ \amp=\$483.54 \end{align*}
   The interest earned is $483.54.
3. \begin{gather*} \frac{483.54}{3{,}483.54} \approx 0.1388 \text{ or } 13.88\% \end{gather*}
   13.88% of the balance is interest.

1. \begin{align*} A\amp=300\left(1+\frac{0.05}{1}\right)^{1*10}\\ \amp\approx \$488.67 \end{align*}
   There will be $488.67 in the account in 10 years.
   Or,

   =FV(0.05/1,1*10,0,300)

2. \begin{align*} I\amp=488.67-300\\ \amp=\$188.67 \end{align*}
   $188.67 of the balance will be interest.
3. \begin{gather*} \frac{188.67}{488.67} \approx 0.3861 \text{ or } 38.61\% \end{gather*}
   The interest makes up 38.61% of the balance.

1. \begin{align*} A\amp=2000\left(1+\frac{0.03}{12}\right)^{12*20}\\ \amp\approx \$3{,}641.51 \end{align*}
   The account will have $3,641.51 in 20 years
   Or,

   =FV(0.03/12,12*20,0,2000)

2. \begin{align*} I\amp=3{,}641.51-2000\\ \amp=\$1{,}641.51 \end{align*}
   The interest will be $1,641.51.
3. \begin{gather*} \frac{1{,}641.51}{3{,}641.51} \approx 0.4508 \rightarrow 45.08\% \end{gather*}
   The interest is 45.08% of the balance.
4. \begin{gather*} \frac{2{,}000}{3{,}641.51} \approx 0.5492 \rightarrow 54.92% \end{gather*}
   The principal is 54.92% of the balance.

1. \begin{align*} A\amp=10000\left(1+\frac{0.04}{52}\right)^{52*25}\\ \amp\approx \$27{,}172.37 \end{align*}
   The balance is $27,172.37.
   Or,

   =FV(0.04/52,52*25,0,10000)

2. \begin{align*} I\amp=27{,}172.37-10{,}000\\ \amp=\$17{,}172.37 \end{align*}
   The interest is $17,172.37.
3. \begin{gather*} \frac{17{,}172.37}{27{,}172.37} \approx 0.632 \rightarrow 63.2\% \end{gather*}
   The percent that is interest is 63.2%.
4. \begin{gather*} \frac{10{,}000}{27{,}172.37} \approx 0.368 \rightarrow 36.8\% \end{gather*}
   The percentage that is the principal is 36.2%.

=PV(0.06/12,12*8,0,6000)

=PV(0.05/4,4*4,0,20000)

1. \begin{align*} A_{Breylan} \amp= 1200\left(1+\frac{0.046}{52}\right)^{52*15}\\ \amp\approx \$2{,}391.73 \end{align*}
   Or, =FV(0.046/52,15*52,0,1200)
   \begin{align*} A_{Angad} \amp= 1200\left(1+\frac{0.0455}{52}\right)^{52*15}\\ \amp\approx \$2{,}373.87 \end{align*}
   Or, =FV(0.0455/52,15*52,0,1200)
   Breylan has an account balance of $2,391.73 and Angad has a balance of $2,373.87
2. \begin{align*} A_{Breylan} \amp= 1200\left(1+\frac{0.046}{52}\right)^{52*30}\\ \amp\approx \$4{,}766.97 \end{align*}
   Or, =FV(0.046,52,30*52,0,1200)
   \begin{align*} A_{Angad} \amp= 1200\left(1+\frac{0.0455}{52}\right)^{52*30}\\ \amp\approx \$4{,}696.06 \end{align*}
   Or, =FV(0.0455/52,30*52,0,1200)
   Breylan has an account balance of $4,766.97 and Angad has a balance of $4,696.06
3. Breylan =EFFECT(0.046,52), which gives 4.71%
   Angad =EFFECT(0.0455,52), which gives 4.65%
   Breylan has an effective rate of 4.71% and Angad has an effective rate of 4.65%.

1. Bill =EFFECT(0.0375,12) \(=3.82\%\) and Ted =EFFECT(0.038,1) =3.8%. Bill has an effective rate of 3.82% and Ted has a rate of 3.8%.
2. \begin{align*} A\amp=6700\left(1+\frac{0.0375}{12}\right)^{12*5}\\ \amp\approx \$8{,}079.38 \end{align*}
   Or, =FV(0.0375/12,5*12,0,6700)
   \begin{align*} A\amp=6500\left(1+\frac{0.038}{1}\right)^{1*5}\\ \amp\approx \$7{,}832.49 \end{align*}
   Or, =FV(0.038,5,0,6500)
   The account balances are $8,079.38 and $7,832.49. So, Bill’s balance is higher.

1. =EFFECT(0.0345,4)= 3.49% and =EFFECT(0.034,365)= 3.46%. The effective rates are 3.49% and 3.46%.
2. \begin{align*} A\amp=5000\left(1+\frac{0.0345}{4}\right)^{4*10}\\ \amp\approx \$7{,}049.51 \end{align*}
   Or, =FV(0.0345/4,10*4,0,5000)
   \begin{align*} A\amp=5000\left(1+\frac{0.034}{365}\right)^{365*10}\\ \amp\approx \$7{,}024.63 \end{align*}
   Or, =FV(0.034/365,10*365,0,5000)
   The account balances are $7,049.51 and $7,024.63.

1. \begin{align*} A\amp=2500e^{0.04*10}\\ \amp\approx \$3{,}729.56 \end{align*}
   Or, 2500*EXP(0.04*10)
   The account balance is $3,729.56.
2. \begin{align*} I \amp= 3{,}729.56-2{,}500\\ \amp= \$1{,}229.56 \end{align*}
   The interest earned is $1,229.56.
3. \begin{gather*} \frac{1{,}229.56}{3{,}729.56} \approx 0.3297 \rightarrow 32.97\% \end{gather*}
   32.97% of the balance is interest.

1. \begin{align*} A\amp= 1000e^{0.0575*15}\\ \amp\approx \$2{,}369.08 \end{align*}
   Or, =1000*EXP(0.0575*15)
   The account balance is $2,369.08
2. \begin{align*} I \amp= 2{,}368.08-1{,}000\\ \amp= \$1{,}368.08 \end{align*}
   The interest earned is $1,368.08
3. \begin{gather*} \frac{1{,}368.08}{2{,}368.08} \approx 0.5777 \rightarrow 57.77\% \end{gather*}
   The interest is 57.77% of the account balance.

1. \begin{align*} A\amp= 5000e^{0.045*5}\\ \amp\approx \$6{,}261.61 \end{align*}
   Or, =5000*EXP(0.045*5)
   The account balance is $6,261.61.
2. \begin{align*} I \amp= 6{,}261.61-5{,}000\\ \amp= \$1{,}261.61 \end{align*}
   The interest earned is $1,261.61
3. \begin{gather*} \frac{1{,}261.61}{6{,}261.61} \approx 0.2015 \rightarrow 20.15\% \end{gather*}
   The interest is 20.15% of the balance.

1. \begin{align*} A_{You}\amp=10000e^{0.055*10}\\ \amp\approx \$17{,}332.53 \end{align*}
   Or, =10000*EXP(0.055*10)
   \begin{align*} A_{Friend} \amp= 10000(1+\frac{0.055}{1})^{1*10}\\ \amp\approx \$17{,}081.44 \end{align*}
   Or, =FV(0.055,10,0,10000)
   Your balance is $17,332.53 and your friend’s balance is $17,081.44.
2. The difference is:

   \begin{gather*} 17{,}332.53-17{,}081.44=\$251.09 \end{gather*}
   You have $251.09 more than your friend.

1. \(\frac{250[(1+\frac{0.065}{12})^{12*35}-1]}{(\frac{0.065}{12})}\)
   Or, FV(0.065/12,12*35,250)
   In 35 years you will have $400,079.05 in your retirement plan.
2. \(400{,}079.05 – 12*35*250\text{.}\) You will have earned $295,079.05 in interest.
3. \(29{,}5079.05 /400{,}079.05\text{.}\) The final balance will be about 73.8% interest.

1. \(\frac{75\left(\left(1+\frac{0.085}{52}\right)^{52*40}-1\right)}{\left(\frac{0.085}{52}\right)}\)
   Or, =FV(0.085/52,52*40,75)
   In 40 years you will have $1,325,130.09 in your retirement plan.
2. \(1{,}325{,}130.09 – 52*40*75\text{.}\) You will have earned $1,169,130.09 in interest.
3. \(1{,}169{,}130.09/1{,}325{,}130.09\text{.}\) The final balance will be about 88.2% interest.

1. \(\frac{750\left(1+\frac{0.0775}{4})^{4*30}-1\right)}{\left(frac{0.0775}{4}\right)}\)
   Or, =FV(0.0775/4,4*30,750)
   In 30 years you will have $348,456.10 in your retirement plan.
2. \(348{,}456.1 – 4*30*750\text{.}\) You will have earned $258,456.10 in interest.
3. \(258{,}456.1 /348{,}456.1\text{.}\) The final balance will be about 74.2% interest.

1. \(\frac{20\left(\left(1+\frac{0.0235}{365}\right)^{365*25}-1\right)}{\left(\frac{0.0235}{365}\right)}\)
   Or, =FV(0.0235/365,365*25,20)
   In 25 years you will have $248,339.80 in your retirement plan.
2. \(248{,}339.8 – 365*25*20\text{.}\) You will have earned $65,839.80 in interest.
3. \(65{,}839.8 /248{,}339.8\text{.}\) The final balance will be about 26.5% interest.

1. In 5 years: \(\frac{130\left(\left(1+\frac{0.09}{12}\right)^{12*5}-1\right)}{\left(\frac{0.09}{12}\right)}\)
   Or, =FV(0.09/12,12*5,130)
   In 25 more years: \(9805.14(1+\frac{0.09}{12})^{12*25}\)
   =FV(0.09/12,12*25,0,9805.14)
   Your final balance will be $92,250.82.
2. \(92{,}250.82 – 130*5*12\text{.}\) You will earn $84,450.82 in interest.
3. \(84{,}450.82/92{,}250.82\text{.}\) The final balance will be about 91.5% interest.

1. \(\frac{200\left(\left(1+\frac{0.05}{12}\right)^{12*20}-1\right)}{\left(\frac{0.05}{12}\right)}\)
   Or, =FV(0.05/12, 12*10, 200)
   After 10 years the balance will be $31,056.46.
2. In 20 more years: \(31056.46\left(1+\frac{0.05}{12}\right)^{12*20}\)
   Or, =FV(0.05/12,12*20,0,31056.46)
   The final balance will be $84,245.00.
3. \(84{,}245 – 200*12*10\text{.}\) You will earn $60,245 in interest.
4. \(60{,}245/84{,}245\text{.}\) Your final balance will be about 71.5%. interest.

1. Jose: \(55000\left(1+\frac{0.056}{12}\right)^{12*25}\)
   Or, =FV(0.056/12,12*25,0,55000)
   Jose’s partner:\(\frac{375\left(\left(1+\frac{0.056}{12}\right)^{12*25}-1\right)}{\left(\frac{0.056}{12}\right)}\)
   Or, =FV(0.056/12,12*25,375)
   Jose will have $222,310.85 and his partner will have $244,447.68.
2. Jose: \(222{,}310.85 – 55{,}000\)
   Jose’s partner: \(24{,}4447.68 – 375*12*25\)
   Jose will earn $167,310.85 and his partner will earn $131,947.68 in interest.
3. Jose: \(167{,}310.85/222{,}310.85\)
   Jose’s partner: \(13{,}1947.68/244{,}447.68\)
   Jose’s final balance will be about 75.3% interest and Jose’s partner’s final balance will be about 54.0% interest.

1. Akiko: \(45000\left(1+\frac{0.078}{12}\right)^{12*30}\)
   Or, =FV(0.078/12,12*30,0,45000)
   Her spouse: \(\frac{200\left(\left(1+\frac{0.078}{12}\right)^{12*30}-1\right)}{\left(\frac{0.078}{12}\right)}\)
   Or, =FV(0.078/12,12*30,200)
   Akiko will have $463,631.61 and her spouse will have $286,243.83.
2. Akiko: \(463{,}631.61 – 45{,}000\)
   Her spouse: \(286{,}243.83 – 200*12*30\)
   Akiko will earn $418,631.61 and her spouse will earn $214,243.83 in interest.
3. Akiko: \(418{,}631.61/463{,}631.61\)
   Her spouse: \(214{,}243.83/286{,}243.83\)
   Akiko’s final balance will be about 90.3% interest and her spouse’s final balance will be about 74.8% interest.

1. \(1000\left(1+\frac{0.045}{12}\right)^{12*10}+\frac{100\left(\left(1+\frac{0.045}{12}\right)^{12*10}-1\right)}{\frac{0.045}{12}}\)
   Or, =FV(0.045/12,12*10,100,1000)
   Sylvin will have a final balance of $16,686.80.
2. \(16{,}686.8 – (1000 + 100*12*10)\text{.}\) Sylvin will earn $3,686.80 in interest.
3. \(3{,}686.8/16{,}686.8\text{.}\) The final balance will be about 22.1% interest.

1. \(5000(1+0.0235)^{20}+\frac{1000\left(\left(1+0.0235\right)^{20}-1\right)}{0.0235}\)
   =FV(0.0235,20,1000,5000)
   Elena’s final balance will be $33,119.05.
2. \(33{,}119.05 – (5000 + 1000*20)\text{.}\) Elena will earn $8,119.05 in interest.
3. \(8{,}119.05/33{,}119.05\text{.}\) The final balance will be about 24.5% interest.

1. \(\frac{100\left(\left(1+\frac{0.04}{12}\right)^{12*25}-1\right)}{\frac{0.04}{12}}\)
   Or, =FV(0.04/12,12*25,100)
   Vanessa will have $51,412.95 when she turns 65.
2. \(\frac{100\left(\left(1+\frac{0.04}{12}\right)^{12*40}-1\right)}{\frac{0.04}{12}}\)
   =FV(0.04/12,12*40,100
   Vanessa would have $118,196.13 if she had started saving when she was 25.

1. \(\frac{1000\left((1+0.05)^{20}-1\right)}{0.05}\)
   Or, =FV(0.05,20,1000)
   Chris will have $33,065.95 after 20 years.
2. \(\frac{1000\left((1+0.05)^{40}-1\right)}{0.05}\)
   Or, =FV(0.05,40,1000)
   Chris will have $120,799.77 after 40 years.

1. \(\frac{50\left(\left(1+\frac{0.035}{52}\right)^{52*18}-1\right)}{\frac{0.035}{52}}\)
   Or, =FV(0.035/52,52*18,50)
   They will have saved $65,164.37 after 18 years.
2. \(\frac{65164.37}{(1+\frac{0.035}{52})^{52*18}}\)
   Or, =PV(0.035/52,52*18,0,65164.37)
   They would have needed an deposit of $34,713.37.

1. \(\frac{100\left(\left(1+\frac{0.045}{12}\right)^{12*20}-1\right)}{\frac{0.045}{12}}\)
   Or, =FV(0.045/12,12*10,100)
   Elisa will have $15,119.81 in 10 years.
2. \(\frac{15119.81}{\left(1+\frac{0.045}{12}\right)^{12*10}}\)
   Or, =PV(0.045/12,12*10,0,15119.81)
   She would have needed an initial deposit of $9,648.93.

1. \(P = \frac{700\left(1-\left(1+\frac{5.5\%}{12}\right)^{-12\times 30}\right)}{\frac{5.5\%}{12}}\text{,}\) which gives \(P \approx \$123{,}285.23\)
   or =PV(0.055/12,12*30,-700) [Note 700 is entered as negative, to signify a payment]
2. \(700 * 12 * 30\) dollars, or $252,000.00 in total payments to the loan company
3. Interest will be the difference between the total payments, and the amount borrowed. So the interest on this loan is \(\$252{,}000.00 - \$123{,}285.23 = \$128{,}714.77\text{.}\)

1. \(P = \frac{250\left(1-\left(1+\frac{7\%}{12}\right)^{-12 \times 5}\right)}{\frac{7\%}{12}}\text{,}\) which gives \(P \approx \$12{,}625.50\)
   or =PV(0.07/12,12*5,-250)
2. \(250 * 12 * 5\) dollars, or $15,000.00 in total payments to the loan company.
3. Interest will be the difference between the total payments, and the amount borrowed. So the interest on this loan is \(\$15{,}000.00 - \$12{,}625.50 = \$2{,}374.50\text{.}\)

1. The loan amount will be 90% of $200,000.00

   \begin{gather*} = (0.9 * \$200{,}000.00)\\ = \$180{,}000.00 \end{gather*}
2. \(d = \frac{180000\left(\frac{5\%}{12}\right)}{\left(1-\left(1+\frac{5\%}{12}\right)^{-12 \times 30}\right)}\text{,}\) which gives \(d \approx \$966.28\)
   or =PMT(0.05/12,12*30,180000)
3. \(d = \frac{180000\left(\frac{6\%}{12}\right)}{\left(1-\left(1+\frac{6\%}{12}\right)^{-12 \times 30}\right)}\text{,}\) which gives \(d \approx \$1{,}079.19\)
   or =PMT(0.06/12,12*30,180000)

1. \(d = \frac{270000\left(\frac{6.5\%}{12}\right)}{\left(1-\left(1+\frac{6.5\%}{12}\right)^{-12 \times 30}\right)}\text{,}\) which gives \(d \approx 1{,}706.58\)
   or =PMT(0.065/12,12*30,270000)
2. The total of all loan payments will be \((12 * 30 * \$1{,}706.58) = \$614{,}368.80\text{.}\)
   So the total interest paid will be \(\$614{,}368.80 – \$270{,}000.00 = \$344{,}368.80\text{.}\)
3. \((\$344{,}368.80 / \$614{,}368.80)\text{,}\) or \(\approx 56.0525\%\)

1. Take the income minus the adjustments \(135000-5600=\$129{,}400\text{.}\) Their adjusted gross income is $129,400.
2. They should take the standard deduction because itemizing saves them less.
3. Taxable income: \(129400-29200=100200\)
   Taxes owed: \(10852+0.22*(100200-94300)=\$12{,}150\)
   They owe $12,150 in taxes.
4. \(12150-15000=-\$2{,}850\) Take the taxes owed minus credits and withholdings. They will receive a refund for $2,850.

• Francis: Taxable Income \(45000-8000-14600 = \$22{,}400\)
• Tax from Table: \(1160+0.12*(22400-11600) = \$2{,}456\)
• Owed/Refund: \(2456 – 3000 - 4000 = -\$4{,}544\)

• Edward: Taxable Income \(50000-4000-14600 = \$31{,}400\)
• Tax from Table: \(1160+0.12*(31400-11600) = \$3{,}536\)
• Owed/Refund: \(3536 - 3500 - 1000 = -\$964\)

• Taxable income: \((45000+50000) – (8000+4000) – 29200 = \$53{,}800\)
• Tax from Table: \(2320 + 0.12*(53800-23200) = \$5{,}992\)
• Owed/Refund: \(5992 – (3000 + 3500) – (4000+1000) = -\$5{,}508\)

1. Gross Income: \(96000+850 = \$96{,}850\)
2. Adjusted Gross Income: \(96850-26000 = \$70{,}850\)
3. She should take itemized deductions since they are greater than the standard deduction for head of household.
4. Taxable Income: \(70850-22600 = \$48{,}250 \)
5. Tax from Table: \(1665 + 0.12*(48250 - 16650) = \$5{,}457\text{.}\) Owed/Refund: \(5457 – 4000 – 5300 = - \$3{,}843\)
   Janice will receive a refund for $3,843.

1. \(I=1525(0.056)(14)\)
   \(A=1525+1525(0.056)(14)\)
   The interest is $1,195.6 and balance is $2,720.6.
2. \(A=1525(1+\frac{0.056}{4})^{4\cdot14}\)
   Or, =FV(0.056/4, 4*14, 0, 1525)
   The interest is $1,796.97 and the balance is $3,321.97.
3. \(A=1525(1+\frac{0.056}{52})^{52\cdot14}\)
   Or, =FV(0.056/52, 52*14, 0, 1525)
   The interest is $1,813.67 and the balance is $3,338.67.
4. \(A=1525e^{0.056\cdot14}\)
   Or, =1525*EXP(0.056*14)
   The interest is $1,815.08 and the balance is $3,340.08.

1. \(A=3.49e^{0.0292\cdot10}\text{.}\) Or, =3.49*EXP(0.0292*10). The cost of gas would be $4.67.
2. \(A=1.99e^{0.0292\cdot10}\text{.}\) Or, =1.99*EXP(0.0292*10). The cost of eggs would be $2.66.
3. \(A=3.29e^{0.0292\cdot10}\text{.}\) Or, =3.29*EXP(0.0292*10). The cost of bread would be $4.41.
4. \(A=79.99e^{0.0292\cdot10}\text{.}\) Or, =79.99*EXP(0.0292*10) The cell phone bill would be $107.11.
5. \(A=1.99e^{0.0292\cdot10}\text{.}\) Or, =1.99*EXP(0.0292*10) The cost of a song download would be $2.66.

1. \(A= \frac{350((1+\frac{0.065}{12})^{12\cdot25}-1)}{\frac{0.065}{12}}\)
   Or, =FV(0.065/12, 12*25, 350, 0)
   The future value is $262,092.78 and the interest earned is $157,092.78.
2. \(A= \frac{500((1+\frac{0.065}{4})^{4\cdot15}-1)}{\frac{0.065}{4}}\)
   Or, =FV(0.065/4, 4*15, 500, 0)
   The future value is $50,168.34 and the interest earned is $20,168.34.
3. \(A= \frac{75((1+\frac{0.045}{52})^{52\cdot30}-1)}{\frac{0.045}{52}}\)
   Or, =FV(0.045/52, 52*30, 75, 0)
   The future value is $247,448.43 and the interest earned is $130,448.43.

1. \(43,000+1,000 = 44,000\text{.}\) Amir’s gross income is $44,000.
2. \(44,000-3,000=41000\text{.}\) Amir’s adjusted gross income is $41,000.

1. \(41,000-14,600= 26,400\text{.}\) Amir’s taxable income is $26,400.
2. \begin{gather*} =1160+0.12(26400-11600)\\ =1160+0.12(14800)\\ =1160+1776\\ =2936 \end{gather*}
   Amir owes $2936 in taxes.

1. All PCC Students
2. 200 students from PCC Cascade campus
3. The collected data is not representative of all PCC students since it only includes responses from students at Cascade campus. This is an example of sampling bias.

1. The intended population is all Washington County residences.
2. 1200 homes in Washington County
3. The collected data is likely representative since residences were selected at random from the entire county.

1. The representatives in a state’s congress.
2. The population size is \(n=106\)
3. The sample size is in \(n=28\)
4. The statistic is \(\frac{14}{28}=0.5\) or 50%
5. The confidence interval is \((45\%, 55\%)\) and tells us that the true percentage of the state congress representatives in support of the new education (the parameter) likely lies between 45% and 55%.

1. All registered voters in the city of Raleigh.
2. The population size is \(n=9500\)
3. The sample size is \(n=350\)
4. The statistic is \(\frac{112}{350}=0.32\) or 32%
5. The confidence interval is \((28.5\%, 35.5\%)\) and tells us that the true percentage of registered voters who will vote for Brown is likely to lie between 28.5% and 35.5%.

1. Stratified
2. Simple Random Sample
3. Systematic

1. Stratified
2. Volunteer
3. Simple Random Sample

1. Volunteer Bias
2. Sampling Bias
3. Response Bias
4. Non Response Bias
5. Response Bias
6. Loaded Question

1. Loaded Question
2. Volunteer Bias
3. Response Bias
4. Volunteer
5. Response Bias
6. Response Vias or Non-response Bias

1. Observational study
2. Experiment
3. Observational study

1. Observational study
2. Experiment
3. Observational study

1. Group 1
2. Group 2
3. Blind because the patients in the study do not know.
4. Controlled experiment

1. Cancer patients
2. No because sampling has variability
3. Stratified
4. Convenient Sample. It does not represent the population.

1. 2nd Group
2. Inert pill group
3. Double Blind because the patients and the advisors do not know who is in each group.
4. Placebo-controlled experiment

1. All students
2. Experiment
3. It is only looking at one class and not all groups that are in the population so Subjects are not randomly sampled from a specified population.

1. 0.05 or 5%
2. (25%, 35%)
3. I am confident that the percentage of college freshmen who prefer morning classes is between 25% to 35%.

1. 3.5% or 0.035
2. (34.5%, 41.5%)
3. I am confident that the percentage of all U.S. Employees are engaged at work is between 34.5% and 41.5%.

1. \((24+36)/2=30\text{.}\) The statistic is 30%.
2. \(30-24=6\text{.}\) The margin of error is 6%.

1. \((44+52)/2=48\text{.}\) The statistic is 48%.
2. \(48-44=4\text{.}\) The margin of error is 4%.

1. Quantitative
2. Categorical
3. Categorical
4. Quantitative
5. Quantitative

1. Quantitative
2. Categorical
3. Quantitative
4. Quantitative
5. Quantitative

1. 2 had 3 children.
2. 15 adults were questioned.
3. 33.33% of the adults questioned had 0 children.

1. 8 movies took 2 days to arrive.
2. She ordered 19 movies total.
3. 21.05% of the movies arrived in one day.

1. These data are categorical.
2. 
3. No, we cannot make a pie chart out of these data. The total of the relative frequencies is 1932, but only 1012 adults were asked. So some adults selected multiple options.

1. These data are categorical.
2. 
3. No, we cannot make a pie chart for these data. The relative frequencies add to 177%, which is more than 100%.

1. 40 households heat their home with firewood.
2. 50% of the households heat their home with natural gas.

1. 54% of the students are below senior class.
2. 9 of the sampled students are freshmen.

1. These data are qualitative.

2. Number of cars in household	Frequency
   0-1	7
   2-3	14
   4-5	3

3. 
4. These data are unimodal and skewed right, with no outliers.

1. These are quantitative data.

2. Math Test Score	Frequency
   40-49	1
   50-59	4
   60-69	3
   70-79	6
   80-89	5
   90-99	2
   100-109	3

3. 
4. These data are unimodal. There are no obvious outliers. The data are either symmetric or skewed left.

1. Normal distribution – The number of heads in 24 sets of 100 coin flips.
2. Positive or right skewed – Distribution of scores on a psychology test.
3. Negative or left skewed – Scores on a 20-point statistics quiz.
4. Bimodal – The frequency of times between eruptions of the Old Faithful geyser.

1. This distribution is unimodal and right skewed. There is a possible outlier between 10 and 11.
2. This distribution is unimodal and symmetric. There are no outliers.
3. This distribution is unimodal and right skewed. There are no obvious outliers.
4. This distribution is multimodal and left skewed. There are no outliers.
5. This distribution is unimodal and left skewed. There are no obvious outliers.

1. 
   
2. The data for patients of both researchers are symmetric. Researcher’s 1 patients’ data appears to be unimodal, but Researcher 2’s patients’ data may be bimodal or multimodal. The data for Researcher 1’s patients does not have any outliers, but the data for Researcher’s 2 may have outliers between 0 and 5 months or 40 and 45 months.

1. In Excel:
   =average(7.50,25,10,10,7.50,8.25,9,5,15,8,7.25,7.50,8,7,12)
   \(=\$9.80\)
   There are 15 amounts shown, so \(n=15\text{.}\) The mean is:
   \begin{align*} \bar{x} \amp= \frac{(7.50+25+10+10+7.50+8.25+9+5+15+8+7.25+7.50+8+7+12)}{15}\\ \amp=\$9.80 \end{align*}
2. In Excel:
   =median(7.50,25,10,10,7.50,8.25,9,5,15,8,7.25,7.50,8,7,12)
   \(=\$8.00\)
   There are 15 times shown, so \(n=15\text{.}\) We start by listing the data in order:
   $5, $7, $7.25, $7.50, $7.50, $7.50, $8, $8, $8.25, $9, $10, $10, $12, $15, $25
   \(Median=\$8.00\)
3. Since the mean is greater than the median, we would expect the distribution will be skewed right.

1. In Excel:
   =average(10,12.75,7,9,9.75,6.5,12.5,12.5,8.75,17,10.5,2)
   \(=9.85\) minutes
   There are 12 times shown, so \(n=12\text{.}\) The mean is:
   \begin{align*} \bar{x} \amp= \frac{(10+12.75+7+9+9.75+6.5+12.5+12.5+8.75+17+10.5+2)}{12}\\ \amp=9.85 \text{ minutes} \end{align*}
2. In Excel:
   =median(10,12.75,7,9,9.75,6.5,12.5,12.5,8.75,17,10.5,2)
   \(=9.88\) minutes
   There are 12 times shown, so \(n=12\text{.}\) We start by listing the data in order:
   2, 6.5, 7, 8.75, 9, 9.75, 10, 10.5, 12.5, 12.5, 12.75, 17
   \begin{align*} Median\amp=\frac{9.75+10}{2}\\ \amp=9.88 \text{ minutes} \end{align*}
3. Because the mean and median are approximately equal, we would expect that the distribution is symmetric.

1. In Excel:
   =average(15.2,18.8,19.3,19.7,20.2,21.8,22.1,29.4)
   \(=20.81\) seconds
   There are 8 times shown, so \(n=8\text{.}\)
   \begin{align*} \bar{x} \amp= \frac{(15.2+18.8+19.3+19.7+20.2+21.8+22.1+29.4)}{15}\\ \amp=20.81 \text{ seconds} \end{align*}
2. In Excel:
   =median(15.2,18.8,19.3,19.7,20.2,21.8,22.1,29.4)
   \(=19.95\) seconds
   There are 8 times shown, so \(n=8\text{.}\) The times are given already in order:
   15.2, 18.8, 19.3, 19.7, 20.2, 21.8, 22.1, 29.4
   \begin{align*} Median\amp=\frac{19.7+20.2}{2}\\ \amp=19.95 \text{ seconds} \end{align*}
3. Since the mean and median are approximately equal, we would expect that the distribution is symmetric.

1. In Excel:
   =average(3.49,3.51,3.51,3.51,3.52,3.54,3.55,3.58,3.61)
   \(=3.536\) grams
   There are 9 weights shown, so \(n=9\text{.}\)
   \begin{align*} \bar{x} \amp= \frac{(3.49+3.51+3.51+3.51+3.52+3.54+3.55+3.58+3.61)}{9}\\ \amp=3.536 \text{ grams} \end{align*}
2. In Excel:
   =median(3.49, 3.51, 3.51, 3.51, 3.52, 3.54, 3.55, 3.58, 3.61)
   \(=3.52\) grams
   There are 9 weights shown, so \(n=9\text{.}\) The weights are already given in order:
   3.49, 3.51, 3.51, 3.51, 3.52, 3.54, 3.55, 3.58, 3.61
   \begin{gather*} Median=3.52 \text{ grams} \end{gather*}
3. Since the mean and median are approximately equal, we would expect the distribution to be symmetric.

1. In GeoGebra Classic, enter the costs into the column A and frequencies into column B of the spreadsheet and use the “One Variable Analysis” function. Then use the “Show Statistics” option.
   \begin{gather*} Mean=33.8 \text{ thousand dollars} \end{gather*}
   The sum of the frequencies is 75, so \(n=75\)
   \begin{align*} \bar{x}\amp=\frac{15\cdot 3+20\cdot 7+25\cdot 10+30\cdot 15+35\cdot 13+40\cdot 11+45\cdot 9+50\cdot 7}{75}\\ \amp=33.8 \text{ thousand dollars} \end{align*}
2. In GeoGebra Classic, enter the costs into the column A and frequencies into column B of the spreadsheet and use the “One Variable Analysis” function. Then use the “Show Statistics” option.
   \begin{gather*} Median=35 \text{ thousand dollars} \end{gather*}
   Since there are 75 values (an odd number), we know that the median will be the single middle data value. Because \(\frac{75}{2}=37.5\text{,}\) we know it will be the 38th value in the list. The 38th value is 35, so the median is 35 thousand dollars.
3. Since the mean is less than the median, we would expect the distribution to be skewed left.

1. In GeoGebra Classic, enter the costs into the column A and frequencies into column B of the spreadsheet and use the “One Variable Analysis” function. Then use the “Show Statistics” option.
   \begin{gather*} Mean=5.3 \text { thousand characters} \end{gather*}
   The sum of the frequencies is 34, so \(n=34\)
   \begin{align*} \bar{x}\amp=\frac{(0\cdot 4+1\cdot 5+2\cdot 2+3\cdot 3+4\cdot 3+5\cdot 1+6\cdot 3+7\cdot 3+9\cdot 3+10\cdot 3+11\cdot 2+14\cdot 2)}{34}\\ \amp=5.3 \text{ thousand characters} \end{align*}
2. In GeoGebra Classic, enter the costs into the column A and frequencies into column B of the spreadsheet and use the “One Variable Analysis” function. Then use the “Show Statistics” option.
   \begin{gather*} Median=4.5 \text{ thousand characters} \end{gather*}
   Since there are 34 values (an even number), we know that the median will be the mean of the two middle values. Because \(\frac{34}{2}=17\text{,}\) we know the two middle values are the 17th and 18th values. The 17th value is 4, and the 18th value is 5, so the median is 4.5 thousand characters.
3. Since the mean is greater than the median, we expect the distribution to be skewed right.

1. For Researcher 1:
   In Excel:
   =average(A1:A40)
   \(=23.6\) months.
   =median(A1:A40)
   \(=24\) months
   The mean for Researcher 1’s patients is 23.6 months, and the median for Researcher’s 1 patients is 24 months.
   For Researcher 2:
   In Excel:
   =average(A1:A40)
   \(=22.8\) months
   =median(A1:A40)
   \(=22\) months
   The mean for Researcher 2’s patients is 22.8 months, and the median is 22 months
2. Both the mean and median for Researcher 1’s patients are greater than the mean and median for Researcher 2’s patients. So, on average, Researcher 1’s patients have a longer life time after starting the cancer treatment than Researcher 2’s patients.

1. For Males:
   In Excel

   =average(53000,70000,12800,30000,4500,42000,48000,60000,108000,11000)

   \(=\$43,930\)

   =median(53000,70000,12800,30000,4500,42000,48000,60000,108000,11000)

   \(=\$45,000\)
   The mean for males is $43,930, and the median for males is $45,000.
   For Females:

   =average(1600,1200,20000,25000,670,29000,44000,30000,5800,50000)

   \(=\$20,727\)

   =median(1600,1200,20000,25000,670,29000,44000,30000,5800,50000)

   \(=\$22,500\)
   The mean for females is $20,727, and the median for females is $22,500
2. Both the mean income and median income for the males in the sample were twice as large as the mean income and median income for the females in the sample. The difference between the mean and median income for the females was slightly larger than the mean and median income for the males, so the distribution of incomes for the females is possibly more skewed than the distribution for the males.

1. 

   

   

2. The mean number of pieces correctly remembered for non-players was 33.65 pieces.
   The mean number of pieces correctly remembered for beginners was 47.6 pieces.
   The mean number of pieces correctly remembered for tournament players was 64.98 pieces.
3. The median number of pieces correctly remembered for non-players was 33.5 pieces.
   The median number of pieces correctly remembered for beginners was 51.3 pieces.
   The median number of pieces correctly remembered for tournament players was 71.1 pieces.
4. The distribution for non-players appears to be uniform. The distribution for beginners looks unimodal and left-skewed. The distribution for tournament players appears bimodal and symmetric.
   The mean and median number of pieces correctly remembered were both greatest for tournament players, with non-players having the smallest mean and median of pieces correctly remembered.

1. 

   

   

   

2. The mean leniency for the false smile group was 5.4.
   The mean leniency for the felt smile group was 4.9.
   The mean leniency for the miserable smile group was 4.9.
   The mean leniency for the neutral control group was 4.1.
3. The median leniency for the false smile group was 5.5.
   The median leniency for the felt smile group was 4.8.
   The median leniency for the miserable smile group was 4.8.
   The median leniency for the neutral control group was 4.0.
4. Answers will vary depending on the graphs created. The shape of the false smile is bimodal, whereas the miserable smile and neutral control groups are both unimodal. The miserable smile and neutral control both appear to be skewed to the right, however, comparing the mean to the median we see that there is not much of a difference. Therefore, none of the graphs are skewed. The felt smile group is the most visually uniform. The measures of center (mean and median) have the center of the false smile near 5.5, the felt smile and miserable smile both near 4.8, and the control near 4

1. There are many possible answers for this problem. Three data sets with 5 values each that have the same mean but different medians are:

   0,	0,	0,	0,	10
   0,	0,	2,	4,	4
   0,	1,	1,	1,	7

2. There are many possible answers for this problem. Three data sets with 5 values that have the same median but different means are:

   10,	10,	10,	10,	10
   0,	0,	10,	15,	20
   1,	5,	10,	10,	10

1. Argument for categorical: Because it is not clear that the shoe sizes represent a measure, the data can be considered categorical.
   Argument for quantitative: Because shoe size is a measurement that corresponds to the length of someone’s foot, it can be treated as quantitative data.
2. Each graph would have frequency along the y-axis. In a bar graph the bars would have spaces between them and each bar would be labeled with the shoe size. In a histogram there would not be any spaces between the bars and the shoe sizes could be the scale on the x-axis.
3. The mean shoe size to be 7.2 and the median shoe size to be 7.
   The mean shoe size is:
   \begin{align*} \bar{x}\amp=\frac{(5\cdot 4+6\cdot 4+7\cdot 6+8\cdot 6+9\cdot 5)}{25}\\ \amp=7.2 \end{align*}
   Since there are 25 values (an odd number), we know that the median will be the single middle data value. Because \(\frac{25}{2}=12.5\text{,}\) we know it will be the 13th value in the list. The 13th value is 7, so the median shoe size is 7.
   The mean and median shoe size might be useful statistics to the store. If shoe size is positively correlated to height, then a shoe store with a comparatively larger mean or median shoe size could determine that their clients are, on average, comparatively taller. (Other answers are possible.)

1. This graph is skewed left.
2. I expect that the mean is less than the median because the graph is skewed left.

1. This graph is symmetric. The graph has a single peak between $65,000 an $66,000, and there are approximately an equal number of data values on either side of this peak.
2. I expect that the mean and median are equal because the graph is symmetric.

1. Incomes are skewed to the right so the mean would be greater than the median.
2. Weights are approximately symmetric so the mean would be about the same as the median.
3. Number of children is skewed to the right so the mean would be greater than the median.
4. Medical costs for all adults are most likely skewed to the right so the mean would be greater than the median.
5. Medical costs for adults 65+ may be symmetric or skewed to the right. Answer according to the shape you chose.

1. In Excel:
   Entering the data values into cells A1 through A15.
   \begin{align*} s\amp=stdev.s(A1:A15)\\ \amp=\$4.82 \end{align*}
   From Exercise 3.3.7.1, the mean is $9.80. There are 15 data values, so \(n=15\text{.}\)
   We will make a table of data values, their deviations from the mean, and the squared deviations:

   Data Value	Deviation	Deviation Squared
   \(7.5\)	\(7.5-9.8=-2.3\)	\((-2.3)^2=5.29\)
   \(25\)	\(25-9.8=15.2\)	\((15.2)^2=231.04\)
   \(10\)	\(10-9.8=0.2\)	\((0.2)^2=0.04\)
   \(10\)	\(10-9.8=0.2\)	\((0.2)^2=0.04\)
   \(7.5\)	\(7.5-9.8=-2.3\)	\((-2.3)^2=5.29\)
   \(8.25\)	\(8.25-9.8=-1.55\)	\((-1.55)^2=2.4\)
   \(9\)	\(9-9.8=-0.8\)	\((-0.8)^2=0.64\)
   \(5\)	\(5-9.8=-4.8\)	\((-4.8)^2=23.04\)
   \(15\)	\(15-9.8=5.2\)	\((5.2)^2=27.04\)
   \(8\)	\(8-9.8=-1.8\)	\((-1.8)^2=3.24\)
   \(7.25\)	\(7.25-9.8=-2.55\)	\((-2.55)^2=6.5\)
   \(7.5\)	\(7.5-9.8=-2.3\)	\((-2.3)^2=5.29\)
   \(8\)	\(8-9.8=-1.8\)	\((-1.8)^2=3.24\)
   \(7\)	\(7-9.8=-2.8\)	\((-2.8)^2=7.84\)
   \(12\)	\(12-9.8=2.2\)	\((2.2)^2=4.84\)

   Next, we add the squared deviations and get \(5.29 + 231.04 + 0.04 + 0.04 + 5.29 + 2.4 + 0.64 + 23.04 + 27.04 + 3.24 + 6.5 + 5.29 + 3.24 + 7.84 + 4.84 = 325.78\) dollars-squared.
   The sample standard deviation is:
   \begin{align*} s\amp=\sqrt{\frac{325.78}{14}}\\ \amp=\$4.82 \end{align*}
2. In GeoGebra:
   In GeoGebra Classic, enter the data values into the column A of the spreadsheet and use the “One Variable Analysis” function. Then use the “Show Statistics” option to find the five-number summary.

   Min	Q1	Median	Q3	Max
   $5	$7.50	$8	$10	$25

   From Exercise 3.3.7.1, the data listed in order is:
   $5.00, $7.00, $7.25, $7.50, $7.50, $7.50, $8.00, $8.00, $8.25, $9.00, $10.00, $10.00, $12.00, $15.00, $25.00
   Also from Exercise 3.3.7.1, there are 15 data values (\(n=15\)), and the median is $8.00. The lower half of the data is:
   $5.00, $7.00, $7.25, $7.50, $7.50, $7.50, $8.00
   The median of the lower half is $7.50, so the lower quartile \(Q_{1}\) is $7.50.
   The upper half of the data is:
   $8.25, $9.00, $10.00, $10.00, $12.00, $15.00, $25.00
   The median of the upper half is $10.00, so the upper quartile \(Q_{3}\) is $10.00.
   The smallest and largest data values are $5.00 and 25.00, respectively, so the min and max are $5.00 and $25.00. The five-number summary is:

   Min	Q1	Median	Q3	Max
   $5	$7.50	$8	$10	$25

3. The range is:
   \begin{align*} Range \amp= Max - Min\\ \amp=25-5\\ \amp=\$20 \end{align*}
   The interquartile range (IQR) is:
   \begin{align*} IQR \amp= Q_{3} - Q_{1}\\ \amp=10-7.5\\ \amp=\$2.50 \end{align*}
4. 

1. In Excel:
   I entered the data values into cells A1 through A12.
   The standard deviation is:
   \begin{align*} s\amp=stdev.s(A1:A12)\\ \amp=3.78 \text{ hours} \end{align*}
2. In GeoGebra:
   In GeoGebra Classic, enter the data values into the column A of the spreadsheet and use the “One Variable Analysis” function. Then use the “Show Statistics” option to find the five-number summary.

   Min	Q1	Median	Q3	Max
   2 hours	7.875 hours	9.875 hours	12.5 hours	17 hours

3. The range is:
   \begin{align*} Range \amp= Max - Min\\ \amp=17-2\\ \amp=15 \text{ hours} \end{align*}
   The interquartile range (IQR) is:
   \begin{align*} IQR \amp= Q_{3} - Q_{1}\\ \amp=12.5-7.875\\ \amp=3.625 \text{ hours} \end{align*}
4. 

1. In Excel:
   I entered the data values into cells A1 through A9.
   The standard deviation is:
   \begin{align*} s\amp=stdev.s(A1:A9)\\ \amp=4.068 \text{ seconds} \end{align*}
2. In GeoGebra:
   In GeoGebra Classic, enter the data values into the column A of the spreadsheet and use the “One Variable Analysis” function. Then use the “Show Statistics” option to find the five-number summary.

   Min	Q1	Median	Q3	Max
   15.2 seconds	19.05 seconds	19.95 seconds	21.95 seconds	29.4 seconds

3. The range is:
   \begin{align*} Range \amp= Max - Min\\ \amp=29.4-15.2\\ \amp=14.2 \text{ seconds} \end{align*}
   The interquartile range (IQR) is:
   \begin{align*} IQR \amp= Q_{3} - Q_{1}\\ \amp=21.95-19.05\\ \amp=2.9 \text{ seconds} \end{align*}
4. 

1. In Excel:
   I entered the data values into cells A1 through A9.
   The standard deviation is:
   \begin{align*} s\amp=stdev.s(A1:A9)\\ \amp=0.039 \text{ grams} \end{align*}
2. In GeoGebra:
   In GeoGebra Classic, enter the data values into the column A of the spreadsheet and use the “One Variable Analysis” function. Then use the “Show Statistics” option to find the five-number summary.

   Min	Q1	Median	Q3	Max
   3.49 grams	3.51 grams	3.52 grams	3.565 grams	3.61 grams

3. The range is:
   \begin{align*} Range \amp= Max - Min\\ \amp=3.61-3.49\\ \amp=0.12 \text{ grams} \end{align*}
   The interquartile range (IQR) is:
   \begin{align*} IQR \amp= Q_{3} - Q_{1}\\ \amp=3.565-3.51\\ \amp=0.055 \text{ grams} \end{align*}
4. 

1. In GeoGebra:
   In GeoGebra Classic, enter the costs into column A and frequencies into column B of the spreadsheet and use the “One Variable Analysis” function. Then use the “Show Statistics” option. The standard deviation is:
   \begin{gather*} s=9.58 \text{ thousand dollars} \end{gather*}
   From Exercise 3.3.7.5, the mean is 33.8 thousand dollars.
   The mean and the standaed deviation together tell us that, on average, the cars at the local dealership are $9,580 from the mean price of $33,800.
2. In GeoGebra:
   In GeoGebra Classic, enter the data values into the column A of the spreadsheet and use the “One Variable Analysis” function. Then use the “Show Statistics” option to find the five-number summary.

   Min	Q1	Median	Q3	Max
   15 thousand dollars	25 thousand dollars	35 thousand dollars	40 thousand dollars	50 thousand dollars

3. The range is:
   \begin{align*} Range \amp= Max - Min\\ \amp=50-15\\ \amp=35 \text{ thousand dollars} \end{align*}
   The interquartile range (IQR) is:
   \begin{align*} IQR \amp= Q_{3} - Q_{1}\\ \amp=40-15\\ \amp=25 \text{ thousand dollars} \end{align*}
4. 

1. In GeoGebra:
   In GeoGebra Classic, enter the costs into column A and frequencies into column B of the spreadsheet and use the “One Variable Analysis” function. Then use the “Show Statistics” option. The standard deviation is:
   \begin{gather*} s=4.20 \text{ thousand characters} \end{gather*}
   From Exercise 3.3.7.6, the mean is 5.3 thousand characters.
   The mean and standard deviation together tell us that on average the emails vary from the mean of 5.3 thousand characters by 4.2 thousand characters.
2. In GeoGebra:
   In GeoGebra Classic, enter the data values into the column A of the spreadsheet and use the “One Variable Analysis” function. Then use the “Show Statistics” option to find the five-number summary.

   Min	Q1	Median	Q3	Max
   0 thousand characters	1 thousand characters	4.5 thousand characters	9 thousand characters	14 thousand characters

3. The range is:
   \begin{align*} Range \amp= Max - Min\\ \amp=14-0\\ \amp=14 \text{ thousand characters} \end{align*}
   The interquartile range (IQR) is:
   \begin{align*} IQR \amp= Q_{3} - Q_{1}\\ \amp=9-1\\ \amp=8 \text{ thousand characters} \end{align*}
4. 

1. In GeoGebra:
   In GeoGebra Classic, enter the data values for Researcher 1 into column A of the spreadsheet, and enter the data values for Researcher 2 into column B of the spreadsheet. Then use the “Multiple Variable Analysis” function. Then use the “Show Statistics” function to display the sample standard deviation for each set of data values.
   The sample standard deviation for Researcher 1 is 11.25 months. The sample standard deviation for Research 2 is 11.38 months.
2. In GeoGebra:
   In GeoGebra Classic, enter the data values for Researcher 1 into column A of the spreadsheet, and enter the data values for Researcher 2 into column B of the spreadsheet. Then use the “Multiple Variable Analysis” function. Then use the “Show Statistics” function to display the sample standard deviation for each set of data values.
   The 5-number summary for Researcher 1 is:

   Min	Q1	Median	Q3	Max
   3 months	15 months	24 months	32.5 months	47 months

   The 5-number summary for Researcher 2 is:

   Min	Q1	Median	Q3	Max
   2 months	16 months	22 months	30 months	44 months

3. The range for Researcher 1 is:
   \begin{align*} Range \amp= Max - Min\\ \amp=47-3\\ \amp=44 \text{ months} \end{align*}
   The interquartile range (IQR) for researcher 1 is:
   \begin{align*} IQR \amp= Q_{3} - Q_{1}\\ \amp=32.5-15\\ \amp=17.5 \text{ months} \end{align*}
   The range for Researcher 2 is:
   \begin{align*} Range \amp= Max - Min\\ \amp=44-2\\ \amp=42 \text{ months} \end{align*}
   The interquartile range (IQR) for Researcher 2 is:
   \begin{align*} IQR \amp= Q_{3} - Q_{1}\\ \amp=30-16\\ \amp=14 \text{ months} \end{align*}
4. In GeoGebra Classic, enter the data values for Researcher 1 into column A of the spreadsheet, and enter the data values for Research 2 into column B of the spreadsheet. Then use the “Multiple Variable Analysis” function. Then select “Stacked BoxPlots” from the drop-down menu.
   
   Researcher 1 has a larger minimum, median, 3rd quartile, and maximum than Researcher 2. For Researcher 1, 50% of the patients live longer than 24 months after treatment, compared to 50% of patients living longer than 22 months after treatment for Researcher 1.
   Researcher 2 has less variation in the life times than Researcher 1, with an IQR of 14 months for Researcher 2, compared to an IQR 16.5 months for Researcher 1.

1. In GeoGebra:
   In GeoGebra Classic, enter the data values for males into column A of the spreadsheet, and enter the data values for females into column B of the spreadsheet. Then use the “Multiple Variable Analysis” function. Then use the “Show Statistics” function to display the sample standard deviation for each set of data values.
   The sample standard deviation for males is $31,530.66. The sample standard deviation for females $18,806.22.
2. In GeoGebra:
   In GeoGebra Classic, enter the data values for males into column A of the spreadsheet, and enter the data values for females into column B of the spreadsheet. Then use the “Multiple Variable Analysis” function. Then use the “Show Statistics” function to display the sample standard deviation for each set of data values.
   The 5-number summary for males is:

   Min	Q1	Median	Q3	Max
   $4,500	$12,800	$45,000	$60,000	$108,000

   The 5-number summary for females is:

   Min	Q1	Median	Q3	Max
   $670	$1,600	$22,500	$30,000	$50,000

3. The range for males is:
   \begin{align*} Range \amp= Max - Min\\ \amp=108000-4500\\ \amp=\$103,500 \end{align*}
   The interquartile range (IQR) for males is:
   \begin{align*} IQR \amp= Q_{3} - Q_{1}\\ \amp=60000-12800\\ \amp=\$47,200 \end{align*}
   The range for females is:
   \begin{align*} Range \amp= Max - Min\\ \amp=50000-670\\ \amp=\$49,330 \end{align*}
   The interquartile range (IQR) for females is:
   \begin{align*} IQR \amp= Q_{3} - Q_{1}\\ \amp=30000-1600\\ \amp=\$28,400 \end{align*}
4. 
   All of the values for the males’ 5-number summary are larger than the corresponding values for the females’ 5-number summary. The maximum income for males is more than twice the maximum income for females. Males have a significantly greater amount of variation in incomes than females, as shown by the larger range, IQR and standard deviation for males compared to females.

1. In GeoGebra:
   In GeoGebra Classic, enter the data values for non-players into column A of the spreadsheet, enter the data values for beginners into column B of the spreadsheet, and enter the data value for tournament players into column C of the spreadsheet. Then use the “Multiple Variable Analysis” function. Then use the “Show Statistics” function to display the sample standard deviation for each set of data values.
   The sample standard deviation for non-players is 8.033 chess pieces. The sample standard deviation for beginners is 9.031 chess pieces. The sample standard deviation for tournament players is 15.622 chess pieces.
2. In GeoGebra:
   In GeoGebra Classic, enter the data values for non-players into column A of the spreadsheet, enter the data values for beginners into column B of the spreadsheet, and enter the data value for tournament players into column C of the spreadsheet. Then use the “Multiple Variable Analysis” function. Then use the “Show Statistics” function to display the sample standard deviation for each set of data values.
   The 5-number summary for non-players is:

   Min	Q1	Median	Q3	Max
   22.1 chess pieces	26.2 chess pieces	32.6 chess pieces	39.7 chess pieces	43.2 chess pieces

   The 5-number summary for beginners is:

   Min	Q1	Median	Q3	Max
   32.5 chess pieces	39.1 chess pieces	48.4 chess pieces	55.7 chess pieces	57.7 chess pieces

   The 5-number summary for tournament palyers is:

   Min	Q1	Median	Q3	Max
   40.1 chess pieces	51.2 chess pieces	64.6 chess pieces	75.9 chess pieces	85.3 chess pieces

3. The range for non-players is:
   \begin{align*} Range \amp= Max - Min\\ \amp=43.2-22.1\\ \amp=21.2 \text{ chess pieces} \end{align*}
   The interquartile range (IQR) for non-players is:
   \begin{align*} IQR \amp= Q_{3} - Q_{1}\\ \amp=39.7-26.2\\ \amp=13.5 \text{ chess pieces} \end{align*}
   The range for beginners is:
   \begin{align*} Range \amp= Max - Min\\ \amp=57.7-32.5\\ \amp=25.2 \text{ chess pieces} \end{align*}
   The interquartile range (IQR) for beginners is:
   \begin{align*} IQR \amp= Q_{3} - Q_{1}\\ \amp=55.7-39.1\\ \amp=16.6 \text{ chess pieces} \end{align*}
   The range tournament players for is:
   \begin{align*} Range \amp= Max - Min\\ \amp=85.3-40.1\\ \amp=45.2 \text{ chess pieces} \end{align*}
   The interquartile range (IQR) for tournament players is:
   \begin{align*} IQR \amp= Q_{3} - Q_{1}\\ \amp=75.9-51.2\\ \amp=24.7 \text{ chess pieces} \end{align*}
4. 
   Tournament players did the best at remembering positions (as shown by all of the numbers of their 5-number summary being larger than the corresponding numbers for the other two groups). However, tournaments players were not completely superior to the other two groups; the best non-players remembered more chess pieces than the worst tournament players. Also tournament players had more variation in how much they.

1. In GeoGebra:
   In GeoGebra Classic, enter the data values for false smile, felt smile, miserable smile, and neutral control into columns A, B, C, and D, respectively, of the spreadsheet. Then use the “Multiple Variable Analysis” function. Then use the “Show Statistics” function to display the sample standard deviation for each set of data values.
   The sample standard deviation for the false smile group is 1.827. The sample standard deviation for the felt smile group is 1.681. The sample standard deviation for tournament players is 1.454. The sample standard deviation for the neutral control is 1.523.
2. In GeoGebra:
   In GeoGebra Classic, enter the data values for false smile, felt smile, miserable smile, and neutral control into columns A, B, C, and D, respectively, of the spreadsheet. Then use the “Multiple Variable Analysis” function. Then use the “Show Statistics” function to display the sample standard deviation for each set of data values.
   The 5-number summary for the false smile group is:

   Min	Q1	Median	Q3	Max
   2.5	3.5	5.5	6.5	9

   The 5-number summary for the felt smile group is:

   Min	Q1	Median	Q3	Max
   2.5	3.5	4.75	6	9

   The 5-number summary for the miserable smile group is:

   Min	Q1	Median	Q3	Max
   2.5	4	4.75	5.5	8

   The 5-number summary for the neutral control group is:

   Min	Q1	Median	Q3	Max
   2	3	4	5	8

3. The range for false smile is:
   \begin{align*} Range \amp= Max - Min\\ \amp=9-2.5\\ \amp=6.5 \end{align*}
   The interquartile range (IQR) for false smile is:
   \begin{align*} IQR \amp= Q_{3} - Q_{1}\\ \amp=6.5-3.5\\ \amp=3 \end{align*}
   The range for felt smile is:
   \begin{align*} Range \amp= Max - Min\\ \amp=9-2.5\\ \amp=6.5 \end{align*}
   The interquartile range (IQR) for felt smile is:
   \begin{align*} IQR \amp= Q_{3} - Q_{1}\\ \amp=6-3.5\\ \amp=2.5 \end{align*}
   The range for miserable smile is:
   \begin{align*} Range \amp= Max - Min\\ \amp=8-2.5\\ \amp=5.5 \end{align*}
   The interquartile range (IQR) for miserable smile is:
   \begin{align*} IQR \amp= Q_{3} - Q_{1}\\ \amp=5.5-4\\ \amp=1.5 \end{align*}
   The range for neutral control is:
   \begin{align*} Range \amp= Max - Min\\ \amp=8-2\\ \amp=6 \end{align*}
   The interquartile range (IQR) for neutral control is:
   \begin{align*} IQR \amp= Q_{3} - Q_{1}\\ \amp=5-3\\ \amp=2 \end{align*}
4. 
   The minimum and median for all the smile groups was larger than the minimum and median for the neutral control, indicating that a smile has some effect on leniency. The false smile seems to have been the most effective because the minimum, median, 3rd quartile, and maximum for the false group are greater than or equal to the corresponding values for any of the other groups.

1. There are many possible answers for this question. For example, the data sets {10, 10, 10, 10, 10} and {9, 9, 10, 11, 11} have the same mean of 10 units, but different standard deviations (0 and 1, respectively).
2. There are many possible answers for this question. For example, the data sets {2, 2, 2, 2, 2} and {9, 9, 9, 9, 9} have the same standard deviation of 0, but different means (2 and 9, respectively).

1. There are many possible answers for this question. The data sets {1, 2, 3, 4, 5, 6, 7} and {11, 12, 13, 14, 15, 16, 17} have the same IQR of 4, but different medians (4 and 14, respectively).
2. There are many possible answers for this question. The data sets {14, 14, 14, 14, 14, 14, 14} and {1, 1, 1, 14, 26, 26, 26} have the same median of 14, but different IQRs (0 and 25, respectively).

1. The 25th, 50th, and 75th percentiles are, respectively, the 1st quartile, median, and 3rd quartile for the data sets. Reading the boxplot for CPAs, the 25th, 50th, and 75th percentiles for CPAs’ salaries are, respectively, $40,000, $75,000, and $90,000. Reading the boxplot for actuaries, the 25th, 50th, and 75th percentiles for actuaries’ salaries are, respectively, $75,000, $90,000, and $94,000
2. Deshawn’s salary (the median salary for an actuary) is $90,000; Kelsey’s salary (the first quartile salary) is also $75,000. So Deshawn makes more than Kelsey, by $15,000.
3. 75% of actuaries make more than the median salary of a CPA ($75,000).
4. 25% of all CPAs earn less than all actuaries.

1. The 25th, 50th, and 75th percentiles for weekly study times for the juniors are 1 hour, 3 hours, and 4 hours, respectively. The 25th, 50th, and 75th percentiles for weekly study times the seniors are 5.5 hours, 6 hours, and 9 hours.
2. Olivia studies more each week than Lucy, by 30 minutes.
3. 50% of juniors study between the minimum and median number of hours for seniors.
4. 100% of seniors study more than the third quartile weekly study time for juniors.

1. \begin{align*} Z\amp=\frac{\text{data value}-\text{mean}}{\text{standard deviation}}\\ \amp=\frac{21.4-25}{1.15}\\ \amp=-3.13 \text{ standard deviations} \end{align*}
2. The \(Z\)-score for the gas mileage of the car is -3.13 standard deviations.

1. \begin{align*} Z\amp=\frac{\text{data value}-\text{mean}}{\text{standard deviation}}\\ \amp=\frac{170-274}{63}\\ \amp=-1.65 \text{ standard deviations} \end{align*}
   The marathon time’s \(Z\)-score is -1.65 standard deviations.
2. Because the marathon finishing time is within 2 standard deviations of the mean finishing time, no, this marathon finishing time is not usually fast.

1. In GeoGebra:
   In GeoGebra Classic, enter the data values into the column A of the spreadsheet and use the “One Variable Analysis” function. Then use the “Show Statistics” option to find the mean and standard deviation.
   The mean is 46.2 hours per year, and the standard deviation is 6.16 hours per year.
2. \begin{align*} Z\amp=\frac{\text{data value}-\text{mean}}{\text{standard deviation}}\\ \amp=\frac{42-46.2}{6.16}\\ \amp=-0.68 \text{ standard deviations} \end{align*}
   The \(Z\)-score for a city with an average delay time of 42 hours per year is -0.68 standard deviations.

1. In GeoGebra:
   In GeoGebra Classic, enter the data values into the column A of the spreadsheet and use the “One Variable Analysis” function. Then use the “Show Statistics” option to find the mean and standard deviation.
   The mean is 122.9 job applicants per job posting, and the standard deviation is 10.91 job applicants per job posting.
2. \begin{align*} Z\amp=\frac{\text{data value}-\text{mean}}{\text{standard deviation}}\\ \amp=\frac{143-122.9}{10.91}\\ \amp=1.84 \text{ standard deviations} \end{align*}
   The \(Z\)-score for a company with 143 job applicants per job posting is 1.84 standard deviations

1. The population is PCC students and 73,000 is the sample size.
2. The sample is 250 students from the 4 main campuses.
3. Stratified Sample
4. Categorical or Qualitative
5. \(\displaystyle 435/1000 = 0.435 = 43.5\%\)

1. The population being studied is American adults.
2. The sample is 500 American adults.
3. Categorical or Qualitative
4. The 62% reported in the problem an example of a statistic because it comes from a sample.
5. \(62\% - 4\% = 58\%\) and \(62\% + 4\% = 66\%\text{.}\) The confidence interval is \((58\%, 66\%)\) and it is in relation to the parameter.
6. We are confident that the true proportion of all adult Americans who favor a law to ban the sale of assault weapons and semi automatic rifles is between 58% and 66%.

1. The population being studied is PCC Students.
2. Categorical or Qualitative
3. 23% is a statistic because it is from a sample.
4. The margin of error is 4%.
5. \(23\% - 4\% = 19\%\) and \(23\% + 4\% = 27\%\text{.}\) The confidence interval is \((19\%, 27\%)\text{.}\)
6. We are confident that the true proportion of all PCC students who prefer to study at the library is between 19% and 27%

1. Systematic
2. Simple Random Sample
3. Convenience
4. Systematic

1. Voluntary Response Bias
2. Sampling Bias
3. Sampling Bias
4. Response bias
5. Perceived lack of anonymity

1. Observational
2. Experiment

1. The treatment group is the group receiving the test medicine for migraines.
2. The control group is the group receiving the inert pill.
3. This is a blind study.
4. This is a placebo-controlled experiment.

1. The treatment group is the group assigned to use the Rewire app.
2. The control group is the group that receives the usual services from the Department of Youth Services.
3. This study is neither blind nor double-blind because the researchers and the youth know whether they are using the Rewire app of receiving the usual services from the Department of Youth Services.
4. This is a controlled experiment.

1. The implied population is United States teenagers.
2. There were 2007 teens surveyed.
3. Qualitative or Categorical
4. 
5. 
6. Answers will vary
7. About 14.3% of teens chose horror as their favorite movie genre.

1. 426 people use TriMet for personal business.
2. 479 people people use TriMet to get to the airport.

1. The mean is $4.78 per gallon. The median is $4.75 per gallon.
2. The mean and median is about the same value therefore the data is symmetric.
3. The standard deviation is $1.34 per gallon.
4. \(z_{\$3.25}=\frac{\$3.25-\$4.78}{\$1.34}=\frac{-\$1.53}{\$1.34}=-1.14\)
   \(z_{\$8.95}=\frac{\$8.95-\$4.78}{\$1.34}=\frac{\$4.17}{\$1.34}=3.11\)
5. 
6. \(\text{Range}=\$8.95-\$3.25=\$5.70\)
   \(\text{IQR}=\$5-\$3.75=\$1.25\)
7. 

1. The mean is about 78.9.
2. The median is 78.
3. The mean is greater than the median therefore the data is left skewed.
4. The standard deviation is about 14.5.
5. The majority of values are between 64.4 and 93.4.
6. \(z=\frac{(99-78.9)}{14.5}=1.39\text{.}\) This 99 is only 1.39 standard deviations above the mean. This is not unusual. Generally, a value should be more than three deviations above or below the mean to be considered unusual.
7. 
8. 
9. 

1. The mean is about $26,200 and the standard deviation is about $10,000.
2. The majority of cars sell for between $16,200 and $36,200.
3. 
4. \(\text{Range} = 45 - 12 = 33\)
   \(\text{IQR}= 32 - 18 = 14\)
5. 

1. The 25th, 50th, and 75th percentile for Team A goals are 2, 4, and 5, respectively.
   The 25th, 50th, and 75th percentile for Team B goals are 6, 8, and 9, respectively.
2. The Median for Team A is 4.
   The Median for Team B is 8.
3. Fifty percent of the goals for Team B exceed the maximum number for goals for Team A.
4. The data for Team A is more symmetric than for Team B.
5. The data for Team B is skewed left.

1. \(\displaystyle z=\frac{(30.4-35)}{1.35}=\frac{-4.6}{1.35} \approx -3.41\)
2. 30.4 mpg is 3.41 standard deviations below the mean so the car is getting unusually low gas mileage.

1. The mean is approximately 50.6 minutes per week. The standard deviation approximately 12.2 minutes per week.
2. \(\displaystyle z=\frac{(42-50.6)}{12.2} \approx -0.70\)
3. 42 minutes per week is 0.70 standard deviations below the mean, so it is not unusual.

1. \begin{align*} \text{P(In morning class)}\amp=\frac{39}{65}\\ \amp= 0.60\text{ or } 60\% \end{align*}
2. \begin{align*} \text{P(Earned a C)}\amp=\frac{25}{65}\\ \amp\approx 0.385 \text{ or } 38.5\% \end{align*}
3. \begin{align*} \text{P(Earned an A and in afternoon class)}\amp=\frac{10}{65}\\ \amp\approx 0.154 \text{ or } 15.4\% \end{align*}
4. \begin{align*} \text{P(Earned an A given in morning class)}\amp=\frac{8}{39}\\ \amp\approx 0.205 \text{ or } 20.5\% \end{align*}
5. \begin{align*} \text{P(In morning class or earned B)}\amp=\frac{43}{65}\\ \amp\approx 0.662 \text{ or } 66.2\% \end{align*}