Each of the twenty solutions names the population and the sample, or shows the proportion and the multiplication that give an estimate. Estimates are marked with the word about. Compare your reasoning with the explanations, especially for the random-versus-biased problems, where several valid wordings can earn full credit.
Problem 1: Population and sample
The population is everyone Mr. Lopez wants to learn about: all \(720\) students.
The sample is the part he actually surveys: the \(45\) students.
Population: the 720 students; sample: the 45 students surveyed
Problem 2: Library books
The population is every item she wants to learn about: all \(4,200\) books in the library.
The sample is the part she actually checks: the \(60\) books.
Population: the 4,200 library books; sample: the 60 books checked
Problem 3: Sample or population?
(a) Population: she wants information only about her class, and all 28 students are included.
(b) Sample: the 30 shoppers are a part of everyone in the town.
(c) Population: every book is checked.
(a) population; (b) sample; (c) population
Problem 4: Random or not random
(a) Random: every student has the same chance of being drawn.
(b) Not random: students who arrive early are favored.
(c) Not random: cooking club members may like different lunches than most students.
Only (a) is random
Problem 5: Which sample is better?
Method B gives every student the same chance of being chosen, so the sample is random and is likely to represent the whole school.
Method A uses one team. Athletes may prefer different seasons than most students, so that sample can be biased.
Method B is more likely to be representative
Problem 6: Explain the bias
People at a Little League game are more likely to like baseball than people in general. Not every town resident had an equal chance of being surveyed, so the sample is not representative.
Biased: the sample favors baseball fans
Problem 7: Bus stop survey
People waiting at a bus stop already ride the bus, so they are more likely to favor a new line than adults in general. Not every adult in town had an equal chance of being surveyed.
Improvement: choose the adults at random, for example by drawing \(50\) names from a list of all adult residents.
Biased: the sample favors bus riders
Problem 8: Estimate a population count
The sample proportion is \(\frac{14}{40}=0.35\).
Estimate: \(0.35\times 600=210\).
About \(210\) students
Problem 9: Reusable bags
The sample proportion is \(\frac{20}{50}=0.4\).
Estimate: \(0.4\times 1500=600\).
About \(600\) shoppers
Problem 10: Estimate a mean
Sum: \(142+150+138+155+147+149+151+144+146+148=1470\).
Mean: \(1470: 10=147\).
About \(147\text{ cm}\)
Problem 11: Voter survey
The sample proportion is \(\frac{52}{80}=0.65\), or \(65\%\).
Estimate: \(0.65\times 2500=1625\).
About \(65\%\) of voters, or about \(1,625\) voters
Problem 12: Homework time
Sum: \(36+42+28+50+45+38+40+41=320\).
Mean: \(320: 8=40\).
Total for the school: \(40\times 640=25,600\).
About \(40\) minutes per student, about \(25,600\) minutes in all
Problem 13: Two samples, two estimates
Sample A: \(\frac{12}{30}=0.4\), and \(0.4\times 900=360\).
Sample B: \(\frac{15}{30}=0.5\), and \(0.5\times 900=450\).
The estimates differ because of sampling variation: each random sample includes different students.
360 students and 450 students
Problem 14: Quality control
The sample proportion is \(\frac{3}{50}=0.06\).
Estimate: \(0.06\times 4000=240\).
About \(240\) defective bulbs
Problem 15: Sample size
Liam: \(\frac{7}{10}=0.70\), and \(0.70\times 1200=840\).
Aisha: \(\frac{58}{100}=0.58\), and \(0.58\times 1200=696\).
Aisha’s estimate is likely more reliable, because a larger random sample varies less.
840 and 696; Aisha’s estimate is more reliable
Problem 16: Does most mean most?
(a) Sample 1: \(\frac{17}{30}\times 800\approx 453\). Sample 2: \(\frac{13}{30}\times 800\approx 347\).
(b) No. One estimate is more than half of \(800\) (more than \(400\)) and the other is less. Random samples vary, so a single small sample does not prove the claim.
(c) Combined: \(\frac{17+13}{30+30}=\frac{30}{60}=0.5\) and \(0.5\times 800=400\).
(a) about 453 and about 347; (b) no; (c) about 400 students, exactly half
Problem 17: Five samples of lunch buyers
(a) Mean: \((9+11+8+12+10): 5=50: 5=10\).
(b) Proportion: \(\frac{10}{20}=0.5\), and \(0.5\times 1200=600\).
(c) Smallest: \(\frac{8}{20}=0.4\), and \(0.4\times 1200=480\). Largest: \(\frac{12}{20}=0.6\), and \(0.6\times 1200=720\).
(a) 10; (b) about 600 students; (c) between 480 and 720 students
Problem 18: Burnt cookies
The sample proportion is \(\frac{6}{80}=0.075\).
Estimate: \(0.075\times 2400=180\) burnt cookies.
Loss: \(180\times 0.75=135\), so about $135.
About \(180\) burnt cookies, a loss of about \(\$135\)
Problem 19: Bass in the pond
Sample A: \(\frac{9}{50}=0.18\), and \(0.18\times 1250=225\).
Sample B: \(\frac{20}{100}=0.20\), and \(0.20\times 1250=250\).
Sample B is likely more reliable: a larger random sample varies less from the true proportion.
About 225 bass and about 250 bass; Sample B is more reliable
Problem 20: New park vote
(a) \(\frac{96}{150}=0.64\), and \(0.64\times 12000=7680\).
(b) Mean: \((96+90+99+93+107): 5=485: 5=97\). Proportion: \(\frac{97}{150}\), and \(\frac{97}{150}\times 12000=97\times 80=7760\).
(c) Smallest: \(\frac{90}{150}\times 12000=7200\). Largest: \(\frac{107}{150}\times 12000=8560\).
(a) about 7,680; (b) about 7,760; (c) between 7,200 and 8,560 residents
See every chapter on the Grade 7 math page.