Sampling, Inferences & Comparing Populations — Practice
You cannot measure everyone, so you measure some. A random sample lets a small group speak honestly for a large one.
What these questions cover
These questions use 12 ideas from the course. Follow any of them for a definition, where it is taught, and what it leads to.
- Rate — A ratio of two different units, like miles per hour or dollars per pound.
- Percent — Parts per hundred. 37% = 37100 = 0.37.
- Population — Every member of the group you care about.
- Sample — The part of the population you actually measure.
- Random sample — Every member has an equal chance of being picked — the only kind you can trust.
- Mean — Two kinds of centre: the balance point, and the middle value.
- Median — Two kinds of centre: the balance point, and the middle value.
- MAD — Mean absolute deviation — the average distance from the mean; a measure of spread.
- Scale — What one grid step is worth — and where the axis starts.
- Outlier — A dot far away from the rest of the pattern.
- Mode — The value that appears most often.
- Mean absolute deviation — The average distance of the values from the mean.
Problem 1
Find the mean of 4, 8, 10, 14.
Worked answer
9
4 + 8 + 10 + 14 = 36
36 ÷ 4 = 9
Problem 2
Find the median of 3, 7, 9, 12, 15.
Worked answer
9
Five values, so the middle is the 3rd
The median is 9.
Problem 3
Find the median of 4, 6, 9, 11.
Worked answer
7.5
Middle two are 6 and 9
6 + 92 = 7.5
Problem 4
In a sample of 50 students, 15 prefer pizza. What percent is that?
Worked answer
30
1550 = 0.30
= 30%
Problem 5
Find the mode of 2, 3, 3, 5, 7, 3, 8.
Worked answer
3
3 appears three times, more than any other value.
Problem 6
In a random sample of 200 voters, 84 support a measure. Estimate how many of 5,000 voters support it.
Worked answer
2100
84200 = 0.42
0.42 × 5000 = 2100
About 2,100 voters.
Problem 7
The mean of 5 numbers is 12. What is their sum?
Worked answer
60
12 × 5 = 60
Problem 8
Four tests average 78. What score on a fifth test would raise the average to 80?
Worked answer
88
Current total: 78 × 4 = 312
Target total: 80 × 5 = 400
400 - 312 = 88
Problem 9
Find the mean absolute deviation of 4, 8, 10, 14.
Worked answer
3
Distances from 9: 5, 1, 1, 5
5 + 1 + 1 + 5 = 12
12 ÷ 4 = 3
Problem 10
You want to know how much TV the average 7th grader watches. Which sample is biased?
- 60 students drawn at random from the full roster
- Every 5th student leaving the building at dismissal
- Students waiting outside the school television studio
- 60 students picked at random from each grade list
Worked answer
Students waiting outside the school television studio
Students at the TV studio are far more interested in television than a typical student.
That makes them over-represent heavy viewers — a biased sample.
The other three give every student a fair chance of selection.
Problem 11
A data set is 12, 15, 15, 18, 140. Which measure better describes a typical value?
- The mean
- The median
- They are equally good
Worked answer
The median
Mean = 2005 = 40
Median = 15
No value is near 40 — the outlier 140 dragged the mean away from the data.
Problem 12
A factory samples 80 items and finds 24 defective. Estimate the number of defective items in a batch of 1,500.
Worked answer
450
2480 = 0.30
0.30 × 1500 = 450
Problem 13
Six numbers have a mean of 20. One of them, 35, is removed. What is the mean of the remaining five?
Worked answer
17
Total: 20 × 6 = 120
After removing: 120 - 35 = 85
85 ÷ 5 = 17
Problem 14
The mean of seven numbers is 15. Every number is increased by 4. What is the new mean?
Worked answer
19
The total rises by 7 × 4 = 28
New total: 105 + 28 = 133
133 ÷ 7 = 19 — shifting every value shifts the mean by the same amount.
Problem 15
Class A has mean 84 and MAD 2. Class B has mean 88 and MAD 3. How large is the mean gap relative to the MADs?
- Twice A's MAD and about 1.33 times B's MAD
- Half of each MAD
- Exactly equal to each MAD
Worked answer
Twice A's MAD and about 1.33 times B's MAD
The gap between means is 88-84=4.
Relative to A: 4/2=2 MADs. Relative to B: 4/3≈1.33 MADs.
This describes the separation; it is not a statistical significance test.
Problem 16
A biologist tags 60 fish and releases them. Later a sample of 45 fish contains 9 tagged ones. Estimate the lake's fish population.
Worked answer
300
Sample fraction tagged: 945 = 15
So the 60 tagged fish are about 15 of the lake
60 × 5 = 300 fish