Random Sampling and Comparing Populations: practice solutions, Grade 7

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Practice solutions Grade 7 : Random Sampling and Comparing Populations — Zyro the alien explorer of Planète Maths

Written solutions to the chapter problems. Check each step, then correct yourself.

1 Population or sample? ★★★

Population: it is the entire group of interest.
Sample: it is a part of the school’s students.
Population: it is every plant in the greenhouse.
Sample: ten plants are only part of the greenhouse.

2 Fair or biased? ★★★

Methods a and d are random, because every student has the same chance of being picked.

Method b is biased: early bus riders may live far away and may prefer different trips. Method c is biased: drama club members may share interests that other students do not.

3 Mean, median, and range ★★★

Sum: $8+12+9+11+10+10=60$, so the mean is $\dfrac{60}{6}=10$ minutes.

In order: 8, 9, 10, 10, 11, 12. The middle values are 10 and 10, so the median is 10 minutes.

Range: $12-8=4$ minutes.

4 Favorite sport ★★★

Sample proportion: $\dfrac{24}{40}=0.6$.

Estimate for the school: $0.6\times 1200=720$.

About 720 students in the school are likely to favor volleyball.

5 Reading a dot plot ★★★

Values: 1, 2, 2, 3, 3, 3, 3, 4, 4, 5.

Mode: 3 books (most dots). Median: the 5th and 6th values are both 3, so the median is 3 books. Range: $5-1=4$ books.

Mean: $\dfrac{1+2+2+3+3+3+3+4+4+5}{10}=\dfrac{30}{10}=3$ books.

6 First MAD ★★★

Mean: $\dfrac{4+6+8+10+12}{5}=8$.

Distances from 8: 4, 2, 0, 2, 4. Their sum is 12.

$\text{MAD}=\dfrac{12}{5}=2.4$.

7 True or false? ★★★

True. With more members, the sample tends to look more like the population.
False. Three people can easily be unusual; 300 randomly chosen people vary much less from sample to sample.
False. Basketball fans are not representative of the whole town, so the sample is biased.

8 Counting maple trees ★★★

Sample proportion: $\dfrac{12}{30}=0.4$.

Estimate: $0.4\times 450=180$ maple trees.

It is an estimate because a different random sample of 30 trees could contain a different number of maples.

9 Two bakeries ★★★

Bakery A: mean $\dfrac{80}{5}=16$ oz. Distances: 1, 0, 0, 0, 1, so $\text{MAD}=\dfrac{2}{5}=0.4$ oz.

Bakery B: mean $\dfrac{80}{5}=16$ oz. Distances: 2, 1, 0, 1, 2, so $\text{MAD}=\dfrac{6}{5}=1.2$ oz.

Both bakeries average 16 oz, but Bakery A has the smaller MAD, so its loaves are more consistent.

10 Five-number summary ★★★

The data are already in order and $n=10$.

Median: $\dfrac{9+11}{2}=10$. Lower half: 3, 5, 6, 8, 9, so $Q_1=6$. Upper half: 11, 12, 14, 15, 17, so $Q_3=14$.

Minimum 3, maximum 17. $\text{IQR}=14-6=8$ minutes.

11 Small sample, big sample ★★★

Lena: $\dfrac{5}{8}=0.625=62.5\%$. Marcus: $\dfrac{36}{80}=0.45=45\%$.

Marcus’s estimate is likely to be more reliable, because it comes from a larger random sample; a sample of only 8 students can easily be far from the true percent.

12 Water bills ★★★

(a) Sum: 516, so the sample mean is $\dfrac{516}{12}=43$. We estimate a mean bill of about $43.

(b) $43\times 2000=86,000$. The total is about $86,000 per month.

13 Gap measured in MADs ★★★

Difference of means: $86-78=8$. Number of MADs: $\dfrac{8}{4}=2$.

The means are 2 MADs apart. That is a fairly noticeable gap compared with the spread, though the two distributions can still overlap.

14 Comparing two box plots ★★★

Group 1 median is 6 and Group 2 median is 8, so Group 2 is greater by 2 thousand steps.
Group 1: $8-4=4$. Group 2: $9-6=3$.
The difference 2 is smaller than both IQRs (4 and 3), so the boxes overlap and the groups are not very different.

15 Spotting and fixing bias ★★★

Plan A is biased because club members may differ from students who leave right away, for example in how they get home. Plan B is a voluntary response sample: people with strong feelings are most likely to answer.

A better plan: number every student from 1 to 600, use a random number generator to pick about 60 numbers, and ask exactly those students. Every student then has an equal chance of being chosen.

16 Same mean, different spread ★★★

Player A: sum 152, mean 19. Median $\dfrac{18+20}{2}=19$. Distances 5, 3, 1, 1, 1, 1, 3, 5 add to 20, so $\text{MAD}=\dfrac{20}{8}=2.5$.

Player B: sum 152, mean 19. Median $\dfrac{19+21}{2}=20$. Distances 11, 7, 3, 0, 2, 4, 7, 8 add to 42, so $\text{MAD}=\dfrac{42}{8}=5.25$.

Both players average 19 points, but Player B’s MAD is about twice Player A’s, so B’s scoring is much less consistent.

17 Defective bulbs ★★★

(a) Mean: $\dfrac{4+6+2+8+6+4}{6}=\dfrac{30}{6}=5\%$. Then $0.05\times 20000=1000$ bulbs.

(b) $0.02\times 20000=400$ and $0.08\times 20000=1600$. A reasonable range is about 400 to 1,600 defective bulbs.

18 Missing score ★★★

(a) The total must be $5\times 82=410$. The known scores add to 333, so the fifth score is $410-333=77$.

(b) Distances from 82: 4, 3, 8, 2, 5, which add to 22. $\text{MAD}=\dfrac{22}{5}=4.4$.

19 Tagging fish ★★★

In the sample, $\dfrac{6}{30}=\dfrac{1}{5}$ of the fish are tagged. If this matches the whole lake, then the 40 tagged fish are $\dfrac{1}{5}$ of all the fish.

So the lake has about $40\times 5=200$ fish.

20 Does fertilizer help? ★★★

No fertilizer: sum 160, mean 16. Distances 4, 2, 1, 1, 0, 0, 1, 2, 2, 3 add to 16, so $\text{MAD}=1.6$.

With fertilizer: sum 190, mean 19. The distances are the same list, so $\text{MAD}=1.6$.

Difference of means: $19-16=3$, and $\dfrac{3}{1.6}=1.875$, about 2 MADs.

The ranges overlap from 15 to 19 cm, so some unfertilized seedlings are taller than some fertilized ones, yet the fertilized group is clearly taller on average.