Explanation:
Let’s go step by step through each part.
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Section A – Dice rolled 200 times
First, fill in the *Relative Frequency* column.
Relative Frequency = Frequency ÷ Total Trials
Total trials = 200.
- For outcome 1: 20 ÷ 200 = 0.10
- Outcome 2: 30 ÷ 200 = 0.15
- Outcome 3: 25 ÷ 200 = 0.125
- Outcome 4: 20 ÷ 200 = 0.10
- Outcome 5: 25 ÷ 200 = 0.125
- Outcome 6: 80 ÷ 200 = 0.40
So relative frequencies:
1 → 0.10
2 → 0.15
3 → 0.125
4 → 0.10
5 → 0.125
6 → 0.40
Now answer the questions:
1) If the dice were rolled 1000 times, how many times would you expect to get a score of 6?
Use the relative frequency from the experiment: 0.40
Expected count = 0.40 × 1000 =
400
2) Do you think the dice is fair? Why?
A fair 6-sided die should have each outcome ~1/6 ≈ 0.1667 relative frequency.
Here, outcome 6 has 0.40 — much higher than expected. Others are lower.
So:
No, it’s not fair — outcome 6 appears far too often.
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Section A – Five-sided spinner spun 120 times
Fill in relative frequencies (total = 120):
- 1: 26 ÷ 120 = 0.2167 (≈ 0.217)
- 2: 22 ÷ 120 = 0.1833 (≈ 0.183)
- 3: 19 ÷ 120 = 0.1583 (≈ 0.158)
- 4: 29 ÷ 120 = 0.2417 (≈ 0.242)
- 5: 24 ÷ 120 = 0.2000
(We’ll keep decimals to 3 places if needed, but for predictions, we can use fractions or decimals.)
3) If the spinner were spun 360 times, how many times would you expect to get a score of 2?
Relative frequency of 2 = 22/120 = 11/60
Expected = (11/60) × 360 = 11 × 6 =
66
4) Do you think the spinner is fair? Why?
A fair 5-sided spinner should give each outcome ~1/5 = 0.20
Observed:
1 → 0.217
2 → 0.183
3 → 0.158
4 → 0.242
5 → 0.200
These are somewhat close to 0.20, especially with only 120 trials. Differences could be due to chance.
But outcome 3 is quite low (0.158), and 4 is high (0.242). Still, not extremely off.
Answer:
Probably fair, but not perfectly — small sample size means some variation is expected.
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Section B – Color of cars passing traffic
Total cars = 13 + 34 + 25 + 10 + 8 =
90
Relative frequencies:
- Red: 13/90 ≈ 0.144
- Silver: 34/90 = 17/45 ≈ 0.378
- Black: 25/90 = 5/18 ≈ 0.278
- White: 10/90 = 1/9 ≈ 0.111
- Blue: 8/90 = 4/45 ≈ 0.089
1) What is the relative frequency of the next passing car being silver?
→ 34/90 =
17/45 or approximately
0.378
2) Estimate number of white cars if 4000 cars passed by?
Relative frequency of white = 10/90 = 1/9
Expected = (1/9) × 4000 = 4000 ÷ 9 ≈
444.44, so about
444 cars (or 444–445; usually round to nearest whole number:
444)
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Section B – Counter from bag (200 trials total)
Trials recorded every 50:
At 50 trials: 8 blue
At 100: 22
At 150: 48
At 200: 59
So total blue counters drawn in 200 trials =
59
Relative frequency after 200 trials = 59/200 =
0.295
3) Best estimate of probability of picking a blue counter?
Use the overall relative frequency:
59/200 = 0.295
4) Estimate probability of taking a blue counter from the bag.
Same as above:
0.295 (or 29.5%)
5) How to get a more accurate estimate?
Do
more trials — the more times you repeat the experiment, the closer the relative frequency gets to the true probability (Law of Large Numbers).
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Now compile all final answers clearly:
Final Answer:
1) 400
2) No, because 6 appears much more often (40%) than expected (≈16.7%).
3) 66
4) Probably fair; results are reasonably close to 20% each, though small sample causes some variation.
1) 34/90 or ≈0.378
2) ≈444
3) 59/200 = 0.295
4) 0.295
5) Increase the number of trials (e.g., repeat 1000 times instead of 200).
Parent Tip: Review the logic above to help your child master the concept of relative frequency worksheet.
