Relative Frequency: Practise Online

Either paper 4 marks

A spinner can land on white, black, orange or purple. The spinner is spun 100 times.

The table gives the results.

Colour White Black Orange Purple

Frequency 19 36 11 34

(a)
Calculate the relative frequency of white. Give your answer as a decimal.

(b)
The spinner is spun 200 more times. Estimate the number of times it will land on white.
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(a) \(0.19\)
(b) \(38\)
The relative frequency is the number of times white came up divided by the total number of spins.

\(\frac{ 19 }{100} = 0.19\)
Use the relative frequency as an estimate of the probability.

\(200 \times 0.19 = 38\)

So expect about 38 of the 200 spins to land on white.
- (a) M1number of white over total spins \(\frac{ 19 }{100}\)
- (a) A1relative frequency \(0.19\)
- (b) M1\(200\) multiplied by the relative frequency \(0.19\)
- (b) A1estimate \(38\)
Either paper 4 marks

A drawing pin is dropped onto a table many times.

Out of \(50\) drops, it lands point up \(14\) times.

(a)
Calculate the relative frequency that it lands point up.

Give your answer as a fraction in its simplest form.

(b)
Calculate the relative frequency as a percentage.
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(a) \(\frac{7}{25}\)
(b) \(28\%\)
Relative frequency is the number of times it happened over the number of trials.

\(\frac{ 14 }{ 50 }\)

Both parts divide by \(2\), so in its simplest form that is

\(\frac{7}{25}\)
A percentage is a share out of \(100\), so multiply the fraction by \(100\).

\(\frac{7}{25} \times 100 = 28\)

So the relative frequency is \(28\%\).
- (a) M1relative frequency as a fraction of the trials, \(\frac{ 14 }{ 50 }\)
- (a) A1simplest form \(\frac{7}{25}\)
- (b) M1multiplying the fraction by 100, \(\frac{7}{25} \times 100\)
- (b) A1\(28\%\)
Either paper 3 marks

Amara wants to test whether a coin is fair. She flips the coin 200 times.

It lands on heads 80 times.

(a)
Work out the relative frequency of heads. Give your answer as a decimal.

(b)
Do you think the coin is fair? Give a reason for your answer.
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(a) \(0.4\)
(b) No — the relative frequency of 0.4 is a long way from 0.5, so the coin looks biased.
The relative frequency is the number of heads divided by the number of flips.

\(\frac{ 80 }{ 200 } = 0.4\)
For a fair coin the theoretical probability of heads is \(\frac{1}{2} = 0.5\).

Over 200 flips the relative frequency was 0.4, which is a long way from 0.5, so the coin looks biased.
- (b) M1theoretical probability for a fair coin, \(\frac{1}{2} = 0.5\)
- (b) A1No — 0.4 is a long way from 0.5, so the coin looks biased
Either paper 3 marks

A bag of counters was tested 340 times.

The relative frequency of green was \(0.35\).

(a)
Work out how many times the counter taken was green.

(b)
Work out how many times it was something else.
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(a) \(119\)
(b) \(221\)
The relative frequency is the number of green results divided by the number of draws, so multiply it back up.

\(340 \times 0.35 = 119\)
\(340 - 119 = 221\)
- (a) M1multiplying the 340 draws by the relative frequency \(0.35\)
- (a) A1number of times \(119\)
Either paper 4 marks
A spinner has sections coloured gold, silver, bronze and grey. It was spun 100 times. The table shows the relative frequency of each colour.

Colour Relative frequency

gold 0.15

silver 0.35

bronze 0.2

grey

(a)
Work out the relative frequency of grey.

(b)
Calculate how many times the spinner landed on grey.
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(a) \(0.3\)
(b) \(30\)
Every spin lands on one of the four colours, so the four relative frequencies add to 1.

\(0.15 + 0.35 + 0.2 = 0.7\)

\(1 - 0.7 = 0.3\)
\(100 \times 0.3 = 30\)
- (a) M1subtracting the known relative frequencies from 1, \(1 - 0.7\)
- (a) A1relative frequency \(0.3\)
- (b) M1multiplying the trials by the relative frequency, \(100 \times 0.3\)
- (b) A1\(30\) times
Either paper 3 marks
Zara and Jack each spin the same biased coin a number of times. The table shows information about the results.

Jack Zara

Number of spins 275 175

Relative frequency of heads 0.16 0.28

(a)
How many more heads did Zara spin than Jack?

(b)
Zara says, "My estimate of the probability of the coin landing on heads must be the best, because 0.28 is greater than 0.16."

Is she correct?

Tick a box. Yes / No

Give a reason for your answer.
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(a) \(5\)
(b) No. Jack spun the coin more times (275 against 175), so Jack's estimate is the better one. A larger relative frequency does not make an estimate better.
Heads = relative frequency \(\times\) number of spins.

Zara: \(0.28 \times 175 = 49\)

Jack: \(0.16 \times 275 = 44\)

\(49 - 44 = 5\)
The more times a coin is spun, the better the estimate of the probability.

Jack spun the coin 275 times and Zara spun it 175 times.

Jack's results come from more spins, so Zara is not correct: the answer is No.
- (a) M1heads found for each person: \(49\) and \(44\)
- (a) A1difference \(5\)
Either paper 4 marks
500 seeds were planted in a greenhouse. The number of seeds that germinated was recorded after every 100 seeds. The table shows some information about the results.

Total number of seeds 100 200 300 400 500

Total number of seeds that germinated 7 32 48 64 65

Relative frequency of seeds that germinated 0.07 0.16

The graph shows the two relative frequencies that are already in the table.
(a)
Complete the table. Give each relative frequency as a decimal.

(b)
Write down the best estimate of the probability that a seed planted in the greenhouse germinates.
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(a) \(0.16,\ 0.16,\ 0.13\)
(b) \(0.13\)
Each relative frequency is the total number of seeds that germinated so far divided by the total number of seeds so far.

\(32 \div 200 = 0.16\)

\(64 \div 400 = 0.16\)

\(65 \div 500 = 0.13\)
The relative frequency is most reliable when it is based on the most seeds, so use the last column, after 500 seeds.

The best estimate is \(0.13\).
- (a) B1first blank cell \(32 \div 200 = 0.16\)
- (a) B1second blank cell \(64 \div 400 = 0.16\)
- (a) B1last cell \(65 \div 500 = 0.13\)
Either paper 4 marks

Isaac and Nadia are testing the same spinner. They each want to estimate the probability that it lands on red.

Isaac spins it 60 times. It lands on red 30 times.

Nadia spins it 40 times. It lands on red 28 times.

(a)
Whose results give the more reliable estimate of the probability? Give a reason for your answer.

(b)
They put all of their results together. Use all 100 spins to work out the best estimate of the probability that the spinner lands on red.
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(a) Isaac — Isaac did 60 spins and Nadia did only 40, and the more trials there are the closer the relative frequency is likely to be to the true probability.
(b) \(0.58\)
Both sets of results give an estimate, but a relative frequency from more trials is more reliable.

Isaac did 60 spins and Nadia did only 40, so Isaac's estimate is the better one.
Altogether the spinner was spun 100 times and landed on red \(30 + 28 = 58\) times.

\(\frac{ 58 }{100} = 0.58\)

Using every result gives the largest number of trials, so this is the best estimate of the three.
- (a) B1Isaac
- (a) B1reason: more trials, 60 against 40
- (b) M1total on red \(30 + 28 = 58\) out of 100 spins
- (b) A1relative frequency \(0.58\)
Either paper 3 marks

A biased dice is rolled a number of times.

It landed on blue \(24\) times.

The relative frequency of landing on blue is \(0.3\).

(a)
Work out how many rolls there were altogether.

(b)
Calculate the relative frequency of not landing on blue.
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(a) \(80\ \text{rolls}\)
(b) \(0.7\)
Relative frequency is the number of times it happened divided by the number of trials, so

\(\frac{ 24 }{\text{number of } rolls } = 0.3\)

Multiplying up and then dividing gives

\(24 \div 0.3 = 80\)

Check it: \(24 \div 80 = 0.3\).
Every one of the \(80\) rolls either landed on blue or did not, so the two relative frequencies add up to \(1\).

\(1 - 0.3 = 0.7\)
- (a) M1dividing the number of times it landed, \(24\), by the relative frequency \(0.3\)
- (a) A1number of rolls \(80\)
Either paper 3 marks

A packing plant checks 700 eggs. Of these, 84 are cracked.

The line is stopped if more than 4% of the eggs are cracked.

Does the line have to be stopped?

You must show your working.
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Yes — the relative frequency is \(0.12\), which is 12%, and that is more than 4%.
The relative frequency of cracked eggs is

\(\frac{ 84 }{ 700 } = 0.12\)

As a percentage that is \(12\%\), and the limit is \(4\%\).

\(12\%\) is more than \(4\%\), so the answer is Yes.
- M1relative frequency, \(\frac{ 84 }{ 700 } = 0.12\)
- M1as a percentage, 12%
- A1Yes — 12% is more than 4%

Total number of seeds	100	200	300	400	500
Total number of seeds that germinated	7	32	48	64	65
Relative frequency of seeds that germinated	0.07		0.16

Colour	Relative frequency
gold	0.15
silver	0.35
bronze	0.2
grey

	Jack	Zara
Number of spins	275	175
Relative frequency of heads	0.16	0.28