Relative Frequency Tables Worksheets | Questions and Revision | MME

# Relative Frequency Worksheets, Questions and Revision

Level 1 Level 2 Level 3

## What you need to know

### Relative Frequency

There are two types of probability you will see:

• Theoretical probability – this is the kind of probability that we have prior understanding of. For example, we know that the chance of rolling a 6 on a fair dice is $\dfrac{1}{6}$.
• Relative frequency – this is the kind of probability that we determine from a survey or experiment.

We calculate relative frequency using the following formula:

$\text{relative frequency }=\dfrac{\text{number of times an outcome happened}}{\text{total number of all outcomes}}$

For example, if you were to flip a coin 100 times and record the results, then you could say

$\text{relative frequency of landing on heads }=\dfrac{\text{number of heads}}{100}$

Having a good knowledge of basic probability is useful for this topic.

### Take Note:

We can use relative frequency to determine expected frequency, which is the number of times we would expect an outcome to happen. It is calculated as such:

$\text{expected frequency }=\text{ probability }\times\text{number of trials}$

For example, if we roll a dice 60 times, the expected frequency of sixes is $\dfrac{1}{6}\times60= 10$. In practice, it probably won’t end up being exactly ten times, i.e. our relative frequency won’t be $\dfrac{1}{6}$. However, if you do more trials – 600 rolls, 6,000 times – the relative frequency will get closer to $\dfrac{1}{6}$

### Bias

One of the key uses of relative frequency is in testing for bias. We say that an experiment is biased when the probability of a particular outcome is unfairly bigger or smaller than it should be.

For example, a biased coin might be cleverly designed so that it’s lands on tails more than 50% of the time. If we suspect a coin of being biased, then we test it by flipping it a load of times and recording the results. Then, if the relative frequency of tails is noticeably more than 50%, we might suspect the coin of being biased.

### Example 1: Relative Frequency

Prim asks some people in her town about their dietary habits and records the results in the table below.

a)Work out the relative frequency of someone in Prim’s town being vegetarian.

b)There are 20,000 people in Prim’s town. Using your answer to part a), find an estimate for the number of people in this town who are vegetarian.

a) Firstly, we need the total number of all outcomes:

$\text{total }=44+23+43=110$

Now, using the formula shown above, we can calculate the relative frequency for vegetarians to be

$\text{relative frequency }=\dfrac{44}{110}=\dfrac{2}{5}$

b) Our answer to part a) suggests that if we picked a random person in Prim’s town, there would be a 2/5 or 0.4 chance that they are vegetarian. Therefore 2/5 of 20,000

$\dfrac{2}{5}\times 20,000=8,000\text{ vegetarians}$

### Example 2: Relative Frequency

Stella and Gary are throwing darts at a target as shown in the table below. By calculating relative frequencies, work out who is more likely to get a hit. Also, state which one of their relative frequencies gives a more accurate statement of their chance of hitting the target and explain why.

To calculate the relative frequencies, we will divide each person’s hits by their total attempts. Gary:

$\dfrac{18}{18+42}=0.3$

Then, Stella:

$\dfrac{45}{45+75}=0.375$

Therefore, Stella is more likely to get a hit. Furthermore, Stella made $45+75=120$ attempts in total, compared to Gary’s $18+42=60$, therefore Stella’s relative frequency is a more accurate representation of her chance of hitting the target than Gary’s is for him.

### Example 3: Relative Frequency and Bias

Amber has a spinner with the numbers 1, 2, 3 and 4 on it. She was told that the spinner is fair, i.e. there is an equal chance of landing on any number when you spin it, but she claims that the spinner is biased.

Amber spins the spinner 150 times and it lands on the number one 54 times. By calculating the relative frequency of 1, comment on Amber’s claim.

Firstly, the relative frequency:

$\text{relative frequency of landing on 1 }=\dfrac{54}{150}=0.36$

So, the relative frequency is 0.36, or 36%. But we know there are 4 equally likely outcomes on the spinner, meaning the theoretical probability of landing on 1 is

$\dfrac{1}{4}=25\%$

25% is significantly smaller than 36%, so it seems that Amber has good to reason to make the claim she does. The results suggests the spinner is biased.

NOTE: in reality we don’t know this for sure. It’s completely possible that a fair spinner could land on the number one 54 times, but it is unlikely, which leads us to believe the spinner is biased. The result could always be due to chance (This is often the answer in exam questions).

We can always do a better experiment and make our conclusion more confident by

increasing the number of trials!

In this case, that would mean spinning the spinner 200, 300, or even 1,000 times. The more, the merrier! If you spin it 1,000 times and still end up getting about 10% more number 1s than you should, then you can be even more confident the spinner is biased.

### Example Questions

a) 12 students out of the total of 62 used public transport so, as a fraction, this can be expressed as:

$\dfrac{12}{62}$

To calculate the relative frequency as a decimal to 3 decimal places, divide the top by the bottom:

$\dfrac{12}{62}=0.194\text{ to 3 decimal places}$

b) 16 students out of the total of 62 walked to school so, as a fraction, this can be expressed as:

$\dfrac{16}{62}$

To calculate the relative frequency as a decimal to 3 decimal places, divide the top by the bottom:

$\dfrac{16}{62}=0.258\text{ to 3 decimal places}$

c) If 9 of the 62 students cycled, then the number who didn’t cycle is:

$62-9=53$

As a fraction, this can be expressed as:

$\dfrac{53}{62}$

To calculate the relative frequency as a decimal to 3 decimal places, divide the top by the bottom:

$\dfrac{53}{62}=0.855\text{ to 3 decimal places}$

In Bev’s experiment, the relative frequency of rolling a 6 is:

$\dfrac{63}{400}=0.1575$

The probability of rolling a 6 is:

$\dfrac{1}{6}=0.166666...$

The two results are fairly similar, in fact in Bev’s experiment she has successfully thrown a 6 slightly less often than would be expected, although not by a significant amount.

So, we can conclude that Bev’s statement is false. The die does not appear to be biased.

a) The relative frequency of a car being black using Hannah’s data can be calculated as follows:

$\dfrac{32}{32+48}=0.4=40\%$.

So, according to her data, her statement is correct.

b) To calculate the relative frequency based on all the data collected, we will need to combine Hannah’s and Mike’s data.

The total number of cars spotted was $32 +41 + 48 + 111 = 232\text{ cars}$.

The total number of black cards spotted was $32+41=73\text{ black cars}$

The relative frequency of a car being black using all the data can be calculated as follows:

$\dfrac{73}{232}=0.315\text{ to 3 decimal places}$

c) To find the expected frequency of black cars in their neighbourhood, we will multiply the probability of a car being black (i.e., the relative frequency calculated in part b) by the total number of cars. The expected frequency of black cars in the neighbourhood is therefore:

$0.315\times5,000=1,575\text{ black cars}.$

a) Bob throws the die 100 times and throws a 7 on 15 occasions. As a fraction this can be written as:

$\dfrac{15}{100}$

The probability of Bob throwing a 7 is the decimal equivalent of the above fraction, which we can calculate by dividing the numerator by the denominator:

$15\div100 = 0.15$

b) Susan throws the die 500 times and throws a 7 on 60 occasions. As a fraction this can be written as:

$\dfrac{60}{500}$

The probability of Susan throwing a 7 is the decimal equivalent of the above fraction, which we can calculate by dividing the numerator by the denominator:

$60\div500 = 0.12$

c) Susan’s results are more accurate because the more times you conduct an experiment, the more accurate the estimate will be.

a) Since the relative frequency for Burger Queen is 0.1, we should be able to turn this into a fraction without too much difficulty:

$0.1 = \dfrac{1}{10}$

We know that 16 people in total opted for Burger Queen, therefore:

$\frac{1}{10} = 16\text{ students}$

If 16 students represents $\frac{1}{10}$ of the total number of students, then the total number of students on the trip can be calculated as follows:

$16\text{ students}\times10 = 160\text{ students}$

b) The relative frequency of Pizza Cottage is 0.4 which, as a fraction is $\frac{4}{10}$. If there are 160 students in total and $\frac{4}{10}$ chose Pizza Cottage, then the number of students who chose Pizza Cottage can be calculated as follows:

$\dfrac{4}{10}\times160\text{ students} = 64\text{ students}$

c) The relative frequency of Dazza’s Fish and Chips is 0.5 which, as a fraction is $\frac{5}{10}$ or $\frac{1}{2}$. If there are 160 students in total and $\frac{1}{2}$ chose Dazza’s Fish and Chips, then the number of students who chose Dazza’s Fish and Chips can be calculated as follows:

$\dfrac{1}{2}\times160\text{ students} = 80\text{ students}$

or

$160\text{ students}\div2 = 80\text{ students}$

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