Chapter 11: Review

The chapter problem, solved

The bus company claims a mean journey time of 25 minutes on the 07:30 service. A passenger group timed 30 journeys. The times, in minutes, are

\begin{aligned} &12,\ 15,\ 16,\ 17,\ 18,\ 18,\ 19,\ 20,\ 20,\ 21,\ 21,\ 22,\ 22,\ 23,\ 23,\\ &23,\ 24,\ 24,\ 25,\ 25,\ 26,\ 27,\ 28,\ 29,\ 30,\ 32,\ 34,\ 36,\ 41,\ 58. \end{aligned}

Step 1. Name the population and the sample

The population is every journey made by that service. The sample is the 30 journeys the group timed. The mean of the sample is an estimate of the mean of the population, not a measurement of it.

Step 2. Draw the picture

The times are grouped for a histogram. The class widths are not all equal, so the bars are drawn with frequency density.

Class (minutes)	Frequency	Class width	Frequency density
$10$ – $20$	7	10	0.7
$20$ – $25$	11	5	2.2
$25$ – $30$	6	5	1.2
$30$ – $40$	4	10	0.4
$40$ – $60$	2	20	0.1

The frequencies add to 30. The tallest bar is the class $20$ – $25$ , so the modal class is $20$ to $25$ minutes. The distribution has a longer tail on the right, so it is positively skewed.

Step 3. Find the mean and the standard deviation

The mean is

\bar{x} = \frac{749}{30} = 24.97 \ (2 \text{ d.p.}).

The standard deviation, dividing by $n$ , is

\sigma = \sqrt{\frac{\sum (x - \bar{x})^2}{30}} = \frac{\sqrt{69389}}{30} = 8.78 \ (2 \text{ d.p.}).

The sample mean of 24.97 minutes is very close to the claimed 25 minutes. The standard deviation of 8.78 minutes is large, so the individual journeys vary a great deal around that mean.

Step 4. Look for an outlier

In order, the 30 values give $Q_1 = 19.75$ and $Q_3 = 28.25$ , so

\text{IQR} = 28.25 - 19.75 = 8.5.

The fences are

19.75 - 1.5(8.5) = 7 \qquad \text{and} \qquad 28.25 + 1.5(8.5) = 41.

The value 58 is above 41, so 58 is a possible outlier. The group should check the original record. If the value is genuine, keep it; a journey can really take 58 minutes. If it is a recording error and the true value is known, correct it.

If 58 is removed, the remaining 29 journeys have mean

\frac{691}{29} = 23.83 \ (2 \text{ d.p.})

and standard deviation

\frac{2\sqrt{8585}}{29} = 6.39 \ (2 \text{ d.p.}).

Removing the outlier lowers the mean by about 1.1 minutes and lowers the standard deviation by about 2.4 minutes. The change must be reported, with the number of values removed.

Step 5. Read the scatter diagram

The group also recorded the distance of each journey. On a scatter diagram of distance against time, the points rise from left to right and lie close to a straight line, so the correlation is positive and strong. Longer journeys tend to take longer, which is expected.

The correlation does not prove that distance causes the time. A crowded bus, road works and the time of day also change the journey time. Distance is one explanatory variable among several.

Step 6. Answer the claim

The sample mean of 24.97 minutes supports the claim of 25 minutes. On this evidence, the group cannot say that the claim is wrong.

The conclusion is limited in four ways. The sample has only 30 journeys. It comes from one week, so a school holiday or an accident would change it. The sample was taken by the group at the stop, which is an opportunity sample, so it may miss passengers who boarded earlier or who gave up waiting. The single value 58 affects the mean and the standard deviation a great deal. A larger random sample over several weeks would give a more reliable estimate.

Mixed practice — choose the method first

For each question, say which method you will use before you calculate: an estimate from a sample, a frequency density, an average or spread, a quartile, a line of best fit, or a method from an earlier chapter. The items mix this chapter with Chapters 2, 3 and 4.

Find the mean of $12, 15, 16, 18, 20, 21$ .
Find the standard deviation of $12, 15, 16, 18, 20, 21$ , dividing by $n$ .
Find the interquartile range of $12, 15, 16, 18, 20, 21$ , using the $(n + 1)$ positions.
A class of width 10 contains 45 values. Find the frequency density.
A population has 450 members in one stratum and 150 in another. A stratified sample of 60 is taken. Find the number from the 450 stratum.
A sample of 40 pupils has a mean of $2.5$ hours of sport each week. Estimate the total number of hours of sport each week for 800 pupils.
A line of best fit passes through $(3, 11)$ and $(7, 23)$ . Find its gradient.
A histogram of 150 values has frequency density $2.5$ in the class $10$ – $30$ . Estimate the probability that a value lies in the class.
A data set has $n = 12$ , $\sum x = 96$ and $\sum x^2 = 1000$ . Find the variance, dividing by $n$ .
A survey runs for 12 weeks. It gets 40 responses in week 1, and each later week gets 6 more responses than the week before. Find the total number of responses over the 12 weeks.
Solve $x^2 - 5x + 6 = 0$ .
A straight line passes through $(1, 4)$ and $(5, 16)$ . Find its gradient and its intercept on the $y$ -axis.
Solve the simultaneous equations $2x + y = 11$ and $x - y = 1$ .
Work out $27^{2/3}$ .

Answers and concise solutions

1. The mean

Answer: The values sum to 102 over 6 values, so the mean is $\dfrac{102}{6} = 17$ .

2. The standard deviation

Answer: The mean is 17. The squared deviations sum to 56, so the variance is $\dfrac{56}{6} = \dfrac{28}{3}$ and $\sigma = \dfrac{2\sqrt{21}}{3} = 3.06$ (3 s.f.).

3. The interquartile range

Answer: The position of $Q_1$ is $\dfrac{7}{4} = 1.75$ , giving $Q_1 = 14.25$ . The position of $Q_3$ is $\dfrac{21}{4} = 5.25$ , giving $Q_3 = 20.25$ . The IQR is $20.25 - 14.25 = 6$ .

4. The frequency density

Answer: Frequency density is frequency divided by class width: $\dfrac{45}{10} = 4.5 = \dfrac{9}{2}$ .

5. The sample size from the 450 stratum

Answer: The population is 600, so the 450 stratum gives $\dfrac{450}{600} \times 60 = 45$ members. The other stratum gives 15, and the two add to 60.

6. The estimated total

Answer: Multiply the sample mean by the population size: $2.5 \times 800 = 2000$ hours. This is an informal inference about the population.

7. The gradient

Answer: The gradient is $\dfrac{23 - 11}{7 - 3} = 3$ , so each extra unit of the explanatory variable is associated with about 3 extra units of the response.

8. The estimated probability

Answer: The frequency is $2.5 \times 20 = 50$ , so the probability is $\dfrac{50}{150} = \dfrac{1}{3}$ . This is the area over the total frequency.

9. The variance

Answer: The mean is $\bar{x} = 8$ . Then the variance is $\dfrac{1000}{12} - 8^2 = \dfrac{250}{3} - 64 = \dfrac{58}{3} = 19.33$ (3 s.f.).

10. The total number of responses

Answer: This is an arithmetic series with first term 40 and common difference 6. The sum is $\dfrac{12}{2}\left(2(40) + 11(6)\right) = 6(80 + 66) = 876$ responses. Sequences come from Chapter 4.

11. The solutions

Answer: Factorise: $(x - 2)(x - 3) = 0$ , so $x = 2$ or $x = 3$ . Quadratics come from Chapter 2.

12. The gradient and intercept

Answer: The gradient is $\dfrac{16 - 4}{5 - 1} = 3$ . Using $y = 3x + c$ at $(1, 4)$ gives $4 = 3 + c$ , so $c = 1$ and the line is $y = 3x + 1$ . Straight lines come from Chapter 3.

13. The solution

Answer: Adding the equations gives $3x = 12$ , so $x = 4$ . Then $y = x - 1 = 3$ . Check: $2(4) + 3 = 11$ . Simultaneous equations come from Chapter 2.

14. The value

Answer: The cube root of 27 is 3, and $3^2 = 9$ . So $27^{2/3} = 9$ . Index laws come from Chapter 2.

What you can now do

explain the difference between a population and a sample, and draw an informal inference about the population;
choose or criticise a simple random, systematic or stratified sample;
read a histogram and use frequency density, remembering that area is frequency;
describe correlation and interpret a line of best fit, including why correlation does not imply causation;
calculate the mean, median, mode, range, interquartile range and standard deviation;
use the mean and the standard deviation to compare two distributions;
recognise and deal with outliers, missing values and errors;
use a large data set and write a conclusion with its limitations.

Your relevance reflection

Return to the prompt in the chapter opener: a claim that used a number from a survey or a sample. Write the claim again. Now add three things you could not have said before this chapter: how the sample was chosen, how much the numbers vary, and what the sample cannot tell you. Finish with one sentence about how you would check the claim if you could collect the data yourself.