Methods of Analysing Data Worked ExamplesIGCSE Maths 0580

(a) As the hours increase, the marks increase steadily, and the points lie close to a straight line.

(b) Find the mean of the $x$ values first.

Now add the $y$ values.

Divide by $8$ for the mean mark.

Confirm the given line passes through the mean point — it should, for a good line of best fit.

(c) Substitute $x = 5.5$ into the equation of the line.

(d) The data only covers $1$ to $8$ hours. Using the line at $x = 20$ is extrapolation — and the line predicts $172$, which is impossible for a test marked out of $100$.

You cannot average the two means, because the classes are different sizes. Work with totals instead. Total $=$ mean $\times$ frequency.

Do the same for class $B$.

Add for the combined total of all the marks.

Divide by the combined number of students.

(a) Convert each mean into a total. The seven numbers add to:

The six remaining numbers add to:

The number removed is the difference between the two totals.

(b) Start again from the original total of $84$. With eight numbers and a mean of $13$:

The added number accounts for the increase.

(a) The mode is the value with the highest frequency — the value itself, not the frequency.

(b) Build a running (cumulative) total to locate positions in the ordered list.

With $40$ values, the median is midway between the $20$th and $21$st. Positions $9$ to $22$ all hold the value $1$.

(c) The lower quartile is at position $\frac{40}{4} = 10$, which also falls in the block of $1$s.

The upper quartile is at position $\frac{3 \times 40}{4} = 30$; positions $23$ to $32$ hold the value $2$.

Subtract for the interquartile range.

(d) For the mean, multiply each value by its frequency and add.

Divide by the total frequency.

(a) The modal class has the highest frequency. Note the classes are not all the same width, but here the largest frequency is also in a $10$-wide class, so no adjustment is needed.

(b) Build cumulative frequencies and find where the $20$th and $21$st values fall.

The $20$th value is the last of the second class and the $21$st is in the third, so the median lies in the third class.

(c) Find the midpoint of each class. The last class runs from $30$ to $50$, so its midpoint is $40$ — not $35$.

Multiply each midpoint by its frequency.

Add these products.

Divide by the total frequency.

(d) The original values are unknown; each is replaced by its class midpoint, which is only an approximation.

(a) Write the total frequency in terms of $x$.

Now write $\sum fx$, keeping the $x$ term.

Evaluate the known products and collect.

Set up the mean equation.

Multiply both sides by $(23 + x)$ and expand.

Collect the $x$ terms on one side.

Check: the total frequency is $40$ and $\sum fx = 65 + 51 = 116$.

(b) With $40$ values the median is between the $20$th and $21$st. Cumulative frequencies are $5, 13, 30, 36, 40$, so positions $14$ to $30$ all hold the score $3$.

(a) Add all eight numbers.

Divide by how many there are.

(b) Put the numbers in order first — this is essential and easy to forget.

With an even count, the median is the mean of the two middle values (the $4$th and $5$th).

(c) The mode is the most frequent value — $3$ appears three times.

(d) The range is the largest minus the smallest.

Read the frequencies from the bar heights and find the total number of customers.

(a) The mode is the size with the tallest bar.

(b) With $25$ customers the median is the $\frac{25 + 1}{2} = 13$th value. Keep a running total along the bars.

The running total reaches $13$ exactly at the end of the size $6$ bar.

(c) Multiply each size by its frequency and add.

Divide by the $25$ customers. Without a calculator, multiply top and bottom by $4$.

(d) A shop orders whole sizes, and the manager wants the size that sells most often.

(a) There are $19$ values, already in order. The median is the $\frac{19 + 1}{2} = 10$th value; counting leaves, $3 + 5 = 8$ takes you to the end of stem $2$, so the $10$th is the second leaf on stem $3$.

The lower quartile is the median of the $9$ values below the median (the $5$th value), and the upper quartile is the median of the $9$ values above it (the $15$th value).

Subtract for the interquartile range.

(b)(i) Add the times as recorded, stem by stem, then adjust for the correction: replacing $58$ by $85$ adds $27$ to the total.

Divide by $19$ and round to $3$ significant figures.

(b)(ii) $58$ was already the largest value and $85$ is still the largest, so every value keeps its position. The $5$th, $10$th and $15$th values are unchanged, so the median and IQR stay the same. Only the largest value moves, so the range changes.

(c) Adding $5$ to every value moves the whole distribution up by $5$: the median rises by $5$, but the gap between the quartiles is unchanged.

(a) $360^{\circ}$ represents all $72$ families, so each family is worth $5^{\circ}$.

Convert the known angles into numbers of families.

(b) Let $p$ families own $1$ car and $q$ families own $2$ cars. The frequencies must total $72$.

A mean of $1.5$ for $72$ families means $1.5 \times 72 = 108$ cars altogether. Write the total number of cars in terms of $p$ and $q$.

Subtract the first equation from the second.

Convert back to angles at $5^{\circ}$ per family, and check the total.

(c) With $72$ families the median is midway between the $36$th and $37$th values. The running totals are $12$, $33$, $63$, $72$.

Write the five numbers in order as $a \le b \le c \le d \le e$. The median is the middle value.

The mode is $8$, so $8$ appears at least twice. As $8$ is less than the median, both $8$s must come before $c$ — and there are only two places there.

The range is the largest value minus the smallest.

A mean of $14$ for $5$ days means a total of $5 \times 14 = 70$. The last value makes up the total.

Check all four facts. $21$ lies between $11$ and $22$, and no other value is repeated, so $8$ is the only mode.

Find the midpoint of each class. The widths are not all equal, so work each one out from its own limits.

Multiply each midpoint by its frequency.

Add the products.

Divide by the total frequency, $74$ — not by $4$, the number of classes.

The answer is not exact, so give it correct to $3$ significant figures.