Write down the total number of people who were surveyed.

Write down the mid-interval value for the $10 < t \leqslant 15$ group.

Find an estimate of the mean time people took to drink their coffee.

The information above has been rewritten as a cumulative frequency table.

Write down the value of $a$ and the value of $b$.

This information is shown in the following cumulative frequency graph.

For the people who were surveyed, use the graph to estimate

(i)     the time taken for the first $40$ people to drink their coffee;

(ii)     the number of people who take less than $8$ minutes to drink their coffee;

(iii)     the number of people who take more than $23$ minutes to drink their coffee.

$60$     (A1)

[1 mark]

$12.5$     (A1)

[1 mark]

$\frac{{3 \times 2.5 + 5 \times 7.5 +  \ldots  + 10 \times 27.5}}{{60}}$     (M1)

Note: Award (M1) for an attempt to substitute their mid-interval values (consistent with their answer to part (b)) into the formula for the mean.

     Award (M1) where a table is constructed with their (consistent) mid-interval values listed along with the frequencies.

$= \frac{{1075}}{{60}}{\text{ }}\left( {\frac{{215}}{{12}},{\text{ 17.9, 17.9166}} \ldots } \right)$     (A1)(ft)(G2)

Note: Follow through from their answer to part (b).

[2 marks]

$a = 34,{\text{ }}b = 60$     (A1)(A1)

[2 marks]

(i)     $\leqslant {\text{21.25 minutes}}$     (A1)

Note: Accept $21.25$.

     Accept any answer between $21$ and $21.5$.

     (Accept 21.5, but do not accept 21.)

(ii)     $5$     (A1)

Note: Accept $< 6$. Do not accept $6$.

     Answer must be an integer.

(iii)     $60 - 45$     (M1)

$= 15$     (A1)(G2)

Notes: Award (M1) for subtraction from $60$. Accept $15 \pm 1$.

     Answer must be an integer.

[4 marks]