IBDP Maths AI: Topic: SL 4.1: Concepts of population, sample: IB style Questions SL Paper 2

Question

200 people were asked the amount of time T (minutes) they had spent in the supermarket. The results are represented in the table below.

State if the data is discrete or continuous.[1]

a.

State the modal group.[1]

b.

Write down the midpoint of the interval 10 < T ≤ 20 .[1]

c.

Use your graphic display calculator to find an estimate for

(i) the mean;

(ii) the standard deviation.[3]

d.

The results are represented in the cumulative frequency table below, with upper class boundaries of 10, 20, 30, 40, 50.

Write down the value of

(i) q;

(ii) r.[2]

e.

The results are represented in the cumulative frequency table below, with upper class boundaries of 10, 20, 30, 40, 50.

On graph paper, draw a cumulative frequency graph, using a scale of 2 cm to represent 10 minutes (T) on the horizontal axis and 1 cm to represent 10 people on the vertical axis.[4]

f.

Use your graph from part (f) to estimate

(i) the median;

(ii) the 90th percentile of the results;

(iii) the number of people who shopped at the supermarket for more than 15 minutes.[6]

g.
Answer/Explanation

Markscheme

continuous     (A1)[1 mark]

a.

20 < T ≤ 30     (A1)[1 mark]

b.

15     (A1)[1 mark]

c.

(i) 21.5     (G2)

(ii) 9.21 (9.20597…)     (G1) [3 marks]

d.

(i) q = 194     (A1)

(ii) r = 200     (A1)[2 marks]

e.

      (A1)(A2)(ft)(A1)

Notes: Award (A1) for scale and axis labels, (A2)(ft) for 5 correct points, (A1)(ft) for 4 or 3 correct points, (A0) for less than 3 correct points, (A1) for smooth curve through their points, starting at (0, 0). Follow through from their answers to parts (e)(i) and (e)(ii).[4 marks]

f.

(i) 22.5 ± 2     (A1) 

(ii) 32 ± 2     (M1)(A1)(ft)(G2)

Note: Award (M1) for lines drawn on graph or some indication of method, follow through from their graph if working is shown.(iii) 44 ± 2     (A1)(ft)

Note: Follow through from their graph if working is shown.

200 − 44 = 156     (M1)(A1)(ft)(G2)

Note: Award (M1) for subtraction from 200, follow through from their graph if working is shown.[6 marks]

g.

Question

As part of his IB Biology field work, Barry was asked to measure the circumference of trees, in centimetres, that were growing at different distances, in metres, from a river bank. His results are summarized in the following table.

State whether distance from the river bank is a continuous or discrete variable.[1]

a.

On graph paper, draw a scatter diagram to show Barry’s results. Use a scale of 1 cm to represent 5 m on the x-axis and 1 cm to represent 10 cm on the y-axis.[4]

b.

Write down

(i)     the mean distance, \(\bar x\), of the trees from the river bank;

(ii)     the mean circumference, \(\bar y\), of the trees.[2]

c.

Plot and label the point \({\text{M}}(\bar x,{\text{ }}\bar y)\) on your graph.[2]

d.

Write down

(i)     the Pearson’s product–moment correlation coefficient, \(r\), for Barry’s results;

(ii)     the equation of the regression line \(y\) on \(x\), for Barry’s results.[4]

e.

Draw the regression line \(y\) on \(x\) on your graph.[2]

f.

Use the equation of the regression line \(y\) on \(x\) to estimate the circumference of a tree that is 40 m from the river bank.[2]

g.
Answer/Explanation

Markscheme

continuous     (A1)[1 mark]

a.

     (A1)(A1)(A1)(A1)

Notes: Award (A1) for labelled axes and correct scales; if axes are reversed award (A0) and follow through for their points. Award (A1) for at least 3 correct points, (A2) for at least 6 correct points, (A3) for all 9 correct points. If scales are too small or graph paper has not been used, accuracy cannot be determined; award (A0). Do not penalize if extra points are seen.[4 marks]

b.

(i)     26 (m)     (A1)

(ii)     65 (cm)     (A1)[2 marks]

c.

point \({\text{M}}\) labelled, in correct position     (A1)(A1)(ft)

Notes: Award (A1)(ft) for point plotted in correct position, (A1) for point labelled \({\text{M}}\) or \((\bar x,{\text{ }}\bar y)\). Follow through from their answers to part (c).[2 marks]

d.

(i)     \(-0.988\;{\text{ }}\left( {-0.988432 \ldots } \right)\)     (G2)

Note: Award (G2) for \(-0.99\). Award (G1) for \(-0.990\).

     Award (A1)(A0) if minus sign is omitted.

(ii)     \(y =  – 0.756x + 84.7\)   \((y =  – 0.756281 \ldots x + 84.6633 \ldots )\)     (G2)

Notes: Award (A1) for \( – 0.756x\), (A1) for \(84.7\). If the answer is not given as an equation, award a maximum of (A1)(A0).[4 marks]

e.

regression line through their \({\text{M}}\)     (A1)((ft)

regression line through their \(\left( {0,85} \right)\) (accept \(85 \pm 1\))     (A1)(ft)

Notes: Follow through from part (d). Award a maximum of (A1)(A0) if the line is not straight. Do not penalize if either the line does not meet the y-axis or extends into quadrants other than the first.

     If \({\text{M}}\) is not plotted or labelled, then follow through from part (c).

     Follow through from their y-intercept in part (e)(ii).[2 marks]

f.

\( – 0.756281(40) + 84.6633\)     (M1)

\( = 54.4{\text{ (cm) }}(54.4120 \ldots )\)     (A1)(ft)(G2)

Notes: Accept \(54.5\) (\(54.46\)) for use of 3 sf. Accept \(54.3\) from use of \(-0.76\) and \(84.7\).

     Follow through from their equation in part (e)(ii) irrespective of working shown; the final answer seen must be consistent with that equation for the final (A1) to be awarded.

     Do not accept answers taken from the graph.[2 marks]

g.
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