Chapter 4 Representation of Data Test Solutions

Chapter 4 – Representation of Data Test Solutions

1.) (a) Mode = 23 B1 1

For Q 1 : = 10.5 Þ 11th observation \ Q 1 = 17 B1

For Q 2 : = 21 Þ = (21st & 22nd) observations
\ Q 2 = = 23.5 M1 A1

For Q 3 : = 31.5 Þ 32nd observation \ Q 3 = 31 B1 4

(c)

Box plot M1

Scale & label M1

Q1, Q2, Q3 A1

11, 43 A1 4

(d) From box plot or M1

Q2 – Q1 = 23.5 – 17 = 6.5

Q3 – Q2 = 31 – 23.5 = 7.5(slight) positive skew B1 1

(e) Back-to-back stem and leaf diagram B1 1

[11]

2.) (a) B1 1

Accept 50% or half or 0.5>
Units not required.

(b) 54 B1 1

Correct answers only.
Units not required.

‘Anomaly’ only award B0
Accept ’85 kg was heaviest instrument on the trip’ or equivalent
for second B1.
Examples of common acceptable instruments; double bass, cello,
harp, piano, drums, tuba
Examples of common unacceptable instruments: violin, viola,
trombone, trumpet, French horn, guitar

(d) Q3 – Q2 = Q2 – Q1 B1
so symmetrical or no skew Dependent – only award if B1 above B1 2

‘Quartiles equidistant from median’ or equivalent award B1 then
symmetrical or no skew for B1
Alternative:
‘Positive tail is longer than negative tail’ or ‘median closer to
lowest value’ or equivalent so slight positive skew.
B0 for ’evenly’ etc. instead or ‘symmetrical’
B0 for ‘normal’ only

(e) P(W < 54) = 0.75 (or p(W > 54) = 0.25) or correctly labelled and M1
shaded diagram
M1B1
s = 13.43..... A1 4

Please note that B mark appears first on ePEN
First line might be missing so first M1 can be implied by second.
Second M1 for standardising with sigma and equating to z value
NB Using 0.7734 should not be awrded second M1
Anything which rounds to 0.67 for B1.
Accept 0.675 if to 3sf obtained by interpolation
Anything that rounds to 13.3. – 13.4 for A1.

[10]

3.) M1

Width / 1 / 1 / 4 / 2 / 3 / 5 / 3 / 12
Freq. Density / 6 / 7 / 2 / 6 / 5.5 / 2 / 1.5 / 0.5

0.5 ×12 or 6 A1

Total area is (1 × 6) + (1 × 7) + (4 × 2) + ..., = 70
(90.5 –78.5) × M1
“70 seen anywhere” B1
Number of runners is 12 A1 5

1st M1 for attempt at width of the correct bar (90.5 – 78.5)
[Maybe on histogram or in table]

1st A1 for 0.5 × 12 or 6 (may be seen on the histogram).
Must be related to the area of the bar above 78.5 – 90.5.

2nd M1 for attempting area of correct bar ×

B1 for 70 seen anywhere in their working

2nd A1 for correct answer of 12.

Minimum working required is 2 × 0.5 × 12 where the 2 should come
from

Beware 90.5 – 78.5 = 12 (this scores M1A0M0B0A0)

Common answer is 0.5×12= 6 (this scores M1A1M0B0A0)

If unsure send to review e.g. 2 × 0.5 × 12 = 12 without 70 being seen

[5]

4.) (a) 18-25 group, area = 7 × 5 = 35 B1
25-40 group, area = 15 × 1 = 15 B1 2

(b) (25 – 20) × 5 + (40 – 25) × 1 = 40 M1A1 2

5 × 5 is enough evidence of method for M1.
Condone 19.5, 20.5 instead of 20 etc.
Award 2 if 40 seen.

Look for working for this question in part (d) too.
Use of some mid-points, at least 3 correct for M1. These may be
tabulated in (d).
Their for M1 and anything that rounds to 18.9 for A1.

(d) alternative OK M1
M1
A1 3

Clear attempt at alternative
for first M1.
They may use their and gain the method mark.
Square root of above for second M1
Anything that rounds to 7.3 for A1.

(e) Q2 = 18 or 18.1 if (n + 1) used B1
Q1 = 10 + × 4 = 13.75 or 15.25 numerator gives 13.8125 M1A1
Q3 = 18 + × 7 = 23 or 25.75 numerator gives 23.15 A1 4

Clear attempt at either quartile for M1
These will take the form ‘their lower limit’ + correct fraction
× ‘their class width’.
Anything that rounds to 13.8 for lower quartile.
23 or anything that rounds to 23.2 dependent upon method used.

(f) 0.376... B1
Positive skew B1ft 2

Anything that rounds to 0.38 for B1 or 0.33 for B1 if (n + 1) used.
Correct answer or correct statement that follows from their
value for B1.