Question 1:
--- 1(a) ---
1(a) | eg 60 people = 1.5 large squares/6 medium squares/150 small squares or
eg 1 person = 0.025 large squares/0.6 medium squares/2.5 small squares or
eg [1 small square =] 0.4 people/[1 medium square =] 10 people/[1 large square =] 40 people
eg a correct f.d. eg 60/(20 – 10) [= 6]
eg a correct frequency, 100, 70, 20, 24 associated with the appropriate bar | B1
8 15 40 30
eg 20 or 24 or 82 or 0.815 or or or 850.4
10 30 2.5 2.5
2150.4 or 16 or 12 or 700.4 | M1
28 people | A1
(3)
(b) | 5 5
Median =[5]+ 37or [10]− 33
70 70 | M1
= 7.642... awrt 7.64 | A1
(2)
(c) | midpointfreq=2.5100+7.570+1560+2520+4524 =3255  | M1
"3255"
Mean =
274 | dM1
= 11.879… awrt 11.9 | A1
(3)
Notes | Total 8
(a) | B1: for finding a ratio between people and area. Allow just the numbers for 1 person or for 1 square ie
0.025, 0.6, 2.5, 0.4, 1.66 or 40 or
calculating f.d. for any bar correctly, fd = 20, 14, 6, 2 or 0.8
Information may be seen on diagram. Must be clear it is a fd either by correct use, seen on axes
associated with correct bar or stated as an fd. May be implied by M1
M1: for a correct method to find the number of people between 22 and 30 km or 30 and 45 km or
between 22 and 45 km
A1: 28
NB An answer of 28 gains 3/3 unless from obvious incorrect working
(b) | 5 Q −n 137−100 33 n−Q 170−137
M1: Allow equivalent for n+ 37 or 2 = or n− 5or 2 = oe
70 5 170−100 70 5 170−100
Q −5 137−170
Allow alternative methods eg 2 =
10−Q 170−137
2
Allow 37.5 for 37, 137.5 for 137, 32.5 for 33
107 215
A1: awrt 7.64 or or allow awrt 7.68 or if using n + 1 Allow awrt 7640 m or 7680 m but must
14 28
have units.
(c) | M1: Attempt at correct expression for midpoint freq - at least 3 products with correct midpoints
added with at least 1 of these products fully correct . Allow for 3255
M1: dep on M1 being awarded for dividing “their sum” by 274
3255
A1: awrt 11.9 or
274
Allow awrt 11900 m but must have units
Qu | Scheme | Marks