Level 4 — ApplicationBasic Data & Probability

Basic Data & Probability

printable — key stays hidden on paper

Level 4 — Application (novel problems, no hints) Time: 60 minutes | Total: 50 marks

Question 1 (10 marks)

A café owner records the number of coffees sold each hour over one 10-hour day:

$14,\ 22,\ 18,\ 22,\ 30,\ 25,\ 22,\ 19,\ 27,\ 21$

(a) Calculate the mean, median, mode and range of the hourly sales. (6)

(b) The owner claims "we sell more than 22 coffees an hour on average." Using two of your averages, decide whether this claim is fair. Justify your answer. (2)

(c) The 30 (a busy hour) was later found to be a recording error and should be 20. State, without full recalculation, what happens to the mode and explain why the median does not change. (2)

Question 2 (11 marks)

A survey of 90 commuters recorded their main travel method. The results are shown as a pie chart with these sector angles:

Method	Angle
Car	$140°$
Bus	$100°$
Bicycle	$?$
Walk	$48°$

(a) Find the missing angle for Bicycle. (2)

(b) Calculate the number of commuters using each method. (4)

(c) The town wants to draw a pictogram where one symbol = 5 commuters. State how many symbols (whole and part) represent the Bus users. (2)

(d) The following year, car use dropped by 25% while total commuters stayed at 90. Find the new pie-chart angle for Car (to the nearest degree). (3)

Question 3 (10 marks)

The table shows the times (in minutes) taken by 40 students to complete a puzzle:

Time $t$ (min)	Frequency
$0 \le t < 5$	6
$5 \le t < 10$	11
$10 \le t < 15$	14
$15 \le t < 20$	9

(a) Estimate the mean time. (4)

(b) Write down the modal class and the class containing the median. (3)

(d) A student states the range is exactly 20 minutes. Explain why this may be wrong and give the best statement you can make about the range. (2)

Question 4 (10 marks)

A bag contains coloured tokens: 5 red, 4 blue, 3 green and some yellow. A token is drawn at random. The probability of drawing yellow is $\dfrac{1}{4}$ .

(a) Show that there are 4 yellow tokens. (3)

(b) Find $P(\text{not blue})$ . (2)

(d) All the green tokens are removed. Write down the new probability of drawing a red token, giving your answer as a fraction in simplest form. (3)

Question 5 (9 marks)

A scientist measures the reaction time (seconds) of 8 people against their age (years):

Age	20	25	30	35	40	45	50	55
Time	0.22	0.25	0.24	0.28	0.30	0.29	0.33	0.36

(a) Describe the type of correlation shown by these data. (2)

(b) One point appears to break the general trend. Identify it and explain. (2)

(c) Use the trend to estimate the reaction time of a 60-year-old, and state one reason your estimate may be unreliable. (3)

(d) Explain why a line graph would not be an appropriate way to display these particular data. (2)

Answer keyMark scheme & solutions

Question 1 (10 marks)

(a) Data: $14,22,18,22,30,25,22,19,27,21$ ; sum $= 220$ , $n=10$ .

Mean $= 220/10 = 22$ (1 sum, 1 mean = 2)
Median: ordered $14,18,19,21,22,22,22,25,27,30$ ; middle two = 5th & 6th = $22,22$ → median $= 22$ (1 ordering, 1 value = 2)
Mode $= 22$ (appears 3 times) (1)
Range $= 30 - 14 = 16$ (1)

(b) Mean $=22$ and median $=22$ , both equal to 22 (not more than). The claim of "more than 22" is not fair — the typical/average value is exactly 22. (1 evidence, 1 conclusion = 2)

(c) Changing 30→20: the value 20 now appears once and 22 still appears three times, so mode stays 22. The median depends on the 5th & 6th ordered values, which are both 22; 30 was the largest value, and replacing it with 20 does not move it past the middle positions, so median stays 22. (1 mode, 1 median reasoning = 2)

Question 2 (11 marks)

(a) Angles sum to $360°$ : $360 - (140+100+48) = 360 - 288 = 72°$ . (2)

(b) Each commuter $= 360/90 = 4°$ . (1 for method)

Car: $140/4 = 35$
Bus: $100/4 = 25$
Bicycle: $72/4 = 18$
Walk: $48/4 = 12$ Total check $35+25+18+12 = 90$ ✓ (3 for values)

(c) Bus = 25 commuters, $25/5 = 5$ symbols (5 whole symbols, no part). (2)

(d) New car users $= 35 \times 0.75 = 26.25$ . Since total stays 90, angle $= 26.25 \times 4 = 105°$ (to nearest degree). (1 drop, 1 method, 1 answer = 3)

(Accept: reasoning that a fractional person is unrealistic but the arithmetic angle is 105°.)

Question 3 (10 marks)

(a) Midpoints: $2.5, 7.5, 12.5, 17.5$ . $\sum fx = 6(2.5)+11(7.5)+14(12.5)+9(17.5) = 15+82.5+175+157.5 = 430$ Mean $= 430/40 = 10.75$ min. (1 midpoints, 1 $\sum fx$ , 1 divide, 1 answer = 4)

(b) Modal class $= 10 \le t < 15$ (highest frequency 14). Median position = $(40+1)/2 = 20.5$ th; cumulative: 6, 17, 31 — the 20th/21st fall in $10 \le t < 15$ . So median class $= 10 \le t < 15$ . (1 modal, 2 median class with reasoning)

(c) We use midpoints (assumed representative), not exact values, so the mean is an estimate. (1)

(d) We only know class limits, not actual smallest/largest values. Data lie between $0$ and $20$ , so the range could be anything up to nearly 20 but cannot be stated exactly — the maximum possible range is 20 min (best statement: range $< 20$ , at most just under 20). (1 why uncertain, 1 best statement)

Question 4 (10 marks)

(a) Let yellow $= y$ . Total $= 12 + y$ . $P(\text{yellow}) = \dfrac{y}{12+y} = \dfrac14$ . Cross-multiply: $4y = 12 + y \Rightarrow 3y = 12 \Rightarrow y = 4$ . ✓ (1 equation, 1 solve, 1 conclusion = 3)

(b) Total $= 16$ . $P(\text{blue}) = 4/16 = 1/4$ , so $P(\text{not blue}) = 1 - 1/4 = 3/4$ . (2)

(c) Red or green $= 5 + 3 = 8$ tokens. $P = 8/16 = 1/2$ . (2)

(d) Remove 3 green: new total $= 16 - 3 = 13$ ; red still 5. $P(\text{red}) = 5/13$ (already simplest). (1 new total, 1 numerator, 1 answer = 3)

Question 5 (9 marks)

(a) As age increases, reaction time generally increases → positive correlation (moderately strong). (2)

(b) The point $(30, 0.24)$ breaks the trend — reaction time dropped from the age-25 value (0.25) instead of rising, going against the general upward pattern. (Accept (25,0.25) vs (30,0.24) discussion.) (1 point, 1 explanation = 2)

(c) Extending the upward trend, a 60-year-old ≈ 0.37–0.40 s (accept any value in this range with reasoning). Unreliable because 60 is outside the range of the data (extrapolation) — the trend may not continue. (1 estimate, 1 range/method, 1 reason = 3)

(d) Line graphs join points to show a continuous change over time; here age is not a continuous time series and different people are measured, so joining points would falsely imply intermediate values. A scatter plot is appropriate. (2)

[
  {"claim":"Q1 mean of hourly sales is 22","code":"data=[14,22,18,22,30,25,22,19,27,21]; result = (sum(data)/len(data))==22"},
  {"claim":"Q1 range is 16","code":"data=[14,22,18,22,30,25,22,19,27,21]; result = (max(data)-min(data))==16"},
  {"claim":"Q2 bicycle angle 72 and car count 35","code":"result = (360-(140+100+48)==72) and (140//4==35)"},
  {"claim":"Q3 estimated mean is 10.75","code":"fx=6*Rational(5,2)+11*Rational(15,2)+14*Rational(25,2)+9*Rational(35,2); result = simplify(fx/40 - Rational(43,4))==0"},
  {"claim":"Q4 yellow=4 gives P(yellow)=1/4","code":"y=symbols('y'); sol=solve(Eq(y/(12+y),Rational(1,4)),y); result = sol==[4]"},
  {"claim":"Q4 P(red) after removing green is 5/13","code":"result = Rational(5,13)==Rational(5,16-3)"}
]