Averages (Mean, Median, Mode and Range): Question 2

Syllabus C9.3, E9.3

Structured Extended 8 marks

A sports coach records the number of goals scored in each match by two of the school's hockey teams during one season.

(a) The table shows the number of goals scored by the Falcons in their 40 matches.

Goals scored (xx) 0 1 2 3 4 5
Number of matches (ff) 6 11 9 8 4 2

Calculate the mean number of goals scored per match by the Falcons. [3]

(b) The table below shows the number of goals scored by the Kestrels. The number of matches in which they scored 3 goals is unknown and is represented by pp.

Goals scored (xx) 0 1 2 3 4
Number of matches (ff) 3 6 12 pp 5

The mean number of goals scored per match by the Kestrels is exactly 2.22.2.

Find the value of pp. [3]

(c) Using your results from parts (a) and (b), calculate the mean number of goals per match for all of the Falcons' and Kestrels' matches combined. [2]

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Worked solution

Part (a): Mean from a frequency table

For data in a frequency table the mean is

xˉ=fxf\bar{x}=\frac{\sum fx}{\sum f}

where fx\sum fx is the total of (value ×\times frequency) and f\sum f is the total frequency (here, the total number of matches).

Step 1: Find fx\sum fx (total number of goals).

fx=(0×6)+(1×11)+(2×9)+(3×8)+(4×4)+(5×2)\sum fx = (0\times 6)+(1\times 11)+(2\times 9)+(3\times 8)+(4\times 4)+(5\times 2)

fx=0+11+18+24+16+10=79\sum fx = 0+11+18+24+16+10 = 79

Step 2: Find f\sum f (total number of matches).

f=6+11+9+8+4+2=40\sum f = 6+11+9+8+4+2 = 40

Step 3: Divide.

xˉ=7940=1.975\bar{x}=\frac{79}{40}=1.975

Mean=1.975 goals per match\boxed{\text{Mean} = 1.975 \text{ goals per match}}


Part (b): Find a missing frequency given the mean

Use the same formula, but now fx\sum fx and f\sum f both contain the unknown pp.

Step 1: Write fx\sum fx in terms of pp.

fx=(0×3)+(1×6)+(2×12)+(3×p)+(4×5)\sum fx = (0\times 3)+(1\times 6)+(2\times 12)+(3\times p)+(4\times 5)

fx=0+6+24+3p+20=50+3p\sum fx = 0+6+24+3p+20 = 50+3p

Step 2: Write f\sum f in terms of pp.

f=3+6+12+p+5=26+p\sum f = 3+6+12+p+5 = 26+p

Step 3: Form an equation using mean =2.2=2.2.

50+3p26+p=2.2\frac{50+3p}{26+p}=2.2

Step 4: Solve. Multiply both sides by (26+p)(26+p):

50+3p=2.2(26+p)50+3p = 2.2(26+p)

50+3p=57.2+2.2p50+3p = 57.2+2.2p

3p2.2p=57.2503p-2.2p = 57.2-50

0.8p=7.20.8p = 7.2

p=7.20.8=9p = \frac{7.2}{0.8}=9

p=9\boxed{p = 9}

Check: with p=9p=9, total goals =50+3(9)=77=50+3(9)=77 and total matches =26+9=35=26+9=35, giving 7735=2.2\dfrac{77}{35}=2.2. ✓

Alternative method: The total number of goals must equal mean×matches=2.2(26+p)\text{mean}\times\text{matches}=2.2(26+p). Setting this equal to the table total 50+3p50+3p gives the same equation 50+3p=2.2(26+p)50+3p=2.2(26+p).


Part (c): Combined mean

When combining two data sets, add the totals; never average the two means:

combined mean=total goals from both teamstotal matches of both teams\text{combined mean}=\frac{\text{total goals from both teams}}{\text{total matches of both teams}}

Step 1: Totals for each team.

  • Falcons: fx=79\sum fx = 79 goals in 4040 matches (from part (a)).
  • Kestrels: fx=50+3(9)=77\sum fx = 50+3(9)=77 goals in 3535 matches (from part (b)), or equivalently 2.2×35=772.2\times 35 = 77.

Step 2: Combine.

combined mean=79+7740+35=15675=2.08\text{combined mean}=\frac{79+77}{40+35}=\frac{156}{75}=2.08

Combined mean=2.08 goals per match\boxed{\text{Combined mean} = 2.08 \text{ goals per match}}

(Note that the simple average 1.975+2.22=2.0875\tfrac{1.975+2.2}{2}=2.0875 is incorrect, because the two teams played different numbers of matches.)