Averages and spread

Range = largest − smallest.

$8 - 2 = 6$

The mode is the value that appears most often.

3 appears four times, more than any other number, so the mode is 3.

(a) Total $= 14+10+10+13+15+9+15+10 = 96$; mean $= 96 \div 8 = 12$ years.

(b) Range $= 15 - 9 = 6$ years.

(c) Two of the 8 children are 15, so $P = \frac{2}{8} = \frac{1}{4}$: mark halfway between 0 and $\frac{1}{2}$.

Add them: $13 + 5 + 4 + 9 + 3 + 8 = 42$

Divide by how many there are: $42 \div 6 = 7$

Multiply each weight by how many people have it:

$40 + 100 + 240 + 350 + 240 + 90 = 1060$ kg

$1060 < 1200$, so the total is less than 1200 kg.

(a) In order: 1, 5, 7, 9. The middle two are 5 and 7, so the median is $(5 + 7) \div 2 = 6$.

(b) Range $= 9 - 1 = 8$

(c) Label each bar with a name, use an evenly spaced scale up the side, and draw bars of heights 7, 9, 1 and 5.

(a) $121 - 19 = 102$

(b) Total $= 143 + 121 + 45 + 19 = 328$

Mean $= 328 \div 4 = 82$

(a) In order: 3, 4, 4, 6, 8, 9. The middle two are 4 and 6, so the median is $\frac{4+6}{2} = 5$

(b) The odd numbers are 9 and 3, so $P(\text{odd}) = \frac{2}{6} = \frac{1}{3}$

(c) Mode 3 means one hidden card is 3. Total $= 5 \times 5 = 25$, so the other card is $25 - 3 - 3 - 8 - 5 = 6$

Smallest age: stem 3, leaf 1, which is 31.

Largest age: stem 7, leaf 4, which is 74.

Range $= 74 - 31 = 43$ years

(a) 15 values, so the median is the 8th: $64$ minutes.

(b) Range $= 81 - 45 = 36$ minutes.

(c) Adults' median 64 < children's median 75, so on average the adults took less time.

Total seats $= 3 \times 380 = 1140$

A and B together: $350 + 250 = 600$

Cinema C: $1140 - 600 = 540$

(a) Use tens as the stem and units as the leaves, in order:

7 | 1 1 4 5 7
8 | 2 4 5 6 7 8 9
9 | 1 4 7 8
10 | 3

Key: 7 | 1 means 71

(b) There are 17 values, so the median is the $\frac{17+1}{2} = 9$th value: 86.

(a) $0\times3 + 1\times57 + 2\times84 + 3\times75 + 4\times81 = 0 + 57 + 168 + 225 + 324 = 774$

(b) There are 300 students, so the median is between the 150th and 151st values.

Running totals: 3, 60, 144, 219. The 150th and 151st are both in the '3' row, so the median is 3.

Change hours to minutes: $25.3 \times 60 = 1518$ minutes.

Mean: $1518 \div 115 = 13.2$ minutes.

(a) 12 values, so the median is halfway between the 6th and 7th: both are 7, so median = 7

(b) $9 - 4 = 5$

(c) Boys' median 7 > girls' 6, so boys' shoe sizes are larger on average.

Boys' range 5 > girls' 3, so boys' sizes are more spread out.

(a) Nobody scored more than 4 points, so the middle value cannot be 5. The 8th and 9th scores are both 2, so the median is 2.

(b) $0 \times 1 = 0$, not 1. The correct total is $0 + 3 + 10 + 12 + 12 = 37$.

(a) The highest frequency is 7, for 2 points, so the mode is 2.

(b) $0\times4 + 1\times3 + 2\times7 + 3\times5 + 4\times6 + 5\times5$

$= 0 + 3 + 14 + 15 + 24 + 25 = 81$

(i) Dresses only come in even sizes, so a size of 15.3 does not exist.

(ii) The mode is 14. It tells the manager which size sells most often, so which to stock more of.

Total weight of 8 parcels: $8 \times 2.5 = 20$ kg

Total of the 3 parcels: $3 \times 2 = 6$ kg

The other 5 weigh $20 - 6 = 14$ kg, so their mean is $14 \div 5 = 2.8$ kg

(a) $\sum fx = 48 + 63 + 150 + 121 + 24 = 406$

Mean $= 406 \div 41 = 9.90... \approx 9.9$ years

(b) No. 15 is how many children there are, not an age. The modal age is 10.

School A has 23 scores, so the median is the 12th: 57.

Range: $79 - 30 = 49$

Median: A (57) is higher than B (56), so A did slightly better on average.

Range: A (49) is higher than B (47), so A's scores were more spread out.

(a) Midpoints 12.5, 17.5, 22.5, 27.5, 32.5.

$\sum fx = 25 + 140 + 292.5 + 577.5 + 195 = 1230$

Mean $\approx 1230 \div 50 = 24.6$ °C

(b) No. The median is about the 25.5th value. $2 + 8 + 13 = 23$, so it is in the $25 < T \le 30$ group, not the middle group.

Midpoints: 155, 165, 175, 185, 195

$155\times8 + 165\times14 + 175\times24 + 185\times30 + 195\times4 = 14080$

$14080 \div 80 = 176$ cm

Known lengths: $3\times5 + 2.5\times8 + 1.5\times14 + 1\times10 = 15 + 20 + 21 + 10 = 66$ m.

Left for 2 m planks: $92 - 66 = 26$ m.

Number of planks: $26 \div 2 = 13$.

Girls: median 165 cm, range $170 - 150 = 20$ cm.

Boys: 15 values, so the median is the 8th: 168 cm. Range $182 - 158 = 24$ cm.

On average the boys are taller (168 > 165), and their heights are more spread out (24 > 20).

Total for all three: $3 \times 80 = £240$

Carl has $240 - 65 - 100 = £75$

Known notes: $3 \times 5 + 20 = £35$

$75 - 35 = £40$, so $40 \div 10 = 4$ notes

(a) 80 plants, so the median is about the 40th–41st value.

Running totals: 7, 20, 34, 46, ... The 40th value is in $40 < h \le 50$.

(b) Plot each frequency at the middle of its class: (15, 7), (25, 13), (35, 14), (45, 12), (55, 16), (65, 18). Join with straight lines.

Buttons already accounted for: $20 \times 7 + 21 \times 3 + 22 \times 1 = 140 + 63 + 22 = 225$

Buttons left: $320 - 225 = 95$

Bags of 19: $95 \div 19 = 5$

(a) 2 students bought 20 to 24 books: $\frac{2}{40} \times 100 = 5\%$

(b) Frequencies 11, 8, 13, 6, 2 and midpoints 2, 7, 12, 17, 22

$11 \times 2 + 8 \times 7 + 13 \times 12 + 6 \times 17 + 2 \times 22 = 22 + 56 + 156 + 102 + 44 = 380$

Mean $\approx 380 \div 40 = 9.5$

Readings: 110, 84, 78, 94, 90, 120

Total $= 576$

Mean $= 576 \div 6 = 96$

The mean equals 96, so the target was met.

Readings: 110, 84, 78, 94, 90, 120

Total $= 576$

Mean $= 576 \div 6 = 96$

The mean equals the target, so yes, the target was met.

(a) The median is about the 60th/61st value. $34 < 60 \le 34 + 29 = 63$, so it is in $100 < w \le 150$.

(b) Midpoints: 75, 125, 175, 225, 275.

$\sum fx = 2550 + 3625 + 4725 + 4275 + 3025 = 18200$

Mean $\approx 18200 \div 120 = 151.67 \approx 152$ g

(a) Total length: $4.2 \times 5 = 21$ cm

Other 4 sticks: $21 - 7 = 14$ cm, so mean $= 14 \div 4 = 3.5$ cm

(b) The other 4 sticks now total $21 - 17 = 4$ cm, so their mean goes down to $4 \div 4 = 1$ cm.

Midpoints: 5, 15, 25, 35, 45

$\sum fx = 40 + 150 + 175 + 70 + 135 = 570$

Mean $\approx 570 \div 30 = 19$ cm

Total for everyone: $50 \times 167.6 = 8380$ cm

Total for men: $20 \times 182 = 3640$ cm

Women: $(8380 - 3640) \div 30 = 4740 \div 30 = 158$ cm

Boys' total $= 18 \times 16.2 = 291.6$

Girls' total $= 27 \times 16.7 = 450.9$

Mean $= \frac{291.6 + 450.9}{45} = \frac{742.5}{45} = 16.5$ years

Red: $4 \times 5 = 20$ kg

Blue: $5 \times 9 = 45$ kg

Total $= 20 + 45 + 6 = 71$ kg

Mean $= 71 \div 10 = 7.1$ kg

$7.1 > 7$, so Donna is not correct.

(a) There are 23 ages, so the median is the 12th value: 59.

The lower quartile is the 6th value: 53. The upper quartile is the 18th value: 66.

(b) Coach A's median, 59, is below coach B's, 70, so yes.

(c) Range: A is $79 - 41 = 38$, B is $85 - 42 = 43$. IQR: A is $66 - 53 = 13$, B is $73 - 54 = 19$. Coach A varies less, so no.

Midpoints: 7.5, 12.5, 17.5, 22.5

$\sum fx = 7.5 + 25 + 122.5 + 180 = 335$

Mean $\approx 335 \div 18 = 18.61... = 18.6$ seconds

Total after 11 games: $33 \times 11 = 363$.

Mean after 10 games was $35$, so total: $35 \times 10 = 350$.

11th game: $363 - 350 = 13$ points, so Jordan is right.

(a) Midpoints: 30, 34, 38, 42.

$30 \times 24 + 34 \times 12 + 38 \times 8 + 42 \times 6 = 720 + 408 + 304 + 252 = 1684$

$1684 \div 50 = 33.68$ inches

(b) $\frac{24}{50} = 0.48$, less than half. Three quarters would be 0.75, far more than the share sold, so the manager is correct.