IB exam questions for topic 9

a)

1. The means are $\bar{m}=40$ and $\bar{p}=16$. We plot the point $\mathrm{M}(40,16)$ on the diagram and label it "M".

b) 2. The line of best fit drawn by eye should pass through M and have approximately the same number of points above and below it, following the increasing trend of the data. 3. A good choice is the line through M that stays close to the points $(10,8)$, $(65,22)$ and $(70,24)$: it runs approximately from $(0,5)$ to $(80,26)$.

c) 4. Going up from $m=20$ on the horizontal axis to the line and reading the height on the vertical axis gives a value between $10$ and $11$. 5. The estimated physics mark is therefore between $10$ and $11$ (the exact value depends on the line drawn by eye).

Answer: a) M$(40,16)$ plotted and labelled; b) line by eye through M, from $(0,5)$ to $(80,26)$ approx.; c) between $10$ and $11$

a)

1. The regression line of $h$ on $a$ always passes through the mean point $(\bar{a},\bar{h})$. As $\bar{a}=50$, we substitute into the equation: $$\bar{h}=0.22\cdot 50+15=11+15=26.$$
2. The people surveyed watch television an average of $26$ hours per week.

b) 3. We calculate two points on the line: for $a=0$, $h=15$; for $a=50$, $h=26$ (the mean point). We draw the line through $(0,15)$ and $(50,26)$, and extend it to $a=80$ (where $h\approx 32.6$).

c) 4. The gradient of the line is positive ($0.22>0$) and the cloud of points rises from left to right: the older the person, the more hours of television. The correlation is positive.

d) 5. The data correspond to ages from $25$ to $75$ years. Diogo's $18$ years lie outside this range, so using the line would be an extrapolation and the estimate would not be reliable.

Answer: a) $26$ hours; b) line through $(0,15)$ and $(50,26)$; c) the correlation is positive; d) $18$ years lies outside the range of the data (extrapolation)

a) 1. We plot the seven points $(\text{pages},\text{cost})$ with $1$ cm per $50$ pages on the horizontal axis and $1$ cm per AUD on the vertical axis.

b) 2. Using the GDC, the linear correlation coefficient is $r=-0.141$.

c) 3. As $|r|=0.141$ is very close to $0$, there is almost no linear correlation between the number of pages and the cost: the points do not lie along a line. 4. Moreover, $50$ pages is at the extreme end of the range of the data. A line of best fit would not give a reliable estimate, so Stephen's method is not appropriate.

Answer: a) scatter diagram; b) $r=-0.141$; c) not appropriate, because $r$ is very close to $0$ (almost no linear correlation)

a) Each mean is found by adding the data and dividing by $5$.

1. (i) Mean height: $$\bar{h}=\frac{182+173+162+178+190}{5}=\frac{885}{5}=177\text{cm}.$$
2. (ii) Mean weight: $$\bar{w}=\frac{73+68+60+66+75}{5}=\frac{342}{5}=68.4\text{kg}.$$

b) On the grid, the weight goes on the horizontal axis and the height on the vertical axis.

1. We mark with a cross the five points (weight, height): $(73,182)$, $(68,173)$, $(60,162)$, $(66,178)$ and $(75,190)$.
2. We also mark the mean point $(68.4;177)$, formed by the two means: the line of best fit always passes through it.
3. We draw an increasing line through the mean point that leaves the points spread on both sides, as close as possible to all of them.

Answer: a) (i) $177$ cm; (ii) $68.4$ kg; b) scatter diagram with the line of best fit passing through $(68.4;177)$

a)

1. Using the graphic display calculator (GDC), we enter the temperatures $t$ as the independent variable and the sales $s$ as the dependent variable, and perform linear regression of $s$ on $t$.
2. The GDC gives $m\approx 3.5644$ and $c\approx -14.613$, so $$s=3.56t-\mathrm{14.6.}$$

b) 3. We substitute $t=24$ using the unrounded coefficients: $$s=3.5644\cdot 24-14.613\approx \mathrm{70.9.}$$ 4. As sales are a whole number of ice creams, we round to the nearest whole number: $71$ ice creams.

Answer: a) $s=3.56t-14.6$; b) $\approx 71$ ice creams

a)

(i) 1. We enter the distances as $x$ and the number of mosquitoes as $y$ in the graphic display calculator (GDC), and perform linear regression. 2. The GDC gives $r=-0.99777\dots$, that is, $r=-0.998$. This is very close to $-1$: a very strong negative linear correlation (the further from the lake, the fewer mosquitoes).

(ii) 3. From the same calculation, the regression line of $y$ on $x$ is $$y=-0.470x+\mathrm{81.7.}$$

b) 4. We substitute $x=28$ using the unrounded coefficients ($-0.46971$ and $81.728$): $$y=-0.46971\cdot 28+81.728\approx \mathrm{68.6.}$$ 5. About $68.6$ mosquitoes (about $69$) are estimated. The distance $28$ m lies within the range of the data ($8\le x\le 82$), so the estimate is reliable.

Answer: a) (i) $r=-0.998$; (ii) $y=-0.470x+81.7$; b) $\approx 68.6$ mosquitoes

a)

1. Using the graphic display calculator (GDC), we enter the Mathematics scores as the list $x$ and the Physics scores as the list $y$.
2. We perform linear regression of $y$ on $x$; the GDC gives Pearson's correlation coefficient: $$r=\mathrm{0.814.}$$

b) 3. From the same linear regression, the GDC gives the gradient and the y-intercept: $$y=0.888x+\mathrm{13.5.}$$

c) 4. We substitute $x=72$ into the line, using the unrounded values of the coefficients: $$y=0.8877\cdot 72+13.49\approx \mathrm{77.4.}$$ 5. As $72$ lies within the range of the data ($32\le x\le 84$), the estimate is an interpolation. The estimated Physics score is about $77$ marks.

Answer: a) $r=0.814$; b) $y=0.888x+13.5$; c) $\approx 77$ marks

a) Using the GDC, linear regression of $y$ on $x$ with the five pairs of data: $a=14.9$ and $b=-80$.

$$y=14.9x-80.$$

b) We substitute $x=17$:

$$y=14.9\cdot 17-80=253.3-80=173.3\approx 173\text{calories}.$$

c) The estimate is reliable: the correlation coefficient is very high ($r\approx 0.989$) and $17$ minutes is within the range of the data ($10$ to $30$ min), so it is an interpolation.

Answer: a) $y=14.9x-80$; b) $173$ calories; c) reliable (very strong correlation and $17$ is within the range of data)

a) 1. Using the GDC, linear regression of $y$ (objects) on $x$ (age): $$y=-0.134x+\mathrm{20.9.}$$

b) 2. We substitute $x=28$ (unrounded values): $$y=-0.13435\cdot 28+20.891\approx \mathrm{17.1.}$$ 3. We estimate about $17$ objects (age $28$ is within the range $15$–$55$).

c) 4. The GDC gives the correlation coefficient: $r=-0.756$.

d) 5. The negative sign indicates that as age increases, the number of objects remembered tends to decrease. As $|r|=0.756$ is greater than $0.5$ but not very close to $1$, the linear correlation is moderately strong and negative.

Answer: a) $y=-0.134x+20.9$; b) $\approx 17$ objects; c) $r=-0.756$; d) moderately strong negative correlation

a) 1. The eighth student did $x=25$ press-ups and $y=52$ sit-ups (from the table). We plot the point $(25,52)$ on the diagram.

b) 2. The line of best fit passes through the mean point $(\bar{x},\bar{y})=(34,40)$, which we plot. We draw a line through it that leaves approximately the same number of points above and below it. The data show a gentle upward trend, so the gradient is positive (approximately $0.4$).

c) 3. From the line we read the $y$-value for $x=60$: about $50$ sit-ups.

Answer: a) point $(25,52)$; b) line through $(34,40)$; c) $\approx 50$ sit-ups

a) 1. As the French mark increases, the German mark also increases, and the points cluster around an upward line. The correlation is strong and positive.

b) 2. The mean point is $M(29,31)$ (the means of the French and German marks). We mark it on the graph.

c) 3. We draw an increasing line through $M$ that leaves approximately the same number of points above and below it.

d) 4. From $32$ on the French axis we move up to the line and read off the German axis: about $33$ marks.

Answer: a) strong positive correlation; b), c) M $(29,31)$ and line of best fit through M; d) $\approx 33$

a)

1. We enter the distances as the list of the independent variable $x$ and the prices as the list of the dependent variable $y$ in the graphic display calculator (GDC).
2. We perform linear regression of $y$ on $x$ (least squares fit). The GDC gives $a\approx 0.38422$ and $b\approx 629.42$, that is, $$y=0.384x+629.$$

b) 3. We substitute $x=2900$ km (using the unrounded coefficients): $$y=0.38422\cdot 2900+629.42\approx 1744.$$ 4. The estimated price is about $1740$ HKD (3 significant figures).

c) 5. The estimate is valid for two reasons:

• $2900$ km lies within the range of distances used to calculate the line ($1700\le x\le 9200$), so this is an interpolation.
• The correlation is linear, positive and strong ($r=0.948$, close to $1$), so the line describes the relationship between distance and price well.

Answer: a) $y=0.384x+629$; b) $\approx 1740$ HKD; c) $2900$ km lies within the range of the data and the correlation is strong ($r=0.948$)

a)

(i) 1. We enter the temperatures as $x$ and the oxygen concentrations as $y$ in the graphic display calculator (GDC), and request linear regression together with its correlation coefficient. 2. The GDC gives $r=-0.9737\dots$, that is, $r=-0.974$. This value is close to $-1$: a strong negative linear correlation (the higher the temperature, the less dissolved oxygen).

(ii) 3. From the same calculation, the regression line of $y$ on $x$ is $$y=-0.365x+\mathrm{17.9.}$$

b) 4. We substitute $x=18$ into the line, using the unrounded coefficients ($-0.36503$ and $17.942$): $$y=-0.36503\cdot 18+17.942\approx \mathrm{11.4.}$$ 5. The estimated concentration is $11.4$ mg l${}^{-1}$. As $18$ °C lies outside the range of the data ($20\le x\le 32$), this is an extrapolation and should be treated with caution.

Answer: a) (i) $r=-0.974$; (ii) $y=-0.365x+17.9$; b) $11.4$ mg l${}^{-1}$

a)

1. Using the graphic display calculator (GDC), we perform linear regression of $y$ on $x$ with the five pairs of data. It gives $$y=15.5x-80.$$
2. Check: the line must pass through the mean point $(20,230)$, and indeed $15.5\cdot 20-80=230$.

b) 3. To draw the line we need only two points:

• The $y$-intercept is $-80$: the line passes through $(0,-80)$.
• It also passes through the mean $(20,230)$.

4. We join both points with a ruler and extend the line across the diagram. It crosses the horizontal axis at $x=80/15.5\approx 5.2$, and at $x=30$ it equals $15.5\cdot 30-80=385$, that is, it reaches close to the last point.

c) 5. The line is only reliable for estimating within the range of temperatures with which it was obtained, which runs from $10$ to $30$ °C. Outside this range we do not know whether the relationship remains linear. 6. Therefore $a=10$ and $b=30$.

Answer: a) $y=15.5x-80$; b) line through $(0,-80)$ and $(20,230)$; c) $a=10$, $b=30$

a)

1. (i) Mean temperature: $$\bar{x}=\frac{7+12+14+15+16+20}{6}=\frac{84}{6}=14.$$

(ii) Mean number of drinks sold: $$\bar{y}=\frac{280+350+380+420+400+450}{6}=\frac{2280}{6}=380.$$

b) 2. Using the graphic display calculator (GDC), we perform linear regression of $y$ on $x$: $$y=13.4x+192.$$ 3. We draw the line: it passes through $(0,192)$ and the mean point $(14,380)$, and reaches about $(20,460)$. It lies very close to the points.

c) 4. We substitute $x=10$ using the unrounded coefficients ($13.404$ and $192.34$): $$y=13.404\cdot 10+192.34\approx 326.$$ 5. About $326$ cold drinks are estimated.

Answer: a) (i) $14$ °C; (ii) $380$ drinks; b) $y=13.4x+192$; c) $\approx 326$ drinks

a)

1. Using the GDC, we enter $n$ as the independent list and $t$ as the dependent list, and carry out linear regression of $t$ on $n$: $$t=-20.1n+205.$$

b) 2. From the same regression, Pearson's correlation coefficient is $r=-0.941$ (strong negative correlation: the more cleaners, the less time).

c) 3. We substitute $n=4$ using the unrounded coefficients ($-20.105$ and $204.756$): $$t=-20.105\cdot 4+204.756\approx 124.$$ 4. Four cleaners take approximately $124$ minutes.

Answer: a) $t=-20.1n+205$; b) $r=-0.941$; c) $124$ minutes

a) The unknown values are read from the diagram, which contains the fifteen points of the table.

1. The points with $x=1$ are $(1,5)$ and $(1,8)$. In the table one is $(1,5)$ and the other is $(1,a)$, so $a=8$.
2. In the table, the point $(b,4)$ appears between $x=4$ and $x=6$; in the diagram the only point with $y=4$ in that area is $(5,4)$, so $b=5$.

Check with the given means: the sum of the $x$ values is $75=15\cdot 5$ and the sum of the $y$ values is $60=15\cdot 4$.

Answer: (i) $a=8$; (ii) $b=5$

b) The line of best fit must pass through the mean point $(5,4)$ and leave approximately the same number of points on each side, following the decreasing trend. A good line is $y=-0.61x+7.04$: it cuts the vertical axis at $7.04$, passes through $(5,4)$ and reaches close to $(10,1)$.

Answer: decreasing line passing through $(5,4)$, for example $y=-0.61x+7.04$

c) We read the line at $x=6.5$:

$$y=-0.61\cdot 6.5+7.04\approx \mathrm{3.09.}$$

Values close to $3$ are accepted for a line drawn by hand.

Answer: a) (i) $a=8$; (ii) $b=5$; b) $y=-0.61x+7.04$; c) $y\approx 3.09$

a) 1. We enter the Physics marks ($x$) and Chemistry marks ($y$) into the GDC and perform linear regression: $r=0.965$.

b) 2. From the same regression, gradient $m=1.1545\dots$ and intercept $c=0.9756\dots$: $$y=1.15x+\mathrm{0.976.}$$

c) 3. We substitute $x=7$ (unrounded values): $$y=1.1545\cdot 7+0.9756\approx \mathrm{9.06.}$$ 4. The estimated mark is about $9.06$.

d) 5. It is valid because $7$ lies within the range of the Physics data ($4\le x\le 12$), so this is an interpolation, and because $r=0.965$ is very close to $1$ (strong linear correlation).

Answer: a) $r=0.965$; b) $y=1.15x+0.976$; c) $\approx 9.06$; d) $7$ lies within the range of the data and $r$ is close to $1$

a) 1. As the mathematics score increases, the game score tends to decrease, and the points follow a downward trend. There is a negative correlation.

b) 2. A line of best fit passes through the mean point $M(56.9;45.9)$. As the question also gives the point $(0,69)$, we draw the line joining the two.

3. Its gradient is $\frac{45.9-69}{56.9-0}\approx -0.41$, so it reaches about $(95,31)$.

c) 4. On the line, with $x=45$ we read off (or calculate): $y\approx 69-0.406\cdot 45\approx 50.7$. 5. Rounding to the nearest integer, Jane would obtain about $51$ points.

Answer: a) negative correlation; b) line through $(0,69)$ and M; c) $\approx 51$ points

To find Spearman's coefficient, we first replace the data by their ranks (from smallest to largest).

1. Ranks of $x$: $1,2,3,4,5$.
2. Ranks of $y$: the values $6,7,8,8,16$ have a tie in the two $8$s, which occupy positions $3$ and $4$; they receive the mean rank $3.5$. Ranks: $1,2,3.5,3.5,5$.
3. As there are ties, we calculate Pearson's product-moment correlation coefficient of the two sets of ranks (using the GDC): $$r_s=\mathrm{0.975.}$$
4. This is very close to $1$: a strong positive correlation (as $x$ increases, $y$ increases).

Answer: $r_s=0.975$ (strong positive correlation)

1. The preference data are already ranks (1 = the most preferred). We order the prices from highest to lowest, so that rank $1$ is the most expensive pizza: $8.50\to 1$; $7.45\to 2$; $6.50\to 3$; $4.50\to 4$; $4.35\to 5$; $3.95\to 6$; $3.50\to 7$; $2.65\to 8$; $1.45\to 9$.
2. Price ranks for A to I: $7,2,3,4,1,8,6,5,9$.
3. Differences $d$ between the preference rank and the price rank: $0,1,1,2,0,1,-4,0,-1$, and $\sum d^2=0+1+1+4+0+1+16+0+1=24$.
4. There are no ties, so $$r_s=1-\frac{6\sum d^2}{n(n^2-1)}=1-\frac{6\cdot 24}{9\cdot 80}=1-0.2=\mathrm{0.8.}$$
5. Comment: the correlation is strong and positive. As rank $1$ is both the most preferred pizza and the most expensive, the pizzas the child likes most tend to be the most expensive ones. (If the prices were ordered from lowest to highest, the sign would be reversed, $-0.8$, with the same strength of relationship.)

Answer: $r_s=0.8$ (strong positive correlation between the preference rank and the price rank, with rank $1$ for the most expensive pizza)

a) The rankings of the players are already given ($1,2,\dots ,8$; rank $1$ is the best). For the aces we assign ranks from highest to lowest, so that rank $1$ is the largest number of aces: $10\to 1$; $7\to 2$; $6\to 3$; $5$ (twice, positions $4$ and $5$) $\to 4.5$; $4\to 6$; $3\to 7$; $1\to 8$. For players $1$ to $8$ the ranks of the aces are $1,2,6,3,4.5,4.5,7,8$.

There is a tie in the $5$ aces, so we calculate Pearson's product-moment correlation coefficient between the two sets of ranks (using the GDC): $$r_s=\mathrm{0.850.}$$

b) $r_s=0.850$ is close to $1$: a strong positive correlation. As rank $1$ corresponds both to the best player and to the player with the most aces, the better the player, the more aces they win. The validity is reasonable but limited: there are only $8$ data points (a small sample, one match per player), there is a tie, and a correlation does not prove causation.

Answer: a) $r_s=0.850$; b) strong positive correlation: the better-ranked players win more aces

a) We assign ranks from smallest to largest in each variable; repeated values receive the mean rank.

1. Heights (already in order, with $153$ and $164$ repeated): $1;2.5;2.5;4;5;6;7;8.5;8.5;10;11;12$.
2. Times: $17.5\to 11$; $18\to 12$; $16.5\to 10$; $16\to 9$; $15.4\to 8$; $13.2$ (twice, positions $4$ and $5$) $\to 4.5$; $14\to 7$; $13.7\to 6$; $12.5\to 3$; $12$ (twice, positions $1$ and $2$) $\to 1.5$. Ranks: $11;12;10;9;8;4.5;7;6;4.5;3;1.5;1.5$.
3. There are ties, so we calculate Pearson's product-moment correlation coefficient between the two sets of ranks (using the GDC): $$r_s=-\mathrm{0.952.}$$

b) $r_s=-0.952$ is very close to $-1$: a strong negative correlation. Taller students tend to take less time to run the $100$ m, that is, they are faster. It is a valid trend in this sample of $12$ students, although the sample is small and a correlation does not prove that height is the cause.

Answer: a) $r_s=-0.952$; b) strong negative correlation: the taller students are the faster ones

a) Ranks from smallest to largest (repeated values receive the mean rank).

1. Age: $11\to 1$; $18\to 2$; $21\to 3$; $24\to 4$; $32\to 5$; $43\to 6$; $66\to 7$; $85\to 8$.
2. Reaction time: $0.15\to 1$; $0.20$ (twice, positions $2$ and $3$) $\to 2.5$; $0.25\to 4$; $0.30\to 5$; $0.35\to 6$; $0.45\to 7$; $0.70\to 8$.

b) With ties, we calculate Pearson's product-moment correlation coefficient of the ranks (using the GDC): $$r_s=\mathrm{0.707.}$$

c) $r_s=0.707$ is positive and relatively close to $1$: there is a fairly strong positive correlation. In this sample, the older the person, the longer the reaction time (the slower they are).

d) Two possible improvements: (1) increase the number of participants (a sample of $8$ is very small); (2) repeat the test several times with each person and use the mean, and/or choose ages spread more evenly with the same number of people in each age band, carrying out the test under identical conditions.

Answer: a) ranks of age $5,4,8,1,6,7,2,3$ and of reaction time $6;2.5;8;5;4;7;1;2.5$; b) $r_s=0.707$; c) fairly strong positive correlation; d) more participants and repeating the test under identical conditions

a) Rank from smallest to largest; repeated values are given the mean rank.

1. Age: $8\to 1$; $14\to 2$; $17\to 3$; $24\to 4$; $28\to 5$; $31\to 6$; $39\to 7$; $40\to 8$; $42$ (appears twice, in positions $9$ and $10$) $\to 9.5$; $45\to 11$; $81\to 12$.
2. Hours: $2\to 1$; $14\to 2$; $15\to 3$; $16\to 4$; $19\to 5$; $20$ (appears twice, in positions $6$ and $7$) $\to 6.5$; $21\to 8$; $25\to 9$; $26\to 10$; $28\to 11$; $30\to 12$.

b) There are ties, so we calculate Pearson's product-moment correlation coefficient of the ranks (using a graphic display calculator): $$r_s=-\mathrm{0.596.}$$

c) $r_s=-0.596$ indicates a moderate negative correlation: in this sample, older people tend to watch less television. However, correlation does not imply causation: we cannot conclude that age causes people to watch less television (other factors may be influential, and the sample is only $12$ people). Therefore, Neeve's claim is not justified by these calculations; we can only conclude that there is a moderate negative association between the two variables.

Answer: a) ranks of age $1;9.5;3;12;11;2;7;9.5;6;8;5;4$ and of hours $6.5;3;12;1;9;11;5;2;4;8;10;6.5$; b) $r_s=-0.596$; c) the claim is not correct: moderate negative correlation, which does not prove causation

a) Using the calculator (one-variable statistics) on the Round 1 values:

1. Mean: $\bar{x}=\frac{869}{12}\approx 72.4$.
2. Standard deviation: ${\sigma }_x\approx 4.77$ (population standard deviation).
3. We calculate the threshold: $\bar{x}+{\sigma }_x\approx 72.42+4.77=77.19$. The scores above this value are $79$ and $82$, that is, 2 golfers.

Answer: (i) $72.4$; (ii) $4.77$; (iii) $2$ golfers

b) Using the calculator (linear regression of $y$ on $x$): $r\approx 0.990$.

Answer: $r=0.990$

c) With the same calculations: $a\approx 1.014$ and $b\approx 0.816$.

$$y=1.01x+\mathrm{0.816.}$$

Answer: $y=1.01x+0.816$

d) We substitute $x=70$ (unrounded coefficients):

$$y=1.01405\cdot 70+0.81618\approx \mathrm{71.8.}$$

Answer: $y\approx 71.8$

e) No, it is not appropriate: the score $89$ is outside the range of the Round 1 data ($66$ to $82$), so using the line would be an extrapolation and the estimate would not be reliable.

Answer: a) (i) $72.4$; (ii) $4.77$; (iii) $2$ golfers; b) $r=0.990$; c) $y=1.01x+0.816$; d) $71.8$; e) no, $89$ is outside the range of the data

a) Using the calculator (one-variable statistics) on the $13$ temperatures:

1. Mean: $\bar{T}\approx 19.2{}^{\circ }\text{C}$ (exactly $19.15\dots$).
2. Standard deviation: ${\sigma }_T\approx 1.45{}^{\circ }\text{C}$.

Answer: (i) $19.2{}^{\circ }\text{C}$; (ii) $1.45{}^{\circ }\text{C}$

b) Using the calculator (linear regression of $n$ on $T$): $r\approx 0.942$.

Answer: $r=0.942$

c) $r=0.942$ is positive and close to $1$: strong positive correlation. The higher the temperature, the more bottles are sold.

Answer: strong positive correlation

d)

1. (i) The calculator gives the line $n\approx 11.5T-100$, so $d=11.5$.
2. (ii) For $T=19.6$: $n=11.5\cdot 19.6-100\approx 125$ bottles (with $d=11.518$ we get $125.4$).

Answer: (i) $d=11.5$; (ii) $125$ bottles

e) The estimate is not reliable because $6^{\circ }$ is far outside the range of temperatures in the data ($17^{\circ }$ to $21^{\circ }$): it would be an extrapolation.

Answer: a) (i) $19.2{}^{\circ }$C; (ii) $1.45{}^{\circ }$C; b) $r=0.942$; c) strong positive correlation; d) (i) $d=11.5$; (ii) $125$ bottles; e) $6^{\circ }$ is outside the range of data

a), c), e) Scatter diagram, mean point and regression line.

1. We plot the ten points $(x,y)$ using the scale indicated (2 cm = $10\%$ on $x$; 2 cm = $10$ heat units on $y$).
2. We plot and label $\mathrm{M}(42,64)$ (part b).
3. We draw the line $y=-0.470x+83.7$: it cuts the vertical axis at $83.7$ and passes through $\mathrm{M}$. As a second check, for $x=90$, $y=-0.470\cdot 90+83.7=41.4$.

Answer: graph with the ten points, $\mathrm{M}(42,64)$ and the line $y=-0.470x+83.7$

b)

1. $\bar{x}=\frac{8+15+22+30+34+45+50+60+74+82}{10}=\frac{420}{10}=42\%$.
2. $\bar{y}=\frac{80+77+74+69+68+61+61+55+50+45}{10}=\frac{640}{10}=64$ heat units.

Answer: (i) $\bar{x}=42\%$; (ii) $\bar{y}=64$

d) Using the calculator (linear regression): $r\approx -0.998$. Very strong negative correlation.

Answer: $r=-0.998$

f) We substitute $x=25$ into the given line:

$$y=-0.470\cdot 25+83.7=-11.75+83.7=71.95\approx 72.0\text{heat units}.$$

Answer: $72.0$ heat units

g) Yes, it is appropriate: the correlation is very strong ($r=-0.998$) and $25\%$ is within the range of the data ($8\%$ to $82\%$), so it is a reliable interpolation.

Answer: b) (i) $\bar{x}=42\%$; (ii) $\bar{y}=64$; d) $r=-0.998$; f) $72.0$ heat units; g) yes, appropriate (strong correlation and $25\%$ is within the range of data)

a) 1. We plot the nine pairs $(x,y)$ on axes with $2$ cm per $10$ m on the horizontal axis and $2$ cm per $10$ species on the vertical axis.

b) 2. As the distance from the shore increases, the number of species decreases and the points follow a fairly good downward line. The correlation is negative (and strong).

c) 3. We enter the data into the GDC and request one-variable statistics:

(i) $\bar{x}=17$ m;

(ii) $\bar{y}=23$ species.

d) 4. We plot $M(17,23)$ on the diagram (figure above). The regression line always passes through this point.

e) 5. Using the GDC, linear regression of $y$ on $x$: $$y=-0.708x+\mathrm{35.0.}$$

f) 6. To draw the line we only need two points: for example $(0,35)$ and $M(17,23)$ (the line passes through $M$). Join them and extend the line across the diagram (figure above).

g) 7. We substitute $x=30$ (using the unrounded regression values): $$y=-0.708\cdot 30+35.04\approx \mathrm{13.8.}$$ 8. As $30$ m is within the range of the data ($2\le x\le 40$), the estimate is reliable. Rounding to a whole number of species: $14$.

Answer: a) scatter diagram; b) negative correlation; c) (i) $\bar{x}=17$; (ii) $\bar{y}=23$; d) point $M(17,23)$; e) $y=-0.708x+35$; f) line through $(0,35)$ and $M$; g) $\approx 14$ species

Using the GDC, we enter the travel times $x$ and the stress levels $y$ and perform linear regression of $y$ on $x$.

(i) The correlation coefficient is $r=0.667$.

(ii) A positive correlation coefficient indicates that as one variable increases, the other tends to increase too: the longer the travel time, the higher the stress level.

(iii) The GDC gives gradient $a=0.181$ and intercept $b=2.22$: $$y=0.181x+\mathrm{2.22.}$$

(iv) Three quarters of an hour is $x=45$ minutes. We substitute (using unrounded values): $$y=0.1810\cdot 45+2.2235\approx \mathrm{10.4.}$$ As the scale goes up to $10$ and the level is a whole number, the estimated stress level is $10$.

(v) It is not reliable, for two reasons:

1. $x=45$ lies outside the range of the data ($6\le x\le 36$): it is an extrapolation.
2. $r=0.667$ is not close to $1$: the correlation is only moderate.

Answer: (i) $r=0.667$; (ii) as the travel time increases, the stress level increases; (iii) $y=0.181x+2.22$; (iv) level $\approx 10$; (v) not reliable (extrapolation and moderate correlation)

a)

1. We draw the axes: the horizontal axis ($x$, temperature difference) from $-0.3$ to $0.5$ with $2$ cm per $0.1$ °C, and the vertical axis ($y$, beats per minute) from $60$ upwards with $1$ cm per $2$ beats.
2. We plot the eight points: $(-0.2;63)$, $(0.3;77)$, $(-0.3;70)$, $(-0.2;74)$, $(-0.1;65)$, $(0;78)$, $(0.2;79)$ and $(0.5;86)$.

b)

(i) 3. The mean of $x$ is the sum of the eight values divided by $8$: $$\bar{x}=\frac{-0.2+0.3-0.3-0.2-0.1+0+0.2+0.5}{8}=\frac{0.2}{8}=\mathrm{0.025.}$$

(ii) 4. The mean of $y$: $$\bar{y}=\frac{63+77+70+74+65+78+79+86}{8}=\frac{592}{8}=74.$$

c) 5. We plot the point $\mathrm{M}(0.025;74)$ on the diagram (almost on the vertical axis, at height $74$) and label it "M".

d)

(i) 6. Using the graphic display calculator (GDC), in the linear regression calculation, we obtain $r=0.8068\dots$, that is, $r=0.807$.

(ii) 7. As $r$ is positive and lies between $0.7$ and $1$, the correlation is positive and moderately strong: the higher the body temperature, the more beats per minute tend to occur.

e) 8. The linear regression of $y$ on $x$ gives $y\approx 21.98x+73.45$, that is, $$y=22.0x+\mathrm{73.5.}$$

f) 9. The line passes through M. We calculate another point to draw it with a ruler: at $x=-0.3$, $y=21.98\cdot (-0.3)+73.45\approx 66.9$; at $x=0.5$, $y\approx 84.4$. We join these points, passing through M.

Answer: a) scatter diagram with the 8 points; b) (i) $\bar{x}=0.025$; (ii) $\bar{y}=74$; c) M$(0.025;74)$ on the diagram; d) (i) $r=0.807$; (ii) positive and moderately strong correlation; e) $y=22.0x+73.5$; f) line drawn through M, from $(-0.3;66.9)$ to $(0.5;84.4)$

a)

(i) 1. We enter the project marks as $x$ and the exam marks as $y$ in the GDC, and request the two-variable statistics. The mean of the project marks is $\bar{x}=14$.

(ii) 2. The mean of the exam marks is $\bar{y}=54$.

(iii) 3. Pearson's correlation coefficient is $r=0.5$.

b)

(i) 4. The GDC gives the regression line of $y$ on $x$ with the exact values $$m=0.875=\frac{7}{8},c=41.75=\frac{167}{4}.$$

(ii) 5. The regression line always passes through the mean point. We check by substituting $x=\bar{x}=14$: $$y=0.875\cdot 14+41.75=12.25+41.75=54=\bar{y}.$$ 6. As the equality holds, $\mathrm{M}(14,54)$ lies on the regression line.

c)

(i) 7. We substitute $x=17$: $$y=0.875\cdot 17+41.75=56.625\approx \mathrm{56.6.}$$

(ii) 8. The value $x=17$ lies within the range of the data ($6\le x\le 18$), so this is an interpolation. 9. However, the correlation is low ($r=0.5$), that is, only moderate: the line describes the relationship between the two marks poorly. Therefore the estimate is not very reliable and it is not entirely valid to use the line.

d) 10. The percentage error compares the estimated value with the real one: $$\text{error}=\left|\frac{56.625-65}{65}\right|\cdot 100=\frac{8.375}{65}\cdot 100\approx 12.9\%.$$

Answer: a) (i) $\bar{x}=14$; (ii) $\bar{y}=54$; (iii) $r=0.5$; b) (i) $m=0.875$, $c=41.75$; (ii) $0.875\cdot 14+41.75=54=\bar{y}$; c) (i) $56.6$; (ii) not very valid: $17$ is within the range but $r=0.5$ is low; d) $12.9\%$

a)

1. The range is the difference between the largest and the smallest mean body weights: $529-3=526$ kg.

b)

(i) 2. We enter the body weights as $x$ and the brain weights as $y$ in the graphic display calculator (GDC) and perform linear regression. It gives $r=0.9218\dots$, that is, $r=0.922$.

(ii) 3. As $r$ is positive and very close to $1$, the correlation is positive and strong: mammals with a greater body weight tend to have a heavier brain.

c) 4. From the same calculation, the GDC gives $m\approx 0.00098584$ and $c\approx 0.092339$: $$y=0.000986x+\mathrm{0.0923.}$$

d) 5. We substitute $x=36$ (using the unrounded values of $m$ and $c$): $$y=0.00098584\cdot 36+0.092339\approx \mathrm{0.128.}$$ 6. The estimated brain weight of a grey wolf is $0.128$ kg.

e) 7. The percentage error is calculated from the estimated and the real value: $$\text{error}=\left|\frac{0.12783-0.120}{0.120}\right|\cdot 100\approx 6.52\%.$$ 8. If the estimate already rounded from d), $0.128$ kg, is used, we get $\left|\frac{0.128-0.120}{0.120}\right|\cdot 100\approx 6.67\%$. To 3 significant figures, the error is approximately $6.5\%$ if the estimate is not rounded and $6.67\%$ if we start from $0.128$.

f) 9. Not valid. The body weight of a mouse, $0.023$ kg, lies outside the range of the data used to obtain the line ($3\le x\le 529$): it would be an extrapolation far beyond the data. Moreover, the line would give $y=0.0923$ kg, a brain heavier than that of the cat ($3$ kg body, $0.026$ kg brain) and greater than the whole body weight of the mouse, which is absurd.

Answer: a) $526$ kg; b) (i) $r=0.922$; (ii) positive and strong; c) $y=0.000986x+0.0923$; d) $0.128$ kg; e) $\approx 6.52\%$ ($6.67\%$ if starting from $0.128$); f) not valid, it lies outside the range of the data

a)

1. We plot the six points $(x,C)$: $(10,35)$, $(15,40)$, $(22,60)$, $(30,72)$, $(28,68)$ and $(21,55)$. On the horizontal axis, $2$ cm represents $5$ (thousand folders); on the vertical axis, $2$ cm represents $10$ (thousand euros).

b) 2. (i) Mean number of folders produced: $$\bar{x}=\frac{10+15+22+30+28+21}{6}=\frac{126}{6}=21\text{ (thousand)}.$$

(ii) Mean production costs: $$\bar{C}=\frac{35+40+60+72+68+55}{6}=\frac{330}{6}=55\text{ (thousand euros)}.$$

c) 3. We mark and label the point $\mathrm{M}(21,55)$ on the diagram.

d) 4. Using the GDC, we perform linear regression of $C$ on $x$: the correlation coefficient is $r=0.990$ (exact value $0.9896$).

e) 5. As $r$ is very close to $1$, the positive linear correlation is very strong: the points are almost in a straight line, so a line describes the relationship well.

f) 6. The GDC gives the regression line of $C$ on $x$: $$C=1.94x+\mathrm{14.2.}$$

g) 7. We draw the line joining, for example, $(0;14.2)$ and $(30;72.4)$. It must pass through the mean point $\mathrm{M}(21,55)$.

h) 8. If $x$ thousand folders are sold at $2.99$ euros each, the revenue is $2.99x$ thousand euros. To exceed the costs: $$2.99x>1.9410x+\mathrm{14.2396.}$$ 9. Solving for $x$: $$1.0490x>14.2396⟹x>\mathrm{13.574.}$$ 10. That is, more than $13573.8$ folders must be sold, so the minimum is $13574$ folders.

Answer: b) (i) $21$ thousand folders; (ii) $55$ thousand euros; d) $r=0.990$; e) very strong linear correlation ($r$ close to $1$); f) $C=1.94x+14.2$; h) $13574$ folders

a)

1. We read the missing points from the diagram: the fourth point has concentration $89$ and is at $x=4.2$ km, so $a=4.2$. The point with $x=8.6$ km has concentration $74$, so $b=74$.

b) 2. We enter the ten pairs of data in the graphic display calculator (GDC) and calculate the statistics:

(i) $\bar{x}=5.91$ km.

(ii) $\bar{y}=88$ micrograms per cubic metre.

(iii) With linear regression, $r=-0.956$ (exact value $-0.9555$).

c) 3. The GDC gives the regression line of $y$ on $x$: $$y=-5.39x+120.$$

d) 4. (i) We substitute $x=14$ using the unrounded coefficients ($-5.3896$ and $119.852$): $$y=-5.3896\cdot 14+119.852\approx \mathrm{44.4.}$$

(ii) The estimate is not reliable: $14$ km lies outside the range of distances in the data ($0.6$ to $12.2$ km), so it is an extrapolation.

Answer: a) $a=4.2$, $b=74$; b) (i) $5.91$ km; (ii) $88$; (iii) $r=-0.956$; c) $y=-5.39x+120$; d) (i) $44.4$; (ii) not reliable: $14$ km lies outside the range of the data