\(\begin{align} & \frac{x}{6.023\times {{10}^{23}}}\times \text{ 313}\text{.53}=\text{ 2000} \\ & \\ & \text{Therefore: }x=\frac{\text{2000}\times 6.023\times {{10}^{23}}}{313.53}=3.84\times {{10}^{24}}\text{ molecules} \\ \end{align} \)

A possible markscheme is presented for some (but not all) questions: The IB markscheme process is a long and rigorous one, the aim here is only to give a rough guideline. Other markschemes are also possible.

M1 rearranging formula correct & substituting in correct numbers (2 or 2000 accepted for the “weight in grams”)

A1 correct answer.

(i)

\(\frac{\text{MAXimum total quantity of I}{{\text{n}}_{2}}{{\text{O}}_{5}}\text{Sn}}{\text{MINimum quantity of I}{{\text{n}}_{2}}{{\text{O}}_{5}}\text{Sn per screen}}=\frac{2.005}{0.00048}=4177.083333=4\text{ }177\text{ (to nearest integer)}\)

(ii)

\(\begin{align} & \frac{\text{MINimum total quantity of I}{{\text{n}}_{2}}{{\text{O}}_{5}}\text{Sn}}{\text{MAXimum quantity of I}{{\text{n}}_{2}}{{\text{O}}_{5}}\text{Sn per screen}}=\frac{1.995}{0.00052}=3836.538462=3\text{ }836\text{ } \\ & \text{rounded down as won }\!\!'\!\!\text{ t be enough for the 3 836 computer screen}\text{.} \\ \end{align} \)

4000 - 3 942 = 58 computers less than the 4000 it anticipated making.

\(\frac{58}{4000}\times 100=1.45%\)

(a) This question can be modelled binomially, given it is a "success or failure" (heating element works or it doesn't) situation and assuming that the chance of an element being faulty is independent of all others. 

p = probability of a heating element being faulty/not working = 0.05, therefore the probability that it does work = 0.95, n = number of trials = 5 kettles per box. 

\(X\sim B(5,0.05)\)

GDC calculation: p(x≥2) - since if 2,3,4 or 5 kettles are NOT working = supermarket will complain:

Lower:2
Upper: 5
Numtrial(n): 5
p : 0.05

Using p=0.95 gives the same answer, working below:

\(X\sim B(5,0.95)\)

p(x ≤ 3 working kettles) = binomcdf(5,3,0.05) or 1 – [5C5  + 5C4 (with their associated probabilities)].

Answer:  0.0226 or 2.26% (to 3 sig fig)

A possible markscheme is suggested below (please do not contact us with questions about markschemes offered on this site. The IB markscheme process is a long and rigorous one, the aim here is only to give a rough guideline. Other markschemes are also possible).

(M1) for correctly identifying n=5 and p=0.05 in a binomial model. 

M1 for correctly modelling the problem using p(x ≤ 3 working) or other valid probability model e.g. 1 - p(x≥4) 

A1 A1 for the correct answer rounded correctly to 3 significant figures.

Conceptually, students will likely find the below probability tree useful to see why there are 26 different options that give either 1, 2 or 3 working kettles, from a pack of 5. A binomial probability is a quick way to work out the number of different ways something can happen and the probabilities associated with each of them.  

(b) Answer: the probability of more than one kettle in each box of five being substandard is 2.26% (to 3 sig fig). This is less than their 3% target, therefore they have met their target. 

[Award A1 if the student has correctly compared their answer (right or wrong) to 3% and come to the correct conclusion, given their answer in part (a)] 

[1]

(a) \(\left( \frac{\mathop{x}_{Marissa}+\mathop{x}_{Hannah}}{2},\frac{\mathop{y}_{Marissa}+\mathop{y}_{Hannah}}{2} \right)=\left( \frac{4+-2}{2},\frac{1+3}{2} \right)=\left( 1,2 \right)\)    Answer:  mid-point = (1,2)

(b) Gradient of line between Hannah and Marissa = \(\frac{rise}{run}=\frac{1-3}{4-(-2)}=\frac{-2}{6}=-\frac{1}{3}\)   (M1)

Therefore gradient of line perpendicular to this  = 3  (since gradient of line x gradient of perpendicular = -1)          (A1)

From part (a) we know the perpendicular bisector MUST pass through the point (1,2) = mid-point between Hannah and Marisa. We can therefore substitue this in to our slope (gradient), intercept form of the linear equation:

\(\begin{align} & x=1,y=2 \\ & y=3x+(y-\text{intercept)} \\ & \text{2 = 3(1) + y-intercept} \\ & \text{y-intercept = 2 - 3 = -1 } \\ \end{align} \)

Equation of the perpendicular bisector:   y = 3x – 1    A1

Straight line drawn correctly through y =  -1 and with gradient = 3    A1 A1

(c) Working out/reasoning shown below, final answer follows in the second diagram down:

> Erase segment DC
> Fill in the dashes between points  B and C.

> Fill in the dashes for segment CG

> Fill in the dashes above point C on the perpendicular bisector for Hannah and Amina

Remaining lines should be left dashed. 

(a) 

A1  for two pairs of crosses and a bunched group of five crosses roughly in line with those shown above.

A1  for the regression line passing very close to the second cross and with three data points above and two below in the final cluster

A1 for the correct regression equation

(b) Using the same process as in part (a) [see GDC 'button-press' screenshots on previous question],  "List3(income)" can be entered as the x-variable (Xlist) and "List2(education)" the y-variable (Ylist). 

The 'ax+b' option now calculates an x(education)-on-y(income) regression. Use the "y-calc" option to predict the mean number of years of education for a woman whose income = $30 000.

(c) In part (a) the y-on-x regression implies that education is the independent variable, and is a suspected 'cause' for changes in income.

In (b) the x-on-y regression would imply that income is the independent variable and is a suspected 'cause' for changes in mean years of education [Note: given most education is obtained before a person begins earning an income, this seems the less likely vector of causation]

(a) 0.0111 x 3600 seconds (in 1h) = 39.96mph \(\approx \) 40

(b) The distance covered between 4^th and 6^th second = 0.0464 miles (to 3 significant figures):

\(\begin{align} & \left( 0.0203+c \right)\times \frac{6\text{secs}-4\text{secs}}{2}=0.0464 \\ & \text{Therefore: c = 0}\text{.0471 - 0}\text{.0203 = 0}\text{.0261 (to 3 s}\text{.f}\text{.)} \\ \end{align} \)

M1 correct equation set-up  A1 answer for ‘c’

If c = 0.0261

Trapezium rule (in formula booklet):

 [0 + 0.0333 + 2(0.0111 + 0.0203 + 0.0261 + 0.0306 + 0.0319 + d + 0.0331 + 0.0331)] = 0.471

2d = 0.471 – 0.0333 - 2(0.0111 + 0.0203 + 0.0261 + 0.0306 + 0.0319 + 0.0331 + 0.0331)

           d =\(\frac{0.0629}{2}=0.03145\)

M1 correct equation set-up  A1 answer for ‘d’

7(a) Casio CG10/20/50 'how to' plot scatter graphs (from Youtuber "eleven bottles") Warning: the numbers used in this video are not the same as in the above question, but the method/how to use your Casio CG10/20/50 is exactly the same):

TiNspire scatter graph (from Youtuber: Nate Murphy), regression (not required in this question), and Pearson's correlation coefficient calculation (which is exactly the same method as for Spearman's only in Spearman's we enter the rank of each number, rather than the raw data). Warning: the numbers used in this video are not the same as in the above question, but the method/how to use your TiNspire is exactly the same):

 

(b) calculating spearmans: rank the data then follow same process as for Pearson's

(c)(i) same r_s value as in (b) since ranks haven't change: r_s = -1

(ii) Spearman's rank correlation coefficient is insensitive to changes in the actual data that do not change the rank order of the data.

(a)

(b) 

(M1) correct diagram (SL3.3) with lengths correctly marked
 M1 correct method and numbers substituted into the formula
 A1 2.3° (no mark if not rounded correctly to 1 d.p.).

\(\begin{align} & \sin \theta =\frac{77.9}{1942} \\ & \theta =\arcsin \left( \frac{77.9}{1942} \right)=2.3{}^\circ \\ \end{align} \)

(a) [2 marks]

M1 working showing candidate has understood that:

thickness of artery = area of outer wall circle – area of inner wall circle.

A1 correct working out and some mention, or working, clearly showing the candidate has understood that\(\text{common factor = }\pi \)

\(\begin{align} & \pi {{\left( 6.1+x \right)}^{2}}\text{ }-\text{ }\pi ({{6.1}^{2}}) \\ & \text{common factor = }\pi \\ & A(x)=\pi \left[ {{\left( 6.1+x \right)}^{2}}-{{6.1}^{2}} \right] \\ \end{align} \)

(b) [2 marks]

\(\begin{align} & \frac{A}{\pi }=~{{\left( 6.1+x \right)}^{2}}\text{ }-\text{ }{{6.1}^{2}} \\ & \frac{A}{\pi }\text{+ }{{6.1}^{2}}=~{{\left( 6.1+x \right)}^{2}}\text{ } \\ & \sqrt{\frac{A}{\pi }\text{+ }{{6.1}^{2}}}=6.1+x \\ & \sqrt{\frac{A}{\pi }\text{+ }{{6.1}^{2}}}-6.1=x \\ & {{A}^{-1}}(x)=\sqrt{\frac{A}{\pi }\text{+ }{{6.1}^{2}}}-6.1\text{ or }\sqrt{\frac{A}{\pi }\text{+ 37}\text{.21}}-6.1 \\ \end{align}\)

c) [2 marks]

(i) Domain \({{A}^{-1}}(x)\)= range of A(x). 

The domain of x was specified in the question as \(0.696\text{ }<\text{ }x\text{ }<\text{ }0.730~\)which gives the range of ‘A’ as:

\(\pi \left( {{\left( 6.1+0.696 \right)}^{2}}\text{ }-\text{ }{{6.1}^{2}} \right)\text{ }<\text{ A }<\text{ }\pi \left( {{\left( 6.1+0.730 \right)}^{2}}\text{ }-\text{ }{{6.1}^{2}} \right)\text{ } \)

Answer:   28.20 < A < 29.65  (where A is measured in mm²)

(ii) \({{A}^{-1}}(x)\) represents the thickness of the artery wall / the difference in radius between the inner and outer artery wall.

(d) [1 mark]

The maximum area of a human artery wall, found in part (b) was 29.65mm². This measurement of 38.72mm² is therefore outside of the upper and lower bound of a human artery wall.

For Casio and Ti84 videos on 'how to' carry out a \(\chi {}^\text{2}\) test for independence, go to the studyib.net free pages and scroll down to video 10.Chi squared independence tests

Casio: https://studyib.net/mathstudies/page/559/casio-calculators
Ti84: https://studyib.net/mathstudies/page/560/ti-84
TiNspire videos to follow . . .  

(a) [2 marks]

TiNspire

(b) [2 marks]

See GDC screenshots above where p-value = 3.3858E-17 = 3.3858 x 10¹⁷ 

p-value = 0.00 000 000 000 000 003 385 8 < 0.05 significance level      A1

Therefore there is evidence at the 5% significance level to reject H₀        A1

The average age of a nation's first time mothers and the country's income level are not independent.

(c)  [2 marks]

Method1 - TiNspire: use the option stat.Expmatrix to view the expected frequencies for each category.

The 2nd row, 1st column of the expected frequency matrix (see screenshots above in part (a)) represents low income nations with an average age of first time mothers between 25.5 and 30.5. The expected number(frequency) of low income countries in this category,
from a sample of 115 countries, is therefore 19.6696 \(\approx \) 20 (to the nearest integer).

Method 2 - Mulitply the probabilities times (independent events) by total number of countries:

If the “average age of first time mothers” and “average income per person” are independent events then:

p(age of first time mother between 25.5 and 30.5) x p(low income family) x total number of countries(115) surveyed = 1 (the actual number observed in the survey data).

\(\frac{58}{115}\times \frac{39}{115}\times 115\approx 19.669....\)

(M1) for showing the correct probabilities multiplied by the correct total number of countries (115)

A1 if average age of first time mothers and income level are independent of each other we would expect 20 low income countries to have an average age for first time mothers between 25.5 and 30.5 years old. [do not award this mark if answer not rounded to nearest integer].

(a) at lower incomes, a larger change in income is required to effect a similar change in demand for private education compared to at higher incomes (or equivalent statement).

(b) M1 correct attempt to sub in three x and y-coordinates

M1 mention and attempt to “solve simultaneously"

A1A1 all coefficients correct given coordiante pairs used by the candidate.

[A1 if only two of three coefficients/parameters are correct]

(c) A1 correct substitution of x=97 into their quadratic model leading to a correct answer (given their model)

A1 correct calculation of the increase: prediction from substituting x=97 into their model minus the 10% (x=90) value from the table.  

(d) Exponential

      (a)  f(x) = 3sin2x

A2 for amplitude = 3

A2 for coefficient (2x) to show they understand the link between 

Working out

Analytical method

\(\begin{align} & \text{If f(x)=sin(x) then af(x)=asin(x) i}\text{.e}\text{. the amplitude = a (vertical stretch)} \\ & \text{Amplitude}=\frac{\max -\min }{2}=\frac{3-(-3)}{2}=\frac{6}{2}=3 \\ \end{align} \)

\(\begin{align} & \text{Period = domain(difference in x-values) of the graph from one maximum to the next maximum } \\ & \text{OR } \\ & \text{from one minimum to the next minimum}\text{.} \\ & \text{In general, for the function sin(bx)} \\ & \text{b = number of complete waves in 360 }{{}^{{}^\circ }}\text{ therefore,} \\ & \text{b = }\frac{{{360}^{{}^\circ }}}{period}\text{ = }\frac{{{360}^{{}^\circ }}}{{{180}^{{}^\circ }}}=2 \\ \end{align} \)

GDC Dynamic Graphs method

The coefficients/model can also be worked out through experimentation using the 'dynamic' graph functions on a GDC - which allows students to quickly vary each coefficient and observe the effects. 

(b) M1 working showing that the candidate has understood the period is now 90°

A1 g(x) = 3sin4x