SL External Assessment - Paper 2
SL Exam Paper 2 - Practice
Below is a full, example Paper 2: extended-response (long) questions (10marks and above each).
A 
Printable version (PDF), with space for student answers, is also available here
A graphics display calculator (GDC) is required for this paper.
All answers should be given exactly e.g. \(\pi,\quad \sqrt { 2 } \), or to three significant figures, unless stated otherwise in the question.
An unannotated copy of the IB Formula Booklete for application and interpretations is required.
This paper contains a total of 80 marks to be completed in 1 hour 30 minutes. On top of this, you will be given an additional 5 minutes reading time. You are not allowed to write anything during this time. We would recommend determining which questions you're not sure you can do and leave them to the end.
Make sure you get the most marks possible by focusing on questions you can do first. Show the examiners what you can do!
[A possible markscheme is presented for some (but not all) questions: 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].
Q1 [Maximum mark: 26]
Two entrepreneurs launched a new drink: “FreshenUp”. They started making and bottling it in their garage. The following function defines the total profits, P(x), resulting from the sale of ‘x’ bottles of the drink. Both P(x), and x, are measured in thousands.
\(Profit\quad function\quad 1:\quad { P }_{ 1 }(x)=0.04{ x }^{ 3 }+0.143x²+0.6x,\quad 0\le x\le 10\)
(a) Sketch the graph on the grid below, showing clearly the coordinates for x=0, 3, 6 and 9
[3]
b. What does the gradient of this graph represent in the context of this question?
[2]
Using only their own time and the space of their garage, they could produce a maximum of 10 000 bottles per week. As their profits were still rising they decided to move production from their garage into a larger factory space and employ two people.
They now had a maximum capacity of 30 000 bottles per week and worked out the below Profit function, P1(x), based on their market research and previous sales:
\(Profit\quad function\quad 2:\quad { P }_{ 2 }(x)=0.04x²+0.88x,\quad 0\le x\le 30\)
Using this new profit function, \({ P }_{ 2 }(x)\)
(c)(i) how many bottles do they need to sell to achieve the same level of profit obtained from selling 10 000 bottles using the previous profit function,\({ P }_{ 1 }(x)\)?
[3]
(ii) Write down the derivative of the function, \({ P }_{ 2 }(x)\)
[2]
(iii) At what quantity of bottles is the profit per bottle highest?
[2]
The entrepreneurs’ aim is to make 40 000€ profit.
(d) Calculate how many bottles of “FreshenUp” they need to sell to achieve this aim?
[2]
They unexpectedly have to replace a broken machine at a cost of 5 000€.
(e) Write down a new function for P1(x) that includes this one off, fixed cost of 5 000€.
[1]
The entrepreneurs’ are now thinking of moving to an even bigger factory space as sales hit 30 000 bottles a week
(f) Give a reason why it would not be appropriate to use their previous profit function, \({ P }_{ 2 }(x)\), when simulating future profits for the new factory.
[1]
One of their competitors sends the below data to the government’s weights and measures authority on the volume of liquid contained in 100 FreshenUp bottles. They claim their data shows that FreshenUp is not meeting the legal requirement that each bottle is normally distributed with a mean of 250ml per bottle and standard deviation of 1.8ml.
(g) Show that, using a normal distribution with mean 250ml and standard deviation 1.8ml, the probability that any given bottle contains between 247ml and 249ml is \(\frac { 6 }{ 25 }\)
[2]
(h) Complete the table below, using your answer from part (g).
[1]
The government inspectors decide to run a \(\chi {}^\text{2}\) goodness of fit test on the above data using a 5% significance level.
(k) Use your graphic display calculator to carry out a \(\chi {}^\text{2}\) goodness of fit test and determine the p-value for this data.
[2]
(l) State the conclusion from this \(\chi {}^\text{2}\) test and interpret it in the context of the question.
[2]
.png)
-(d).png)
Q2 [Maximum mark: 24 marks]
- Use the year and temperature (in °C) coordinate pairs: (1708, 8.73) and (1958, 9.45) from the table below to work out the equation of a line, in gradient-intercept form, passing through both points
[3]
(b) Use your model to predict the Earth’s average temperature in 2018 and 2019.
[2]
(c) Calculate:
(i) the Pearson’s product-moment correlation coefficient, r, for this data
[2]
(ii) the equation of the regression line y on x
(d) Use the equation of the regression line y on x to estimate the Earth’s average temperature in 2018.
[2]
(e) Compare your prediction in (d) with your prediction in (b). Give one reason which you think is likely to give the more accurate prediction.
[1]
(i) Plot this data on the graph below
[2]
(f) Give two reasons why Pearson’s correlation coefficient is not an appropriate technique to use for this data.
[2]
(g) Find the interquartile range for this data and use it to decide whether or not the 1708, 50 year average temperature of 8.73°C is an outlier.
[2]
It is found that the records from which the 50 year average for 1708 was calculated, are not reliable. Consequently, a new data table is produced, not including 1708, and replacing the actual year with a 50 year scale: 1 unit = 50 years after 1758, 2 units = 100 years after 1758 etc.
(h) Using the table above write down the constant term, c, of the quadratic: y = ax² + bx + c that models this data
[1]
(i) Using the coordinates: (0,9.22), (2, 9.12) and (5,9.76) set up three equations to find a quadratic model for this data.
[2]
(i2) Solve your simultaneous equations to find a quadratic model for this data:
[2]
(j) Use this quadratic model to predict the Earth’s average temperature in 2018.
[1]
The Earth’s average temperature in 2018 was 10.1°C.
(k) Work out the percentage different between:
(i) the actual 2018 temperature and the linear model’s prediction [from part c(ii)]
(ii) the actual 2018 temperature and the quadratic model’s prediction
[2]
Q2 [Maximum mark: 24]
- Use the year and temperature (in °C) coordinate pairs: (1708, 8.73) and (1958, 9.45) from the table below to work out the equation of a line, in gradient-intercept form, passing through both points.
Method1: slope-point form [3]
\(\begin{align} & \frac{9.45-8.73}{1958-1708}=0.00288 \\ & \frac{y-9.45}{x-1958}=0.00288\text{ }\to \text{ }y-9.45=0.00288(x-1958) \\ & y=0.00288x+0.00288(-1958)+9.45 \\ & y=0.00288x+3.81096 \\ & \text{Answer: }y=0.00288x+3.81\text{ (to 3 s}\text{.f}\text{.)} \\ \end{align} \)
Method2: slop-intercept form [3]
\(\begin{align} & 9.45=0.00288(1958)+b \\ & 9.45-0.00288(1958)=b\approx \text{ 3}\text{.81096 }\approx 3.81\text{ (to 3 s}\text{.f}\text{.)} \\ & \text{Answer: }y=0.00288x+3.81\text{ (to 3 s}\text{.f}\text{.)} \\ \end{align} \)
Using the point (1708, 8.73) gives, of course (as both points were used to derive the gradient), the same answer.
Year (x) | 1708 | 1758 | 1808 | 1858 | 1908 | 1958 | 2008 |
Temp °C (y) | 8.73 | 9.22 | 9.10 | 9.12 | 9.13 | 9.45 | 9.76 |
(b) Use your model to predict the Earth’s average temperature in 2018 and 2019.
\(\begin{align} & y=0.00288(2018)+3.81\text{ }\approx \text{ 9}\text{.62184} \\ & \text{Answer: Temperature in 2018 }\approx \text{ 9}\text{.62 (to 3 s}\text{.f}\text{.)} \\ & \\ & y=0.00288(2019)+3.81\text{ }\approx \text{ 9}\text{.62472} \\ & \text{Answer: Temperature in 2019 }\approx \text{ 9}\text{.62 (to 3 s}\text{.f}\text{.)} \\ \end{align} \)
- Calculate:
(i) the Pearson’s product-moment correlation coefficient, r, for this data.png)
(ii) the equation of the regression line y on x
y = 0.00256x + 4.46 (to 3 s.f.)
(d) Use the equation of the regression line y on x to estimate the Earth’s average temperature in 2018.
\(\begin{align} & y=0.00256(2018)+4.46\text{ }\approx \text{ 9}\text{.62608} \\ & \text{Answer: Temperature in 2018 }\approx \text{ 9}\text{.63 (to 3 s}\text{.f}\text{.)} \\ \end{align} \)
(e) Compare your prediction in (d) with your prediction in (b). Give one reason which you think is likely to give the more accurate prediction.
There is little difference between the two, but the least square’s regression calculator value is more likely to be accurate as it uses all the data to calculate the regression equation whereas the calculation in (a) used only two coordinate pairs.
(i) Plot this data on the graph below
.png)
(f) Give two reasons why Pearson’s correlation coefficient is not an appropriate technique to use for this data.
- Pearson’s correlation coefficient calculation is distorted by more extreme data points, such as 8.73, in this data set. If the ‘r’ value is worked out without it: 0.729 with it: 0.861 (see GDC calculations below).
- From the scatter graph, the relationship between years and temperature is non-linear
(g) Find the interquartile range for this data and use it to decide whether or not the 1708, 50 year average temperature of 8.73°C is an outlier.
In GDC IQR = Q3 – Q1 = 9.45 – 9.1 = 0.35
Outlier is defined, in syllabus (item SL4.1) as “1.5 × interquartile range (IQR) from the nearest quartile”.
1.5 x IQR = 1.5 x 0.35 = 0.525
Q1 – 0.525 = 9.1 – 0.525 = 8.575. 8.73 >8.575 and therefore, by this definition, is not an outlier.
[2]
It is found that the records from which the 50 year average for 1708 was calculated, are not reliable. Consequently, a new data table is produced, not including 1708, and replacing the actual year with a 50 year scale: 1 unit = 50 years after 1758, 2 units = 100 years after 1758 etc.
Year (x) | 0 | 1 | 2 | 3 | 4 | 5 |
Temp °C (y) | 9.22 | 9.10 | 9.12 | 9.13 | 9.45 | 9.76 |
(h) Using the table above write down the constant term, c, of the quadratic: y = ax² + bx + c that models this data
When x=1708 = 0 (zero lots of 50yrs after 1708), y = 9.22 in the table.
Substituting x=0 into the general quadratic gives: a(0)² + b(0) + 9.22 = y
answer: c = 9.22
i.e. when x=0, the y-value = y-intercept = ‘c’, the constant term i.e. doesn’t vary as ‘x’ varies, it is unchanging = “constant”.
[1]
(i1) & (i2) Using the coordinates: (0,9.22), (2, 9.12) and (5,9.76) set up three equations to find a quadratic model for this data. [2]
Solve your simultaneous equations to find a quadratic model for this data:
[2]
Therefore a ≈ 0.0527 (3 sf), b ≈ -0.155 c = 9.22
y = 0.0527x² - 0.155x + 9.22
(j) Use this quadratic model to predict the Earth’s average temperature in 2018.
5 lots of 50 years “after 1758” = 1758+250 = 2008. 2008 to 2018 is another 10years
= 10yrs/50yrs = 0.2 lots of 50years more.
Therefore 2018 would scale to 5.2 lots of 50years after 2018 (1758 + 5.2(50) = 2018)
Substitute x=5.2 into the quadratic model:
2018 Temperature = 0.0527(5.2)² - 0.155(5.2) + 9.22 ≈ 9.839008 ≈ 9.84
[1]
The Earth’s average temperature in 2018 was 10.1°C.
(k) Work out the percentage difference between the actual 2018 temperature and:
(i) the linear regression model’s prediction for 2018
Linear model, answer from c(ii): \(y=0.00256(2018)+4.46\text{ }\approx \text{ 9}\text{.62608}\approx \text{ 9}\text{.63 (to 3 s}\text{.f}\text{.)}\)
\(\frac{9.63-10.1}{101.}\times 100=-4.65%\)
(ii) the quadratic model’s prediction for 2018
(ii) Quadratic model: use the answer from part (j) above= 0.0527(5.2²) - 0.155(5.2) + 9.22 ≈ 9.839008 ≈ 9.84
\(\frac{9.84-10.1}{101.}\times 100=-2.57%\)
Q3) [Maximum mark:17]
A couple are expecting a baby. They initially think to invest
[2]
(a)
(b) The couple take the time to estimate that a four year college education, using an average yearly course fee (plus expenses) estimate of
[2]
Accept also an answer of
(c) They plan to put 'x' dollars into a saving account each year, at a 3% per year interest rate, so as to have
[4]
(d) After college, their daughter decides to buy a flat. She cannot find anything that suits her needs for less than
Option A: 15 year loan, compounded semi-annually, at 4% per year.
Option B: 25 year loan, compounded monthly, at 3% per year.
(i) For which option will she pay the least interest?
[7]
(ii) How much will she save by choosing the cheaper option?
[1]
(iii) For what reason might someone choose to take the option with the higher total interest?
[1]
(i) Answer: Option A
(ii) Answer:
(iii) The payment per year is lower for option B than Option A, so they may not be able to afford to make the repayments on option A.
Q4) [Maximum mark: 13]
A new medicine has been developed that is thought to enhance people's memory.
One test group (A) were not given the drug whereas a second group (B) were.
A memory test: the Rey Auditory Verbal Learning test, of 15 random words were read at one second intervals. At the end of the reading, participants had to say out loud as many of the 15 words as they could remember.
The results are shown below. There were 10 participants in each group.
GroupA | 7 | 10 | 10 | 7 | 6 | 6 | 10 | 10 | 11 | 6 |
GroupB | 9 | 8 | 7 | 10 | 12 | 11 | 9 | 12 | 11 | 11 |
(a)(i) State null and alternative hypothesis
[2]
(ii) Calculate the p-value for this test.
[3]
(iii) State, giving a reason, whether or not, at the 5% sig level, this data/experiment supports the null hypothesis.
[2]
(i) \({ H }_{ 0 }\quad :\quad { \mu }_{ A }-{ \mu }_{ B }=0\\ { H }_{ 1 }\quad :\quad { \mu }_{ A }-{ \mu }_{ B }<0\)
(iii)
.png)
(iii)
0.0295 < 0.05, there is evidence, at the 5% significance level, to reject the null hypothesis and accept the alternative hypothesis: that the mean number of words retained by group B, having taken the memory enhancing medicine, is higher than that of group A .
A second test was carried out to check if there was a 'placebo' effect in the first trial. This time one group, (C), were each given a sweet, but told that it was a memory enhancing medicine, the other group, (D), was given the same, real memory enhancing drug as used in part (a) above.
(b)(i) State the null and alternative hypothesis.
[2]
The test data gave the following ‘p’ value: 0.091 (to 3 s.f.).
(ii) At the 5% significance level, we would say that (circle the most appropriate response from the four options below):[2]
> There is no evidence, at the 5% significance level, to reject the null hypothesis
> The medicine has no significant effect on memory
> The alternative hypothesis is true
> At the 5% significance level, there is evidence to suggest that the medicine significantly improves memory
[2]
b(i)
\({ H }_{ 0 }\quad :\quad { \mu }_{ C }-{ \mu }_{ D }=0\\ { H }_{ 1 }\quad :\quad { \mu }_{ C }-{ \mu }_{ D }<0\)
b(ii)
(M1) for choosing either “no significant effect on memory” or “no evidence, at the 5% significance level, to reject the null hypothesis” -> as the candidate has correctly deduced that 0.091 > 0.05 [one-tailed test]
(A1) for choosing: ”no evidence, at the 5% significance level, to reject the null hypothesis”. Significance tests do not provide evidence that the null hypothesis is true, only that we don’t have sufficient evidence to reject it. As a helpful analogy: evidence that someone is not guilty does not mean they are innocent. Juries and courts cannot say, for definite, if a person is guilty or innocent. They decide, given the evidence presented to the court/in front of the jury, whether or not there is sufficient evidence to reject the assumption of “not guilty”. This is an important point, particulary in light of serious discussions within the scientific community around “significance levels” (Amrhein, Valentin, et al. “Scientists Rise up against Statistical Significance” Nature, vol. 567, no. 7748, 2019, pp. 305–307., doi:10.1038/d41586-019-00857-9).
(c) The Intelligence Quotient (IQ) test scales people’s results to a mean of 100 and standard deviation of 10. One of the above two groups all had IQs above 120. The other group's IQs were normally distributed with mean 100 and a standard deviation of 10. State two problems this might cause for the ‘pooled’ t-test used above?
[2]
(1) the underlying distribution of the variables must be normal for the t-test to be valid. If one of the groups had a minimum IQ of 120, the distribution is likely to be skewed/non-normal.
(2) a ‘pooled’ t-test requires the variance of both groups to be equal, this is unlikely/not necessarily the case.