Find the remainder when $p\left(x\right)$ is divided by $\left(x-2\right)$.

Find the remainder when $p\left(x\right)$ is divided by $\left(x-3\right)$.

Prove that $p\left(x\right)$ has only one real zero.

Write down the transformation that will transform the graph of $y=p\left(x\right)$ onto the graph of $y=8x^3-12x^2+16x-24$.

The random variable $X$ follows a Poisson distribution with a mean of $\lambda$ and $6\text{P}\left(X=3\right)=3\text{P}\left(X=2\right)-2\text{P}\left(X=1\right)+3\text{P}\left(X=0\right)$.

Find the value of $\lambda$.

$p\left(2\right)=8-12+16-24$       (M1)

Note: Award M1 for a valid attempt at remainder theorem or polynomial division.

= −12     A1

remainder = −12

[2 marks]

$p\left(3\right)=27-27+24-24$ = 0      A1 

remainder = 0

[1 mark]

$x=3$ (is a zero)     A1

Note: Can be seen anywhere.

EITHER

factorise to get $\left(x-3\right)\left(x^2+8\right)$      (M1)A1

$x^2+8\ne 0$ (for $x\in \mathbb{R}$) (or equivalent statement)      R1

Note: Award R1 if correct two complex roots are given.

OR

$p^{\prime }\left(x\right)=3x^2-6x+8$    A1

attempting to show $p^{\prime }\left(x\right)\ne 0$       M1

eg discriminant = 36 – 96 < 0, completing the square

no turning points       R1

THEN

only one real zero (as the curve is continuous)      AG

[4 marks]

new graph is $y=p\left(2x\right)$     (M1)

stretch parallel to the $x$-axis (with $x=0$ invariant), scale factor 0.5    A1

Note: Accept “horizontal” instead of “parallel to the $x$-axis”.

[2 marks]

$\frac{6{\lambda }^3{e}^{-\lambda }}{6}=\frac{3{\lambda }^2{e}^{-\lambda }}{2}-2\lambda e^{-\lambda }+3e^{-\lambda }$     M1A1

Note: Allow factorials in the denominator for A1.

$2{\lambda }^3-3{\lambda }^2+4\lambda -6=0$    A1

Note: Accept any correct cubic equation without factorials and $e^{-\lambda }$.

EITHER

$4\left(2{\lambda }^3-3{\lambda }^2+4\lambda -6\right)=8{\lambda }^3-12{\lambda }^2+16\lambda -24=0$       (M1)

$2\lambda =3$      (A1)

OR

$\left(2\lambda -3\right)\left({\lambda }^2+2\right)=0$       (M1)(A1)

THEN

$\lambda$ = 1.5    A1

[6 marks]

Use the data in the second table to find the value of $m$ and the value of $b$ for the regression line, $\ln x=m(\ln d)+b$.

Assuming that the model found in part (a) remains valid, estimate the percentage of trees in stock when $d=25$.

$m=-0.695 \left(-0.695383\dots \right); b=4.63 \left(4.62974\dots \right)$                  A1A1

[2 marks]

$\ln x=-0.695\left(\ln 25\right)+4.63$                  M1

$\ln x=2.39288\dots$                  (A1)

$x=10.9\%$                  A1

[3 marks]

Those candidates who did this question were often successful. There were a number, however, who found an equation of a line through two of the points instead of using their technology to find the equation of the regression line. A common problem was to introduce rounding errors at various stages throughout the problem. Some candidates failed to find the value of $x$ from that of $\ln x$.

Those candidates who did this question were often successful. There were a number, however, who found an equation of a line through two of the points instead of using their technology to find the equation of the regression line. A common problem was to introduce rounding errors at various stages throughout the problem. Some candidates failed to find the value of $x$ from that of $\ln x$.

Find the mean weight of a bag of apples.

Find the standard deviation of the weights of these bags of apples.

Find the probability that a bag selected at random weighs more than $1 \text{kg}$.

$158\times 6=948 \left(\text{g}\right)$           (M1)A1

[2 marks]

variance $6\times {13}^2$           (M1)

$\text{SD}=31.8 \left(\text{g}\right) \left(13\sqrt{6}, 31.8433\dots \right)$           A1

[2 marks]

$X~\text{N}\left(948, 31.8433{\dots }^2\right)$

$\text{P}\left(X>1000\right)=0.0512 \left(0.0512350\dots \right)$           (M1)A1

Note: Accept $0.0510 \left(0.0510014\dots \right)$ if $3$ sf value $31.8$ is used.
Award (M1)A1FT if the answer is correct for their SD, even if no working is shown. e.g. If the SD is $78$ then accept $0.252$.

[2 marks]

Find an unbiased estimate for the mean number $\left(\mu \right)$ of chocolates per packet.

Use the formula $s_{n-1}^2=\frac{\text{Σ}{x}^2-{\displaystyle \frac{{\left(\text{Σ}x\right)}^2}{n}}}{n-1}$ to determine an unbiased estimate for the variance of the number of chocolates per packet.

Find a $95\%$ confidence interval for $\mu$. You may assume that all conditions for a confidence interval have been met.

Suggest, with justification, a valid conclusion that Talha could make.

$\overline{x}=\frac{\text{Σ}x}{n}=\frac{2506}{30}=83.5 \left(83.5333\dots \right)$        A1

[1 mark]

$\left(s_{n-1}^2=\frac{\text{Σ}{x}^2-{\displaystyle \frac{{\left(\text{Σ}x\right)}^2}{n}}}{n-1}=\right) \frac{209738-{\displaystyle \frac{{2506}^2}{30}}}{29}$           (M1)

$=13.9 \left(13.9126\dots \right)$           A1

[2 marks]

$\left(82.1, 84.9\right) \left(82.1405\dots , 84.9261\dots \right)$       A2

[2 marks]

$85$ is outside the confidence interval and therefore Talha would suggest that the manufacturer’s claim is incorrect         R1

Note: The conclusion must refer back to the original claim.

          Allow use of a two sided $t$-test giving a $p$-value rounding to $0.04<0.05$ and therefore Talha would suggest that the manufacturer’s claims in incorrect.

[1 mark]

Identify the type of sampling used by the restaurant manager.

Write down the null and alternative hypotheses.

Find the $p$-value for the test.

State the conclusion of the test. Give a reason for your answer.

Convenience                 A1 

[1 mark]

${\text{H}}_0\mathrm{:}$ $1\%$ of the toys produced are faulty               A1 

${\text{H}}_1\mathrm{:}$ More than $1\%$ are faulty               A1 

[2 marks]

$X~\text{B}\left(200, 0.01\right)$               (M1) 

$\text{P}\left(X\ge 4\right)=0.142$               A1 

Note: Any attempt using Normal approximation to find $p$-value is awarded M0A0.

[2 marks]

$14\%>10\%$               R1 

so there is insufficient evidence to reject ${\text{H}}_0$.               A1 

Note: Do not award R0A1. Accept “fail to reject ${\text{H}}_0$” or “accept ${\text{H}}_0$”.

[2 marks]

Explain why $28-T$ can be modeled by an exponential function.

Find the equation of the least squares exponential regression curve for $28-T$.

Write down the coefficient of determination, $R^2$.

Interpret what the value of $R^2$ tells you about the model.

Hence predict the temperature of the water after 3 minutes.

Rearranging the model gives $28-T=a{b}^t$       A1

So $28-T$ can be modeled by an exponential function.        AG

[2 marks]

       (A1)

$28-T=22.7{\left(0.925\right)}^t$        M1A1

[3 marks]

$R^2=0.974$         A1

[1 mark]

Since the value of $R^2$ is close to +1, the model is a good fit for the data.         R1

[1 mark]

$T=28-\left(22.69\dots \right){\left(0.9250\dots \right)}^3=10.0$ minutes         M1A1

[2 marks]

Find an estimate for how many copies the vendor expects to sell each day.

State the null and alternative hypotheses for this test.

Write down the degrees of freedom for this test.

Write down the conclusion to the test. Give a reason for your answer.

$\left(\frac{74+97+91+86+112}{5}\right)=92$             A1

[1 mark]

${\text{H}}_0\text{:}$ The data satisfies the model             A1

${\text{H}}_1\text{:}$ The data does not satisfy the model             A1

Note: Do not accept “${\text{H}}_0\text{:}$ The same number of copies will be sold each day” but accept a similar statement if the word ‘expect’ or ‘expected’ is included. Similarly for ${\text{H}}_1$.           

[2 marks]

$4$             A1

[1 mark]

${{\chi }^2}_{\mathrm{calc}}=8.54 \left(8.54347\dots \right)$  OR  $p$-value $=0.0736 (0.0735802\dots )$         A2

$8.54<9.49$  OR  $0.0736>0.05$             R1

therefore there is insufficient evidence to reject ${\text{H}}_0$        A1

(i.e. the data satisfies the model)

Note: Do not award R0A1. Accept “accept” or “do not reject” in place of “insufficient evidence to reject”.
          Award the R1 for comparing their $p$-value with $0.05$ or their ${\chi }^2$ value with $9.49$ and then FT their final conclusion.

[4 marks]

Find the value of a and the value of b.

Find the expected value of T.

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

$a=\frac{3}{16}$ and $b=\frac{5}{16}$     (M1)A1A1

[3 marks]

Note: Award M1 for consideration of the possible outcomes when rolling the two dice.

$\text{E}\left(T\right)=\frac{1+6+15+28}{16}=\frac{25}{8}\left(=3.125\right)$     (M1)A1

Note: Allow follow through from part (a) even if probabilities do not add up to 1.

[2 marks]

Find the probability that more than $7$ people arrive at the ride before Shunsuke.

Find the probability there will be space for him on the $9.01$ car.

* This sample question was produced by experienced DP mathematics senior examiners to aid teachers in preparing for external assessment in the new MAA course. There may be minor differences in formatting compared to formal exam papers.

Let $X$ be the number of people who arrive between $9.00$ am and $9.01$ am

$X~\text{Po}\left(9\right)$

$\text{P}\left(X>7\right)=\text{P}\left(X\ge 8\right)$         (M1)

$0.676 \left(0.67610\dots \right)$        A1

[2 marks]

Mean number of people arriving each $30$ seconds is $4.5$        (M1)

Let $X_1$ be the number who arrive in the first $30$ seconds and $X_2$ the number who arrive in the second $30$ seconds.

$\text{P}$(Shunsuke will be able to get on the ride)

$=\text{P}\left(X_1\le 4\right)\times \text{P}\left(X_2\le 3\right)+\text{P}\left(X_1=5\right)\times \text{P}\left(X_2\le 2\right)+\text{P}\left(X_1=6\right)\times \text{P}\left(X_2\le 1\right)+\text{P}\left(X_1=7\right)\times \text{P}\left(X_2=0\right)$        M1M1

Note: M1 for first term, M1 for any of the other terms.

null        (A1)(A1)

Note: (A1) for one correct value, (A1)(A1) for four correct values.

$=0.221 \left(0.220531\dots \right)$        A1

[6 marks]

Show that the probability that Chloe wins the game is $\frac{3}{8}$.

Determine the mean of X.

Determine the variance of X.

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

METHOD 1

number of possible “deals” $=4!=24$     A1

consider ways of achieving “no matches” (Chloe winning):

Selena could deal B, C, D (ie, 3 possibilities)

as her first card     R1

for each of these matches, there are only 3 possible combinations for the remaining 3 cards     R1

so no. ways achieving no matches $=3\times 3=9$     M1A1

so probability Chloe wins $=\frac{9}{23}=\frac{3}{8}$     A1AG

METHOD 2

number of possible “deals” $=4!=24$     A1

consider ways of achieving a match (Selena winning)

Selena card A can match with Chloe card A, giving 6 possibilities for this happening     R1

if Selena deals B as her first card, there are only 3 possible combinations for the remaining 3 cards. Similarly for dealing C and dealing D     R1

so no. ways achieving one match is $=6+3+3+3=15$     M1A1

so probability Chloe wins $=1-\frac{15}{24}=\frac{3}{8}$     A1AG

METHOD 3

systematic attempt to find number of outcomes where Chloe wins (no matches)

(using tree diag. or otherwise)     M1

9 found     A1

each has probability $\frac{1}{4}\times \frac{1}{3}\times \frac{1}{2}\times 1$     M1

$=\frac{1}{24}$     A1

their 9 multiplied by their $\frac{1}{24}$     M1A1

$=\frac{3}{8}$     AG

[6 marks]

$X\sim \text{B}\left(50,\frac{3}{8}\right)$     (M1)

$\mu =np=50\times \frac{3}{8}=\frac{150}{8}\left(=\frac{75}{4}\right)(=18.75)$     (M1)A1

[3 marks]

${\sigma }^2=np(1-p)=50\times \frac{3}{8}\times \frac{5}{8}=\frac{750}{64}\left(=\frac{375}{32}\right)(=11.7)$     (M1)A1

[2 marks]

Find the probability that the company fails the inspection.

A statistician in the company suggests it would be fairer if the company passes the inspection when the mean weight of five randomly chosen bags is greater than $950 \text{g}$.

Find the probability of passing the inspection if the statistician’s suggestion is followed.

let $X$ be the weight of sugar in the bag

$\text{P}\left(X<950\right)=0.308537\dots \approx 0.309$         (M1)A1

[2 marks]

METHOD 1

let $\overline{X}$ be the mean weight of $5$ bags of sugar

$\text{E}\left(\overline{X}\right)=1000$         (A1)

use of $\text{Var}\left(\overline{X}\right)=\frac{{\sigma }^2}{n}$         (M1)

$\text{Var}\left(\overline{X}\right)=\frac{{100}^2}{5} \left(=2000\right)$         (A1)

$\overline{X}~\text{N}\left(1000, 2000\right)$

$\text{P}\left(\overline{X}>950\right)=0.868223\dots \approx 0.868 \left(86.8\%\right)$        A1

METHOD 2

let $T$ be the total weight of $5$ bags of sugar

$\text{E}\left(T\right)=5000$         (A1)

use of $\text{Var}\left(X_1+X_2\right)=\text{Var}\left(X_1\right)+\text{Var}\left(X_2\right)$ for independent random variables         (M1)

$\text{Var}\left(T\right)=5\times {100}^2 \left(=50 000\right)$         (A1)

$T~\text{N}\left(5000, 50 000\right)$

$\text{P}\left(T>4750\right)=0.868223\dots \approx 0.868 \left(86.8\%\right)$         A1

[4 marks]

Part (a) was straightforward, and a good number of candidates showed their knowledge in achieving a correct answer. Candidates are advised to not use calculator notation, as examiners cannot be familiar with all variations of GDC syntax; instead, correct mathematical notation and/or a written commentary will ensure the method is communicated to the examiner. Rounding errors once again caused problems for some. Good answers to part (b) were much less common and this was a challenging question for many. A few understood how to use the central limit theorem to find the sampling distribution of the sample mean and a few used the mean and variance of the sum of independent random variables.

Find the value of p.

Find μ, the expected value of X.

Find P(X > μ).

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

equating sum of probabilities to 1 (p + 0.5 − p + 0.25 + 0.125 + p³ = 1)       M1

p³ = 0.125 = $\frac{1}{8}$

p= 0.5      A1

[2 marks]

μ = 0 × 0.5 + 1 × 0 + 2 × 0.25 + 3 × 0.125 + 4 × 0.125       M1

= 1.375 $\left(=\frac{11}{8}\right)$     A1

[2 marks]

P(X > μ) = P(X = 2) + P(X = 3) + P(X = 4)      (M1)

= 0.5       A1

Note: Do not award follow through A marks in (b)(i) from an incorrect value of p.

Note: Award M marks in both (b)(i) and (b)(ii) provided no negative probabilities, and provided a numerical value for μ has been found.

[2 marks]

Find the probability of making a Type I error when weighing a male cuttlefish.

Find the probability of making a Type II error when weighing a female cuttlefish.

Find the probability of making an error using the zoologist’s method.

$\text{P}$(Type I error) $=\text{P}$(stating female when male) 

$=\text{P}\left(W_{Male}>11.5\right)$         (M1)

$=0.00135 \left(0.00134996\dots \right)$          A1

[2 marks]

$\text{P}$(Type II error) $=\text{P}$(stating male when female) 

$=\text{P}\left(W_{Female}<11.5\right)$         (M1)

$=0.309 \left(0.308537\dots \right)$          A1

[2 marks]

attempt to use the total probability           (M1)

$\text{P}$(error) $=0.9\times 0.00134996\dots +0.1\times 0.308537\dots$

$=0.0321 \left(0.0320687\dots \right)$          A1

[2 marks]

This was a straightforward problem on Type I and Type II errors which some candidates answered successfully in a couple of lines but many candidates were unable to do the correct calculations.

By drawing a Venn diagram, or otherwise, find $\text{P}\left(A\cup B\right)$.

Show that the events $A$ and $B$ are not independent.

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

        (M1)

Note: Award M1 for a Venn diagram with at least one probability in the correct region.

EITHER

$\text{P}\left(A\cap B^{\prime }\right)=0.3$     (A1)

$\text{P}\left(A\cup B\right)=0.3+0.4+0.1=0.8$     A1

OR

$\text{P}\left(B\right)=0.5$     (A1)

$\text{P}\left(A\cup B\right)=0.5+0.4-0.1=0.8$     A1

[3 marks]

METHOD 1

$\text{P}\left(A\right)\text{P}\left(B\right)=0.4\times 0.5$        (M1)

= 0.2      A1

statement that their $\text{P}\left(A\right)\text{P}\left(B\right)\ne \text{P}\left(A\cap B\right)$      R1

Note: Award R1 for correct reasoning from their value.

⇒ $A$, $B$ not independent     AG

METHOD 2

$\text{P}\left(A|B\right)=\frac{\text{P}\left(A\cap B\right)}{\text{P}\left(B\right)}=\frac{0.1}{0.5}$        (M1)

= 0.2      A1

statement that their $\text{P}\left(A|B\right)\ne \text{P}\left(A\right)$      R1

Note: Award R1 for correct reasoning from their value.

⇒ $A$, $B$ not independent     AG

Note: Accept equivalent argument using $\text{P}\left(B|A\right)=0.25$.

[3 marks]

Explain why this graph indicates that $P$ is inversely proportional to $d^n$.

Find the equation of the least squares regression line of ${\log }_{10} P$ against ${\log }_{10} d$.

Use your answer to part (b) to write down the value of $n$ to the nearest integer.

Find an expression for $P$ in terms of $d$.

* This sample question was produced by experienced DP mathematics senior examiners to aid teachers in preparing for external assessment in the new MAA course. There may be minor differences in formatting compared to formal exam papers.

a straight line with a negative gradient      A1A1

[2 marks]

$\log P=-2.040\dots \log d-0.12632\dots \approx -2.04 \log d-0.126$     A1A1

Note: A1 for each correct term.

[2 marks]

$n=2$     A1

[1 mark]

$P={10}^{-0.126\dots }{d}^{-2}$          (M1)

$\approx 0.748d^{-2}$          A1

[2 marks]

State a suitable null and alternative hypotheses for Tom’s test.

Find the probability of a Type I error.

The average number of fish caught in an hour is actually 2.5.

Find the probability of a Type II error.

${\text{H}}_0\text{:}m=3$,         A1

Note: Accept equivalent statements in words.

[1 mark]

(let $X$ be the number of fish caught)

$\text{P}\left(X⩽1|m=3\right)=0.199$      M1A1

[2 marks]

$\text{P}\left(X⩾2|m=2.5\right)\left(=1-\text{P}\left(X⩽1|m=2.5\right)\right)$     M1A1

Note: Award M1 for using $m$ = 2.5 to evaluate a probability, award A1 for also having $X$ ≥ 2 .

= 0.713       A1

[3 marks]

The number of cars passing a certain point in a road was recorded during 80 equal time intervals and summarized in the table below.

Carry out a ${\chi }^2$ goodness of fit test at the 5% significance level to decide if the above data can be modelled by a Poisson distribution.

H₀ : The data can be modeled by a Poisson distribution.

H₁ : The data cannot be modeled by a Poisson distribution.

$\sum f=80\text{,}\frac{\sum fx}{\sum f}=\frac{0\times 4+1\times 18+2\times 19+\dots +5\times 8}{80}=\frac{200}{80}=2.5$        A1

Theoretical frequencies are

$f\left(0\right)=8.0{\text{e}}^{-2.5}=6.5668$        (M1)(A1)

$f\left(1\right)=\frac{2.5}{1}\times 6.5668=16.4170$        A1

$f\left(2\right)=\frac{2.5}{2}\times 16.4170=20.5212$

$f\left(3\right)=\frac{2.5}{3}\times 20.5212=17.1010$

$f\left(4\right)=\frac{2.5}{4}\times 17.1010=10.6882$        A1

Note:    Award A1 for $f\left(2\right)$, $f\left(4\right)$, $f\left(4\right)$.

$f$(5 or more) $=80-\left(6.5668+16.4170+20.5212+17.1010+10.6882\right)$        A1

          $=8.7058$

${\chi }^2=\frac{{\left(4-6.5668\right)}^2}{6.5668}+\frac{{\left(18-16.4170\right)}^2}{16.4170}+\frac{{\left(19-20.5212\right)}^2}{20.5212}+\frac{{\left(20-17.1010\right)}^2}{17.1010}+\frac{{\left(11-10.6882\right)}^2}{10.6882}+\frac{{\left(8-8.7058\right)}^2}{8.7058}$

      $=1.83$  (accept 1.82)        (M1)(A1)

      $v=4$  (six frequencies and two restrictions)        (A1)

      ${\chi }^2\left(4\right)=9.488$ at the 5% level.        A1

       Since 1.83 < 9.488 we accept H₀ and conclude that the distribution can be modeled by a Poisson distribution.        R1    N0

[11 marks]

Calculate the mean number of eggs laid by these birds.

Write down appropriate hypotheses.

Carry out a test at the 1% significance level, and state your conclusion.

Mean $=\frac{1\times 19+2\times 34+\dots +5\times 4}{100}$       (M1)

$=2.16$         A1  N2

[2 marks]

H₀ : Poisson law provides a suitable model          A1

H₁ : Poisson law does not provide a suitable model          A1

[2 marks]

The expected frequencies are

        A1A1A1A1A1A1

Note: Accept expected frequencies rounded to a minimum of three significant figures.

${\chi }^2=\frac{{\left(10-11.533\right)}^2}{11.533}+\dots +\frac{{\left(4-6.824\right)}^2}{6.824}$          (M1)(A2)

$=5.35$   (accept 5.33 and 5.34)       A2

$v=4$   (6 cells − 2 restrictions)          A1

Note: If candidates have combined rows allow FT on their value of $v$.

Critical value ${\chi }^2=13.277$

Because 5.35 < 13.277, the Poisson law does provide a suitable model.          R1  N0

[14 marks]

Product research leads a company to believe that the revenue ($R$) made by selling its goods at a price ($p$) can be modelled by the equation.

$R\left(p\right)=cp{\text{e}}^{dp}$,  $c$, $d\in \mathbb{R}$

There are two competing models, A and B with different values for the parameters $c$ and $d$. 

Model A has $c$ = 3, $d$ = −0.5 and model B has $c$ = 2.5, $d$ = −0.6.

The company experiments by selling the goods at three different prices in three similar areas and the results are shown in the following table.

The company will choose the model with the smallest value for the sum of square residuals.

Determine which model the company chose.

(Model A)

$R=3p{\text{e}}^{-0.5p}$      M1

predicted values

   (A1)

$S{S}_{res}={\left(1.8196-1.5\right)}^2+{\left(2.2073-1.8\right)}^2+{\left(2.0082-1.5\right)}^2$       (M1)

= 0.5263…       A1

(Model B)

$R=2.5p{\text{e}}^{-0.6p}$

predicted values

      (A1)

$S{S}_{res}=$ 0.170576…       A1

chose model B       A1

Note: Method marks can be awarded if seen for either model A or model B. Award final A1 if it is a correct deduction from their calculated values for A and B.

[7 marks]

Find the probability he will choose a female student 8 times.

The Head of Year, Mrs Smith, decides to select a student at random from the year group to read the notices in assembly. There are 80 students in total in the year group. Mrs Smith calculates the probability of picking a male student 8 times in the first 20 assemblies is 0.153357 correct to 6 decimal places.

Find the number of male students in the year group.

P(X = 8)       (M1)

Note: Award (M1) for evidence of recognizing binomial probability. eg, P(X = 8), X ∼ B$\left(20,\frac{6}{15}\right)$.

= 0.180 (0.179705…)    A1

[2 marks]

let $x$ be the number of male students

recognize that probability of selecting a male is equal to $\frac{x}{80}$       (A1)

$\left(\text{set up equation}{}^{20}{\text{C}}_8{\left(\frac{x}{80}\right)}^8{\left(\frac{80-x}{80}\right)}^{12}=\right)0.153357$          (M1)

number of male students = 37         (M1)A1

Note: Award (M1)A0 for 27.

[4 marks]

State the name of this test for reliability.

State a possible disadvantage of using this test for reliability.

Calculate Pearson’s product moment correlation coefficient for this data.

Hence determine, with a reason, if the survey is reliable.

Parallel Forms         A1

[1 mark]

EITHER

The two sets of questions might not be of equal difficulty         R1

OR

It is time consuming to create two sets of questions           R1        

[1 mark]

$r=0.958$           A2

[2 marks]

Since the value of $r$ is close to +1,           R1

The survey is reliable.           A1

[2 marks]

Find the probability that a sack is under its labelled weight.

Find the lower quartile of the weights of the sacks of potatoes.

The sacks of potatoes are transported in crates. There are $10$ sacks in each crate and the weights of the sacks of potatoes are independent of each other.

Find the probability that the total weight of the sacks of potatoes in a crate exceeds $500 \text{kg}$.

let $X$ be the random variable “the weight of a sack of potatoes”

$\text{P}\left(X<50\right)$                 (M1)

$=0.588 \text{kg} \left(0.587929\dots \right)$                 A1

[2 marks]

$\text{P}\left(X<l\right)=0.25$                 (M1)

$49.2 \text{kg} \left(49.1929\dots \right)$                 A1

[2 marks]

attempt to sum $10$ independent random variables                 (M1)

$Y=\underset{i=1}{\overset{10}{\text{Σ}}}{X}_i~\text{N}\left(498, 10\times 0.9^2\right)$                 (A1)

$\text{P}\left(Y>500\right)=0.241$                 A1

[3 marks]

The first part of the question was often answered well but there were a number of candidates who interpreted finding $P\left(X<50\right)$ by finding $P\left(X<49.9\right)$ or something similar. Not all candidates, however, understood that the lower quartile is given by $P\left(X<l\right)=0.25$. Part (c) was less well understood. Attempts to sum $10$ independent random variables correctly involved multiplication of the mean by $10$ but the standard deviation and not the variance was incorrectly multiplied by $10$.

The first part of the question was often answered well but there were a number of candidates who interpreted finding $P\left(X<50\right)$ by finding $P\left(X<49.9\right)$ or something similar. Not all candidates, however, understood that the lower quartile is given by $P\left(X<l\right)=0.25$. Part (c) was less well understood. Attempts to sum $10$ independent random variables correctly involved multiplication of the mean by $10$ but the standard deviation and not the variance was incorrectly multiplied by $10$.

The first part of the question was often answered well but there were a number of candidates who interpreted finding $P\left(X<50\right)$ by finding $P\left(X<49.9\right)$ or something similar. Not all candidates, however, understood that the lower quartile is given by $P\left(X<l\right)=0.25$. Part (c) was less well understood. Attempts to sum $10$ independent random variables correctly involved multiplication of the mean by $10$ but the standard deviation and not the variance was incorrectly multiplied by $10$.

Calculate the value of $r$, the product moment correlation coefficient of the sample.

Assuming that the distribution of $X, Y$ is bivariate normal with product moment correlation coefficient $\rho$, calculate the $p$-value of your result when testing the hypotheses $H_0 : \rho =0 ; H_1 : \rho >0$.

State whether your $p$-value suggests that $X$ and $Y$ are independent.

Given a further value $x=5.2$ from the distribution of $X$, $Y$, predict the corresponding value of $y$. Give your answer to one decimal place.

use of

$r=\frac{\sum xy-n\overline{x} \overline{y}}{\sqrt{\left(\sum x^2-n{\overline{x}}^2\right)\left(\sum y^2-n{\overline{y}}^2\right)}}$        M1

$=\frac{495.4-12\times {\displaystyle \frac{76.3}{12}}\times {\displaystyle \frac{72.2}{12}}}{\sqrt{\left(563.7-{\displaystyle \frac{12\times 76.3^2}{{12}^2}}\right)\left(460.1-{\displaystyle \frac{12\times 72.2^2}{{12}^2}}\right)}}$        A1

$=0.809$        A1

Note: Accept any answer that rounds to $0.81$.

[3 marks]

$t=0.80856\dots \sqrt{\frac{10}{1-0.80856{\dots }^2}}$         (M1)

$=4.345\dots$        A1

$p$-value $=7.27\times {10}^{-4}$        A1

Note: Accept any answer that rounds to $7.2$ or $7.3\times {10}^{-4}$.

Note: Follow through their $p$-value

[3 marks]

this value indicates that $X,Y$ are not independent       A1

[1 mark]

use of

$y-\overline{y}=\frac{\sum xy-n\overline{x} \overline{y}}{\sum x^2-n{\overline{x}}^2}\left(x-\overline{x}\right)$       M1

$y-\frac{72.2}{12}=\left(\frac{495.4-12\times {\displaystyle \frac{76.3}{12}}\times {\displaystyle \frac{72.2}{12}}}{563.7-12\times {\displaystyle \frac{76.3^2}{{12}^2}}}\right)\left(x-\frac{76.3}{12}\right)$       A1

putting  $x=5.2$  gives  $y=5.5$       A1

[3 marks]

Find the value of $k$.

By considering the graph of f write down the mean of $X$;

By considering the graph of f write down the median of $X$;

By considering the graph of f write down the mode of $X$.

Show that $P(0⩽X⩽2)=\frac{1}{4}$.

Hence state the interquartile range of $X$.

Calculate $P(X⩽4|X⩾3)$.

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

attempt to equate integral to 1 (may appear later)     M1

$k\underset{0}{\overset{6}{\int }}\sin \left(\frac{\pi x}{6}\right)\text{d}x=1$

correct integral     A1

$k{\left[-\frac{6}{\pi }\cos \left(\frac{\pi x}{6}\right)\right]}_0^6=1$

substituting limits     M1

$-\frac{6}{\pi }(-1-1)=\frac{1}{k}$

$k=\frac{\pi }{12}$ A1

[4 marks]

mean $=3$     A1

Note:     Award A1A0A0 for three equal answers in $(0,6)$.

[1 mark]

median $=3$     A1

Note:     Award A1A0A0 for three equal answers in $(0,6)$.

[1 mark]

mode $=3$     A1

Note:     Award A1A0A0 for three equal answers in $(0,6)$.

[1 mark]

$\frac{\pi }{12}\underset{0}{\overset{2}{\int }}\sin \left(\frac{\pi x}{6}\right)\text{d}x$    M1

$=\frac{\pi }{12}{\left[-\frac{6}{\pi }\cos \left(\frac{\pi x}{6}\right)\right]}_0^2$     A1

Note:     Accept without the $\frac{\pi }{12}$ at this stage if it is added later.

$\frac{\pi }{12}\left[-\frac{6}{\pi }\left(\cos \frac{\pi }{3}-1\right)\right]$     M1

$=\frac{1}{4}$     AG

[4 marks]

from (c)(i) $Q_1=2$     (A1)

as the graph is symmetrical about the middle value $x=3\Rightarrow Q_3=4$     (A1)

so interquartile range is

$4-2$

$=2$     A1

[3 marks]

$P(X⩽4|X⩾3)=\frac{P(3⩽X⩽4)}{P(X⩾3)}$

$=\frac{\frac{1}{4}}{\frac{1}{2}}$     (M1)

$=\frac{1}{2}$     A1

[2 marks]

State the null and alternative hypotheses for the test.

Find the probability that Sheila will make a type I error in her test conclusion.

Sheila finds $126$ coffees were sold during the $5$-hour period.

State Sheila’s conclusion to the test. Justify your answer.

${\text{H}}_0: m=110, {\text{H}}_1: m>110$             A1

Note: Accept other appropriate variables for the mean.
         Accept $22$ in place of $110$.

[1 mark]

$\text{P}\left(X\ge 128\right)=0.05024$          (M1)(A1)

$\text{P}\left(X\ge 129\right)=0.04153$                (M1)

(probability of making a type I error is)  $0.0415$           A1

Note: If other probabilities are seen, the final A1 cannot be awarded unless $0.0415$ is clearly identified as the final answer.

[4 marks]

$X~\text{Po}\left(110\right)$

$\text{P}\left(X\ge 126\right)=0.072>0.05$  OR  recognizing $126<129$ or $\le 128$           R1

so there is insufficient evidence to reject ${\text{H}}_0$           A1

(ie there is insufficient evidence to suggest that the number of coffees being sold has increased)

Note: Accept ‘Accept ${\text{H}}_0$’.
          Do not award R0A1.

[2 marks]

Calculate unbiased estimates of the population mean and the population variance.

State suitable hypotheses.

Calculate the value of the ${\chi }^2$ statistic and state your conclusion using a 10% level of significance.

$\overline{x}=\frac{412.11}{241}=1.71$            A1

$s^2=\frac{705.5721}{240}-\frac{{412.11}^2}{240\times 241}=0.0036$        M1A1

[3 marks]

H₀: Data can be modelled by a normal distribution

H₁: Data cannot be modelled by a normal distribution           A1

[1 mark]

The expected frequencies are

       A1A1A1A1A1A1

${\chi }^2=\frac{5^2}{8.04}+\frac{{34}^2}{30.19}+\dots +\frac{{12}^2}{16.10}-241=3.30\text{/}3.29$       M1A1

Degrees of freedom = 3       A1

Critical value = 6.251 or p-value = 0.35       A1

The data can be modelled by a normal distribution.       R1

[11 marks]

Determine the transition matrix for this graph.

If the mouse was left to wander indefinitely, use your graphic display calculator to estimate the percentage of time that the mouse would spend at point $\text{F}$.

Comment on your answer to part (b), referring to at least one limitation of the model.

transition matrix is        M1A1A1

Note: Allow the transposed matrix.
          Award M1 for a $6\times 6$ matrix with all values between $0$ and $1$, and all columns (or rows if transposed) adding up to $1$, award A1 for one correct row (or column if transposed) and A1 for all rows (or columns if transposed) correct.


[3 marks]

attempting to raise the transition matrix to a large power             (M1)

steady state vector is $\left(\begin{array}{c}\left(0.157\right)\\ \left(0.0868\right)\\ \left(0.256\right)\\ \left(0.241\right)\\ \left(0.0868\right)\\ 0.173\end{array}\right)$             (A1)

so percentage of time spent at vertex $\text{F}$ is $17.3\%$                   A1

Note: Accept $17.2\%$.


[3 marks]

the model assumes instantaneous travel from junction to junction,             R1
and hence the answer obtained would be an overestimate             R1

OR

the mouse may eat the sugar over time             R1
and hence the probabilities would change             R1


Note: Accept any other sensible answer.

[3 marks]

Calculate the mean number of brown eggs in a box.

Hence estimate $p$, the probability that a randomly chosen egg is brown.

By calculating an appropriate ${\chi }^2$ statistic, test, at the 5% significance level, whether or not the binomial distribution gives a good fit to these data.

Note: Candidates may obtain slightly different numerical answers depending on the calculator and approach used. Use discretion in marking.

Mean $=\frac{1\times 29+\dots +6\times 1}{100}=1.98$       (A1)

[1 mark]

Note: Candidates may obtain slightly different numerical answers depending on the calculator and approach used. Use discretion in marking.

$\hat{p}=\frac{1.98}{6}=0.33$      (A1)

[1 mark]

Note: Candidates may obtain slightly different numerical answers depending on the calculator and approach used. Use discretion in marking.

The calculated values are

$f_0$               $f_e$        ${\left(f_0-f_e\right)}^2$
10            9.046         0.910
29          26.732          5.14       (M1)
31          32.917         3.675      (A1)
18          21.617       13.083      (A1)
12            9.688         5.345      (A1)

Note: Award (M1) for the attempt to calculate expected values, (A1) for correct expected values, (A1) for correct ${\left(f_0-f_e\right)}^2$ values, (A1) for combining cells.

${\chi }^2=\frac{0.910}{9.046}+\dots +\frac{5.345}{9.688}=1.56$      (A1)

OR

${\chi }^2=1.56$      (G5)

Degrees of freedom = 3; Critical value = 7.815

(or p-value = 0.668 (or 0.669))      (A1)(A1)

We conclude that the binomial distribution does provide a good fit.      (R1)

[8 marks]

Show that $\text{ln}\left(T-25\right)=bt+\text{ln}a$.

Find the equation of the regression line of $\text{ln}\left(T-25\right)$ on $t$.

find the value of $a$ and of $b$.

predict the temperature of the metal rod after 3 minutes.

$\text{ln}\left(T-25\right)=\text{ln}\left(a{\text{e}}^{bt}\right)$       M1

$\text{ln}\left(T-25\right)=\text{ln}a+\text{ln}\left({\text{e}}^{bt}\right)$            A1

$\text{ln}\left(T-25\right)=bt+\text{ln}a$            AG

[2 marks]

$\text{ln}\left(T-25\right)=-0.00870t+3.89$      M1A1A1

[3 marks]

$b=-0.00870$       A1

$a=e^{\mathrm{3.89...}}=49.1$     M1A1

[3 marks]

$T\left(180\right)=49.1e^{-0.00870\left(180\right)}+25=35.2$    M1A1

[2 marks]

Complete the following transition diagram to represent this information.

Katie works for $180$ days in a year.

Find the probability that Katie cycles to work on her final working day of the year.

           A1A1

[2 marks]

$\mathit{A}=\left(\begin{array}{cc}0.8& 0.6\\ 0.2& 0.4\end{array}\right)$               (A1)

${\mathit{A}}^{180}=\left(\begin{array}{cc}0.75& 0.75\\ 0.25& 0.25\end{array}\right)$               (M1)

$0.75$               A1

[3 marks]

This question was often done well. Many textbooks teach the method of multiplying the transition matrix by an initial state vector. This was often seen in candidates’ responses. eg ${\left(\begin{array}{cc}0.8& 0.6\\ 0.2& 0.4\end{array}\right)}^{180}\left(\begin{array}{c}1\\ 0\end{array}\right)=\left(\begin{array}{c}0.75\\ 0.25\end{array}\right)$. Errors were often due to the figures being incorrectly placed in the transition matrix; not just the transpose, but other combinations of the four values as well.

This question was often done well. Many textbooks teach the method of multiplying the transition matrix by an initial state vector. This was often seen in candidates’ responses. eg ${\left(\begin{array}{cc}0.8& 0.6\\ 0.2& 0.4\end{array}\right)}^{180}\left(\begin{array}{c}1\\ 0\end{array}\right)=\left(\begin{array}{c}0.75\\ 0.25\end{array}\right)$. Errors were often due to the figures being incorrectly placed in the transition matrix; not just the transpose, but other combinations of the four values as well.

Find an unbiased estimate of the population mean of $d$.

Find an unbiased estimate of the population variance of $d$.

State the alternative hypothesis.

Given that all assumptions for this test are satisfied, carry out an appropriate hypothesis test. State and justify your conclusion. Use a $5\%$ significance level.

$\overline{x}=4.63 \left(4.62686\dots \right)$           A1

[1 mark]

$s_{n-1}=1.098702$           (A1)

$s_{n-1}^2=1.21 \left(1.207146\dots \right)$          A1

Note: Award A0A0 for an answer of $1.19$ from biased estimate.

[2 marks]

$H_1 : \mu >4.4$          A1

[1 mark]

METHOD 1

using a $z$-test         (M1)

$p=0.0454992\dots$          A1

$p<0.05$          R1

reject null hypothesis          A1

(therefore there is significant evidence that the IB HL math students know more digits of $\mathrm{\pi }$ than the population in general)

Note: Do not award R0A1. Allow R1A1 for consistent conclusion following on from their $p$-value.

METHOD 2

using a $t$-test         (M1)

$p=0.0478584\dots$          A1

$p<0.05$          R1

reject null hypothesis          A1

(therefore there is significant evidence that the IB HL math students know more digits of $\mathrm{\pi }$ than the population in general)

Note: Do not award R0A1. Allow R1A1 for consistent conclusion following on from their $p$-value.

[4 marks]

In parts (a) and (b), candidates used the 1-Var Stats facility to find the estimates of mean and variance although some forgot to include the frequency list so that they just found the mean and variance of the numbers 2, 3, …6, 7. Candidates who looked ahead realized that the answers to parts (a) and (b) would be included in the output from using their test. In part (c), the question was intended to use the t-test (as the population variance was unknown), however since the population could not be assumed to be normally distributed, the Principal Examiner condoned the use of the z-test (with the estimated variance from part (b)). As both methods could only produce an approximate p-value, either method (and the associated p-value) was awarded full marks.

Find $s_{n-1}$ for this sample.

Find a 95 % confidence interval for the population mean, giving your answer to 4 significant figures.

The bags are labelled as being 1.5 kg mass. Comment on this claim with reference to your answer in part (b).

$s_{n-1}=\sqrt{\frac{10}{9}}\times 0.0196=0.02066\dots$  (M1)A1

[2 marks]

(1.463, 1.493)          (M1)A1

Note: If $s_n$ used answer is (1.464, 1.492), award M1A0.

[2 marks]

95 % of the time these results would be produced by a population with mean of less than 1.5 kg, so it is likely the mean mass is less than 1.5 kg           R1

[1 mark]

State the name of this test for validity.

Calculate Pearson’s product moment correlation coefficient for this data.

Hence determine, with a reason, if the new exam is a valid indicator of future performance.

criterion-related          A1

[1 mark]

$r=0.414$         A2

[2 marks]

Since the value of $r$ is low (closer to 0 than +1),         R1

The new exam is not a valid indicator of future performance.         A1

[2 marks]

In a coffee shop, the time it takes to serve a customer can be modelled by a normal distribution with a mean of 1.5 minutes and a standard deviation of 0.4 minutes.

Two customers enter the shop together. They are served one at a time.

Find the probability that the total time taken to serve both customers will be less than 4 minutes.

Clearly state any assumptions you have made.

let $T$ be the time to serve both customers and $T_i$ the time to serve the $i$th customer

assuming independence of $T_1$ and $T_2$      R1

$T$ is normally distributed and $T=T_1+T_2$       (M1)

$E\left(T\right)=1.5+1.5=3$     A1

$\text{Var}\left(T\right)={0.4}^2+{0.4}^2=0.32$      M1A1

       A1

[6 marks]

Show that $\lambda =1$ is always an eigenvalue for M and find the other eigenvalue in terms of $a$ and $b$.

Find the steady state probability vector for M in terms of $a$ and $b$.

        M1A1

$\Rightarrow {\lambda }^2-\left(a+b\right)\lambda +a+b-1=0\Rightarrow \left(\lambda -1\right)\left(\lambda +\left(1-a-b\right)\right)=0$         A1