Determine the value of $m$.

Given that $P\left(\left|X-m\right|\le a\right)=0.3$, determine the value of $a$.

recognises that $\underset{0}{\overset{m}{\int }}\text{arccos} x dx=0.5$                     (M1)

$m \text{arccos} m-\sqrt{1-m^2}-\left(0-\sqrt{1}\right)=0.5$

$m=0.360034\dots$

$m=0.360$                     A1

[2 marks]

METHOD 1

attempts to find at least one endpoint (limit) both in terms of $m$ (or their $m$) and $a$                     (M1)

$\text{P}\left(m-a\le X\le m+a\right)=0.3$                   

$\underset{0.360034\dots -a}{\overset{0.360034\dots +a}{\int }}\text{arccos} x dx=0.3$                     (A1)

Note: Award (A1) for $\underset{m-a}{\overset{m+a}{\int }}\text{arccos} x dx=0.3$.

${\left[x \text{arccos} x-\sqrt{1-x^2}\right]}_{0.360034\dots -a}^{0.360034\dots +a}$

attempts to solve their equation for $a$                     (M1)

Note: The above (M1) is dependent on the first (M1).

$a=0.124861\dots$

$a=0.125$                       A1

METHOD 2

$\underset{-a}{\overset{a}{\int }}\text{arccos\hspace{0.05em}}\overline{)x-0.360034\dots \overline{) dx \left(=0.3\right)}}$                     (M1)(A1)

Note: Only award (M1) if at least one limit has been translated correctly.

Note: Award (M1)(A1) for $\underset{-a}{\overset{a}{\int }}\text{arccos\hspace{0.05em}}\overline{)x-m\overline{) dx \left(=0.3\right)}}$.

attempts to solve their equation for $a$                     (M1)

$a=0.124861\dots$

$a=0.125$                       A1

METHOD 3

EITHER 

$\underset{-a}{\overset{a}{\int }}\text{arccos\hspace{0.05em}}\left(x+0.360034\dots \right) dx \left(=0.3\right)$                     (M1)(A1)

Note: Only award (M1) if at least one limit has been translated correctly.

Note: Award (M1)(A1) for $\underset{-a}{\overset{a}{\int }}\text{arccos\hspace{0.05em}}\left(x+m\right) dx \left(=0.3\right)$.

OR

$\underset{2\left(0.360034\dots \right)-a}{\overset{2\left(0.360034\dots \right)+a}{\int }}\text{arccos\hspace{0.05em}}\left(x-0.360034\dots \right) dx \left(=0.3\right)$                     (M1)(A1)

Note: Only award (M1) if at least one limit has been translated correctly.

Note: Award (M1)(A1) for $\underset{2m-a}{\overset{2m+a}{\int }}\text{arccos\hspace{0.05em}}\left(x-m\right) dx \left(=0.3\right)$.

THEN

attempts to solve their equation for $a$                     (M1)

Note: The above (M1) is dependent on the first (M1).

$a=0.124861\dots$

$a=0.125$                       A1

[4 marks]

Find, in terms of $a$, the probability that $X$ lies between 1 and 3.

Sketch the graph of $f$. State the coordinates of the end points and any local maximum or minimum points, giving your answers in terms of $a$.

$a$.

$\text{E}\left(X\right)$.

the median of $X$.

       (M1)(A1)(A1)

$=3a+\frac{11}{3}a$

$=\frac{20}{3}a\left(=6.67a\right)$        A1

[4 marks]

        A4

award A1 for (0, 3$a$), A1 for continuity at (2, 3$a$), A1 for maximum at (3, 4$a$), A1 for (5, 0)

Note: Award A3 if correct four points are not joined by a straight line and a quadratic curve.

[4 marks]

$\text{P}\left(0⩽X⩽5\right)=6a+a{\int }_2^5-x^2+6x-5\text{d}x$       (M1)

$=15a$       (A1)

$15a=1$       (M1)

$\Rightarrow a=\frac{1}{15}\left(=0.0667\right)$       A1

[4 marks]

$\text{E}\left(X\right)=\frac{1}{5}{\int }_0^{2}x\text{d}x+\frac{1}{15}{\int }_2^5-x^3+6x^2-5x\text{d}x$       (M1)(A1)

= 2.35       A1

[3 marks]

attempt to use ${\int }_0^{m}f\left(x\right)\text{d}x=0.5$       (M1)

$0.4+a{\int }_2^m-x^2+6x-5\text{d}x=0.5$       (A1)

$a{\int }_2^m-x^2+6x-5\text{d}x=0.1$

attempt to solve integral using GDC and/or analytically       (M1)

$\frac{1}{15}{\left[-\frac{1}{3}{x}^3+3x^2-5x\right]}_2^m=0.1$

$m=2.44$       A1

[4 marks]

Show that $\text{E}\left(X\right)=\frac{n+1}{n+2}$.

Show that $\text{Var}\left(X\right)=\frac{n+1}{{\left(n+2\right)}^2\left(n+3\right)}$.

$\text{E}\left(X\right)=\left(n+1\right)\underset{0}{\overset{1}{\int }}{x}^{n+1}dx$                       M1

$=\left(n+1\right){\left[\frac{x^{n+2}}{n+2}\right]}_0^1$                         A1

leading to $\text{E}\left(X\right)=\frac{n+1}{n+2}$                   AG

[2 marks]

METHOD 1

use of $\text{Var}\left(X\right)=\text{E}\left(X^2\right)-{\left[\text{E}\left(X\right)\right]}^2$                  M1

$\text{Var}\left(X\right)=\left(n+1\right)\underset{0}{\overset{1}{\int }}{x}^{n+2}dx-{\left(\frac{n+1}{n+2}\right)}^2$

$=\left(n+1\right){\left[\frac{1}{n+3}{x}^{n+3}\right]}_0^1-{\left(\frac{n+1}{n+2}\right)}^2$

$=\frac{n+1}{n+3}-{\left(\frac{n+1}{n+2}\right)}^2$                       A1

$=\frac{\left(n+1\right){\left(n+2\right)}^2-{\left(n+1\right)}^2\left(n+3\right)}{{\left(n+2\right)}^2\left(n+3\right)}$                  M1

EITHER

$=\frac{\left(n+1\right)\left(n^2+4n+4-\left(n^2+4n+3\right)\right)}{{\left(n+2\right)}^2\left(n+3\right)}$                       A1

OR

$=\frac{\left(n^3+5n^2+8n+4\right)-\left(n^3+5n^2+7n+3\right)}{{\left(n+2\right)}^2\left(n+3\right)}$                       A1

THEN

so $\text{Var}\left(X\right)=\frac{n+1}{{\left(n+2\right)}^2\left(n+3\right)}$                       AG

METHOD 2

use of $\text{Var}\left(X\right)=\text{E}{\left(X-\text{E}\left(X\right)\right)}^2$                  M1

$\text{Var}\left(X\right)=\left(n+1\right)\underset{0}{\overset{1}{\int }}{\left(x-\frac{n+1}{n+2}\right)}^2{x}^{n}dx$

$=\left(n+1\right){\left[\frac{1}{n+3}{x}^{n+3}-\frac{2\left(n+1\right)}{{\left(n+2\right)}^2}{x}^{n+2}+\frac{n+1}{{\left(n+2\right)}^2}{x}^{n+1}\right]}_0^1$

$=\frac{n+1}{n+3}-{\left(\frac{n+1}{n+2}\right)}^2$                       A1

$=\frac{\left(n+1\right)\left({\left(n+2\right)}^2-\left(n+1\right)\left(n+3\right)\right)}{{\left(n+2\right)}^2\left(n+3\right)}$                  M1

EITHER

$=\frac{\left(n+1\right)\left(n^2+4n+4-\left(n^2+4n+3\right)\right)}{{\left(n+2\right)}^2\left(n+3\right)}$                       A1

OR

$=\frac{\left(n^3+5n^2+8n+4\right)-\left(n^3+5n^2+7n+3\right)}{{\left(n+2\right)}^2\left(n+3\right)}$                       A1

THEN

so $\text{Var}\left(X\right)=\frac{n+1}{{\left(n+2\right)}^2\left(n+3\right)}$                       AG

[6 marks]

Find the probability that a runner selected at random will complete the marathon in less than 3 hours.

Calculate $T_1$.

Find the standard deviation of the times taken by female runners.

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

$T\sim N(196,{24}^2)$

     (M1)A1

[2 marks]

     (M1)

$T_1=157$     A1

[2 marks]

$F\sim N(210,{\sigma }^2)$

     (M1)

$\frac{235-210}{\sigma }=0.806421$ or equivalent     (M1)(A1)

$\sigma =31.0$     A1

[4 marks]

Show that $\text{P}(X=3)=0.001$ and $\text{P}(X=4)=0.0027$.

Find the values of the constants $a$ and $b$.

Deduce that $\frac{\text{P}(X=n)}{\text{P}(X=n-1)}=\frac{0.9(n-1)}{n-3}$ for $n>3$.

(i)     Hence show that $X$ has two modes $m_1$ and $m_2$.

(ii)     State the values of $m_1$ and $m_2$.

Determine the minimum value of $x$ such that the probability Kati receives at least one free gift is greater than 0.5.

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

$\text{P}(X=3)=(0.1{)}^3$    A1

$=0.001$    AG

$\text{P}(X=4)=\text{P}(VV\overline{V}V)+\text{P}(V\overline{V}VV)+\text{P}(\overline{V}VVV)$    (M1)

$=3\times (0.1{)}^3\times 0.9$ (or equivalent)     A1

$=0.0027$    AG

[3 marks]

METHOD 1

attempting to form equations in $a$ and $b$     M1

$\frac{9+3a+b}{2000}=\frac{1}{1000}(3a+b=-7)$    A1

$\frac{16+4a+b}{2000}\times \frac{9}{10}=\frac{27}{10000}(4a+b=-10)$    A1

attempting to solve simultaneously     (M1)

$a=-3,b=2$    A1

METHOD 2

$\text{P}(X=n)=\left(\begin{array}{c}n-1\\ 2\end{array}\right)\times {0.1}^3\times {0.9}^{n-3}$    M1

$=\frac{(n-1)(n-2)}{2000}\times {0.9}^{n-3}$    (M1)A1

$=\frac{n^2-3n+2}{2000}\times {0.9}^{n-3}$    A1

$a=-3,b=2$    A1

Note: Condone the absence of ${0.9}^{n-3}$ in the determination of the values of $a$ and $b$.

[5 marks]

METHOD 1

EITHER

$\text{P}(X=n)=\frac{n^2-3n+2}{2000}\times {0.9}^{n-3}$    (M1)

OR

$\text{P}(X=n)=\left(\begin{array}{c}n-1\\ 2\end{array}\right)\times {0.1}^3\times {0.9}^{n-3}$    (M1)

THEN

$=\frac{(n-1)(n-2)}{2000}\times {0.9}^{n-3}$    A1

$\text{P}(X=n-1)=\frac{(n-2)(n-3)}{2000}\times {0.9}^{n-4}$    A1

$\frac{\text{P}(X=n)}{\text{P}(X=n-1)}=\frac{(n-1)(n-2)}{(n-2)(n-3)}\times 0.9$    A1

$=\frac{0.9(n-1)}{n-3}$    AG

METHOD 2

$\frac{\text{P}(X=n)}{\text{P}(X=n-1)}=\frac{\frac{n^2-3n+2}{2000}\times {0.9}^{n-3}}{\frac{{(n-1)}^2-3(n-1)+2}{2000}\times {0.9}^{n-4}}$    (M1)

$=\frac{0.9(n^2-3n+2)}{(n^2-5n+6)}$    A1A1

Note: Award A1 for a correct numerator and A1 for a correct denominator.

$=\frac{0.9(n-1)(n-2)}{(n-2)(n-3)}$    A1

$=\frac{0.9(n-1)}{n-3}$    AG

[4 marks]

(i)     attempting to solve $\frac{0.9(n-1)}{n-3}=1$ for $n$     M1

$n=21$    A1

   R1

   R1

$X$ has two modes     AG

Note: Award R1R1 for a clearly labelled graphical representation of the two inequalities (using $\frac{\text{P}(X=n)}{\text{P}(X=n-1)}$).

(ii)     the modes are 20 and 21     A1

[5 marks]

METHOD 1

$Y\sim \text{B}(x,0.1)$    (A1)

attempting to solve $\text{P}(Y⩾3)>0.5$ (or equivalent eg $1-\text{P}(Y⩽2)>0.5$) for $x$     (M1)

Note: Award (M1) for attempting to solve an equality (obtaining $x=26.4$).

$x=27$    A1

METHOD 2

$\sum _{n=0}^x\text{P}(X=n)>0.5$    (A1)

attempting to solve for $x$     (M1)

$x=27$    A1

[3 marks]

Complete the given probability tree diagram for Iqbal’s three attempts, labelling each branch with the correct probability.

Calculate the probability that Iqbal passes at least two of the papers he attempts.

Find the probability that Iqbal passes his third paper, given that he passed only one previous paper.

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

     A1A1A1

Note: Award A1 for each correct column of probabilities.

[3 marks]

probability (at least twice) =

EITHER

$\left(0.6\times 0.7\times 0.8\right)+\left(0.6\times 0.7\times 0.2\right)+\left(0.6\times 0.3\times 0.6\right)+\left(0.4\times 0.6\times 0.7\right)$       (M1)

OR

$\left(0.6\times 0.7\right)+\left(0.6\times 0.3\times 0.6\right)+\left(0.4\times 0.6\times 0.7\right)$       (M1)

Note: Award M1 for summing all required probabilities.

THEN

= 0.696     A1

[2 marks]

P(passes third paper given only one paper passed before)

$=\frac{\text{P}\left(\text{passes third AND only one paper passed before}\right)}{\text{P}\left(\text{passes once in first two papers}\right)}$      (M1)

$=\frac{\left(0.6\times 0.3\times 0.6\right)+\left(0.4\times 0.6\times 0.7\right)}{\left(0.6\times 0.3\right)+\left(0.4\times 0.6\right)}$      A1

= 0.657     A1

[3 marks]

Find $\mu$ and $\sigma$.

Find .

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

   (M1)

either      (M1)

$\frac{30.31-\mu }{\sigma }=\underset{-1.1850\dots }{\underbrace{{\mathrm{\Phi }}^{-1}(0.1180)}}$    (A1)

$\frac{42.52-\mu }{\sigma }=\underset{0.5072\dots }{\underbrace{{\mathrm{\Phi }}^{-1}(0.6940)}}$    (A1)

attempting to solve simultaneously     (M1)

$\mu =38.9$ and $\sigma =7.22$     A1

[6 marks]

(or equivalent eg. )     (M1)

$=0.770$    A1

Note: Award (M1)A1 for .

[2 marks]

both a sandwich and a cake.

only a sandwich.

Find the expected number of cakes sold on a typical day.

Find the probability that more than 100 cakes will be sold on a typical day.

A customer is selected at random. Find the probability that the customer is male and buys a sandwich.

A female customer is selected at random. Find the probability that she buys a sandwich.

use of formula or Venn diagram       (M1)

0.72 + 0.45 − 1       (A1)

= 0.17       A1

[3 marks]

0.72 − 0.17 = 0.55      A1

[1 mark]

200 × 0.45 = 90      A1

[1 mark]

let X be the number of customers who order cake

X ~ B(200,0.45)        (M1)

P(X > 100) = P(X ≥ 101)(= 1 − P(X ≤ 100))    (M1)

= 0.0681      A1

[3 marks]

0.46 × 0.8 = 0.368    A1

[1 mark]

METHOD 1

$0.368+0.54\times \text{P}\left(S|F\right)=0.72$       M1A1A1 

Note: Award M1 for an appropriate tree diagram. Award M1 for LHS, M1 for RHS.

$\text{P}\left(S|F\right)=0.652$     A1  

METHOD 2

$\text{P}\left(S|F\right)=\frac{\text{P}\left(S\cap F\right)}{\text{P}\left(F\right)}$       (M1)

$=\frac{0.72-0.368}{0.54}$       A1A1

Note: Award A1 for numerator, A1 for denominator.

$\text{P}\left(S|F\right)=0.652$     A1 

[4 marks]

Given that $\mu =253$ and $\sigma =1.5$ find the probability that a randomly chosen packet of biscuits is underweight.

Calculate the new value of $\mu$ giving your answer correct to two decimal places.

Calculate the new value of $\sigma$.

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

     (M1)A1

[2 marks]

$\frac{250-\mu }{1.5}=-2.878\dots$     (M1)(A1)

$\Rightarrow \mu =254.32$     A1

Notes:     Only award A1 here if the correct 2dp answer is seen. Award M0 for use of ${1.5}^2$.

[3 marks]

$\frac{250-253}{\sigma }=-2.878\dots$     (A1)

$\Rightarrow \sigma =1.04$     A1

[2 marks]

Show that $\sqrt{16+k}-\sqrt{k}=\sqrt{k}\sqrt{16+k}$.

Find the value of $k$.

recognition of the need to integrate $\frac{x}{\sqrt{{\left(x^2+k\right)}^3}}$       (M1)

$\int \frac{x}{\sqrt{{\left(x^2+k\right)}^3}}dx\left(=1\right)$

EITHER

$u=x^2+k\Rightarrow \frac{du}{dx}=2x$ (or equivalent)       (A1)

$\int \frac{x}{\sqrt{{\left(x^2+k\right)}^3}}dx=\frac{1}{2}\int u^{-\frac{3}{2}}du$

$=-u^{-\frac{1}{2}}\left(+c\right) \left(=-{\left(x^2+k\right)}^{-\frac{1}{2}}\left(+c\right)\right)$        A1

OR

$\int \frac{x}{\sqrt{{\left(x^2+k\right)}^3}}dx=\frac{1}{2}\int \frac{2x}{\sqrt{{\left(x^2+k\right)}^3}}dx$       (A1)

$=-{\left(x^2+k\right)}^{-\frac{1}{2}}\left(+c\right)$        A1

THEN

attempt to use correct limits for their integrand and set equal to $1$        M1

${\left[-u^{-\frac{1}{2}}\right]}_k^{16+k}=1$  OR  ${\left[-{\left(x^2+k\right)}^{-\frac{1}{2}}\right]}_0^4=1$

$-{\left(16+k\right)}^{-\frac{1}{2}}+k^{-\frac{1}{2}}=1\left(\Rightarrow \frac{1}{\sqrt{k}}-\frac{1}{\sqrt{16+k}}=1\right)$        A1

$\sqrt{16+k}-\sqrt{k}=\sqrt{k}\sqrt{16+k}$        AG

[5 marks]

attempt to solve $\sqrt{16+k}-\sqrt{k}=\sqrt{k}\sqrt{16+k}$      (M1)

$k=0.645038\dots$

$=0.645$        A1

[2 marks]

Find the value of a and the value of b.

Find the value of p and the value of q.

Jennifer was absent for the first test but scored 29 marks on the second test. Use an appropriate regression equation to estimate Jennifer’s mark on the first test.

a = 1.29 and b = −10.4      A1A1

[2 marks]

recognising both lines pass through the mean point       (M1)

p = 28.7, q = 30.3       A2

[3 marks]

substitution into their $x$ on $y$ equation        (M1)

$x$ = 1.29082(29) − 10.3793

$x$ = 27.1      A1

Note: Accept 27.

[2 marks]

Find Pearson’s product-moment correlation coefficient, $r$, for these data.

The relationship between the variables can be modelled by the regression equation $D=ah+b$. Write down the value of $a$ and the value of $b$.

One of these eight students was disappointed with her result and wished she had practised more. Based on the given data, determine how her score could have been expected to alter had she practised an extra five hours per week.

Lucy asserts that the number of hours a student practises has a direct effect on their final diploma result. Comment on the validity of Lucy’s assertion.

Lucy suspected that each student had not been practising as much as they reported. In order to compensate for this, Lucy deducted a fixed number of hours per week from each of the students’ recorded hours.

State how, if at all, the value of $r$ would be affected.

use of GDC to give                          (M1)

$r=0.883529\dots$

$r=0.884$                         A1

Note: Award the (M1) for any correct value of $r$, $a$, $b$ or $r^2=0.780624\dots$ seen in part (a) or part (b).

[2 marks]

$a=1.36609\dots , b=64.5171\dots$

$a=1.37 , b=64.5$                       A1

[1 mark]

attempt to find their difference                       (M1)

$5\times 1.36609\dots$  OR  $1.36609\dots \left(h+5\right)+64.5171\dots -\left(1.36609\dots h+64.5171\dots \right)$

$6.83045\dots$

$=6.83$  $\left(6.85 \mathrm{from} 1.37\right)$

the student could have expected her score to increase by $7$ marks.                       A1

Note: Accept an increase of $6, 6.83$ or $6.85$.

[2 marks]

Lucy is incorrect in suggesting there is a causal relationship.

This might be true, but the data can only indicate a correlation.                     R1

Note: Accept ‘Lucy is incorrect as correlation does not imply causation’ or equivalent.

[1 mark]

no effect                 A1

[1 mark]

Find E(T).

Given that Var(X) = 0.8419, find Var(T).

METHOD 1

attempting to use the expected value formula      (M1)

$\text{E}\left(X\right)=\left(1\times 0.60\right)+\left(2\times 0.30\right)+\left(3\times 0.03\right)+\left(4\times 0.05\right)+\left(5\times 0.02\right)$

$\text{E}\left(X\right)=1.59$($)     (A1)

use of $\text{E}\left(1.20X+2.40\right)=1.20\text{E}\left(X\right)+2.40$      (M1)

$\text{E}\left(T\right)=1.20\left(1.59\right)+2.40$

$=4.31$($)      A1

METHOD 2

attempting to find the probability distribution for T      (M1)

     (A1)

attempting to use the expected value formula       (M1)

$\text{E}\left(T\right)=\left(3.60\times 0.60\right)+\left(4.80\times 0.30\right)+\left(6.00\times 0.03\right)+\left(7.20\times 0.05\right)+\left(8.40\times 0.02\right)$

$=4.31$($)      A1

[4 marks]

METHOD 1

using Var(1.20 X  + 2.40) = (1.20)² Var(X) with Var(X) = 0.8419      (M1)

Var(T) = 1.21      A1

METHOD 2

finding the standard deviation for their probability distribution found in part (a)    (M1)

Var(T) = (1.101…)² 

= 1.21    A1

Note: Award M1A1 for Var(T) = (1.093…)² = 1.20.

[2 marks]

The mean number of squirrels in a certain area is known to be 3.2 squirrels per hectare of woodland. Within this area, there is a 56 hectare woodland nature reserve. It is known that there are currently at least 168 squirrels in this reserve.

Assuming the population of squirrels follow a Poisson distribution, calculate the probability that there are more than 190 squirrels in the reserve.

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

X is number of squirrels in reserve
X ∼ Po(179.2)      A1

Note: Award A1 if 179.2 or 56 × 3.2 seen or implicit in future calculations.

recognising conditional probability     M1

P(X > 190 | X ≥ 168)

$=\frac{\text{P}\left(X>190\right)}{\text{P}\left(X⩾168\right)}=\left(\frac{0.19827\dots }{0.80817\dots }\right)$       (A1)(A1)

= 0.245      A1

[5 marks]

Find the probability that a wolf selected at random is at least 5 years old.

Eight wolves are independently selected at random and their ages recorded.

Find the probability that more than six of these wolves are at least 5 years old.

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

P(L ≥ 5) = 0.910      (M1)A1

[2 marks]

X is the number of wolves found to be at least 5 years old recognising binomial distribution      M1

X ~ B(8, 0.910…)

P(X > 6) = 1 − P(X ≤ 6)      (M1)

= 0.843       A1

Note: Award M1A0 for finding P(X ≥ 6).

[3 marks]

A biased coin is weighted such that the probability, $p$, of obtaining a tail is $0.6$. The coin is tossed repeatedly and independently until a tail is obtained.

Let $E$ be the event “obtaining the first tail on an even numbered toss”.

Find $\text{P}\left(E\right)$.

* 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.

METHOD 1

$E_n$ is the event “the first tail occurs on the $2$nd, $4$th, $6$th, …, $2n$th toss”

$\text{P}\left(E\right)=\underset{n=1}{\overset{\infty }{\text{Σ}}}\text{P}\left(E_n\right)$         (A1)

Note: Award A1 for deducing that either $1$ head before a tail or $3$ heads before a tail or $5$ heads before a tail etc. is required. In other words, deduces $\left(2n-1\right)$ heads before a tail.

$\text{P}\left(E\right)=0.4\times 0.6+{\left(0.4\right)}^3\times 0.6+{\left(0.4\right)}^5\times 0.6+\dots$         M1A1

Note: Award M1 for attempting to form an infinite geometric series.

Note: Award A1 for $\text{P}\left(E\right)=\underset{n=1}{\overset{\infty }{\text{Σ}}}{\left(0.4\right)}^{2n-1}\left(0.6\right)$.

uses $S_{\infty }=\frac{u_1}{1-r}$ with $u_1=0.6\times 0.4$ and $r={\left(0.4\right)}^2$         (M1)

Note: Award M1 for using $S_{\infty }=\frac{u_1}{1-r}$ with $u_1=0.4$ and $r={\left(0.4\right)}^2$

$=\frac{0.6\times 0.4}{1-{\left(0.4\right)}^2}$        A1

$=0.286 \left(=\frac{2}{7}\right)$        A1

METHOD 2

let $T_1$ be the event “tail occurs on the first toss”

uses $\text{P}\left(E\right)=\text{P}\left(E \overline{)T_1}\right)\text{P}\left(T_1\right)+\text{P}\left(E \overline{)T_1\text{'}}\right)\text{P}\left(T_1\text{'}\right)$         M1

concludes that $\text{P}\left(E \overline{)T_1}\right)=0$ and so $\text{P}\left(E\right)=\text{P}\left(E \overline{)T_1\text{'}}\right)\text{P}\left(T_1\text{'}\right)$         R1

$\text{P}\left(E \overline{)T_1\text{'}}\right)=\text{P}\left(E\text{'}\right)\left(=1-\text{P}\left(E\right)\right)$         A1

Note: Award A1 for concluding: given that a tail is not obtained on the first toss, then $\text{P}\left(E \overline{)T_1\text{'}}\right)$ is the probability that the first tail is obtained after a further odd number of tosses, $\text{P}\left(E\text{'}\right)$.

$\text{P}\left(T_1\text{'}\right)=0.4$

$\text{P}\left(E\right)=0.4\left(1-\text{P}\left(E\right)\right)$         A1

attempts to solve for $\text{P}\left(E\right)$         (M1)

$=0.286 \left(=\frac{2}{7}\right)$         A1

[6 marks]

Find the least possible value of n.

It is further given that P(X ≤ 1) = 0.09478 correct to 4 significant figures.

Determine the value of n and the value of p.

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

np = 3.5      (A1)

p ≤ 1 ⇒ least n = 4       A1

[2 marks]

(1 − p)ⁿ + np(1 − p)ⁿ⁻¹ = 0.09478     M1A1

attempt to solve above equation with np = 3.5     (M1)

n = 12,  p = $\frac{7}{24}$ (=0.292)     A1A1

Note: Do not accept n as a decimal.

[5 marks]

Show that $a=\frac{2}{3}$.

Find .

Given that , and that 0.25 < s < 0.4 , find the value of s.

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

$a\left[{\int }_0^{0.5}3x\text{d}x+{\int }_{0.5}^2\left(2-x\right)\text{d}x\right]=1$     M1

Note: Award the M1 for the total integral equalling 1, or equivalent.

$a\left(\frac{3}{2}\right)=1$     (M1)A1

$a=\frac{2}{3}$     AG

[3 marks]

EITHER

${\int }_0^{0.5}2x\text{d}x+\frac{2}{3}{\int }_{0.5}^1\left(2-x\right)\text{d}x$     (M1)(A1)

$=\frac{2}{3}$     A1

OR

$\frac{2}{3}{\int }_1^2\left(2-x\right)\text{d}x=\frac{1}{3}$     (M1)

so       (M1)A1

[3 marks]

     M1A1

$={\left[x^2\right]}_s^{0.5}+0.27$

$0.25-s^2+0.27$     (A1)

     A1

$=\frac{2}{3}{\left[2x-\frac{x^2}{2}\right]}_{2s}^{0.8}$

$\frac{2}{3}\left(1.28-\left(4s-2s^2\right)\right)$

equating

$0.25-s^2+0.27=\frac{4}{3}\left(1.28-\left(4s-2s^2\right)\right)$     (A1)

attempt to solve for s      (M1)

s = 0.274      A1

[7 marks]

Show that $a=32$ and $b=\frac{1}{12}$.

Find $\text{E}(X)$.

Find $\text{Var}(X)$.

Find the median of $X$.

Find $\text{E}(Y)$.

Find $\text{P}(Y⩾3)$.

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

$\underset{0}{\overset{4}{\int }}\left(\frac{x^2}{a}+b\right)\text{d}x=1\Rightarrow {\left[\frac{x^3}{3a}+bx\right]}_0^4=1\Rightarrow \frac{64}{3a}+4b=1$    M1A1

$\underset{2}{\overset{4}{\int }}\left(\frac{x^2}{a}+b\right)\text{d}x=0.75\Rightarrow \frac{56}{3a}+2b=0.75$    M1A1

Note:    $\underset{0}{\overset{2}{\int }}\left(\frac{x^2}{a}+b\right)dx=0.25\Rightarrow \frac{8}{3a}+2b=0.25$ could be seen/used in place of either of the above equations.

evidence of an attempt to solve simultaneously (or check given a,b values are consistent)     M1

$a=32,b=\frac{1}{12}$     AG

[5 marks]

$\text{E}\left(X\right)=\underset{0}{\overset{4}{\int }}x\left(\frac{x^2}{32}+\frac{1}{12}\right)\text{d}x$    (M1)

$\text{E}(X)=\frac{8}{3}(=2.67)$     A1

[2 marks]

$\text{E}\left(X^2\right)=\underset{0}{\overset{4}{\int }}{x}^2\left(\frac{x^2}{32}+\frac{1}{12}\right)\text{d}x$     (M1)

$\text{Var}(X)=\text{E}(X^2)-[\text{E}(X){]}^2=\frac{16}{15}(=1.07)$     A1

[2 marks]

$\underset{0}{\overset{m}{\int }}\left(\frac{x^2}{32}+\frac{1}{12}\right)\text{d}x=0.5$    (M1)

$\frac{m^3}{96}+\frac{m}{12}=0.5(\Rightarrow m^3+8m-48=0)$     (A1)

$m=2.91$     A1

[3 marks]

$Y\sim B(8,0.75)$     (M1)

$\text{E}(Y)=8\times 0.75=6$     A1

[2 marks]

$\text{P}(Y⩾3)=0.996$     A1

[1 mark]

Rachel and Sophia are competing in a javelin-throwing competition.

The distances, $R$ metres, thrown by Rachel can be modelled by a normal distribution with mean $56.5$ and standard deviation $3$.

The distances, $S$ metres, thrown by Sophia can be modelled by a normal distribution with mean $57.5$ and standard deviation $1.8$.

In the first round of competition, each competitor must have five throws. To qualify for the next round of competition, a competitor must record at least one throw of $60$ metres or greater in the first round.

Find the probability that only one of Rachel or Sophia qualifies for the next round of competition.

Rachel: $R~\text{N}\left(56.5,3^2\right)$

$\text{P}\left(R\ge 60\right)=0.1216\dots$          (A1)

Sophia: $S~\text{N}\left(57.5,1.8^2\right)$

$\text{P}\left(S\ge 60\right)=0.0824\dots$          (A1)

recognises binomial distribution with $n=5$          (M1)

let $N_R$ represent the number of Rachel’s throws that are longer than $60$ metres

$N_R~\text{B}\left(5,0.1216\dots \right)$

either $\text{P}\left(N_R\ge 1\right)=0.4772\dots$  or  $\text{P}\left(N_R=0\right)=0.5227\dots$          (A1)

let $N_S$ represent the number of Sophia’s throws that are longer than $60$ metres

$N_S~\text{B}\left(5,0.0824\dots \right)$

either $\text{P}\left(N_S\ge 1\right)=0.3495\dots$  or  $\text{P}\left(N_S=0\right)=0.6504\dots$          (A1)

EITHER

uses $\text{P}\left(N_R\ge 1\right)\text{P}\left(N_S=0\right)+\text{P}\left(N_S\ge 1\right)\text{P}\left(N_R=0\right)$          (M1)

$\text{P}\left(\text{one of Rachel or Sophia qualify}\right)=\left(0.4772\dots \times 0.6504\right)+\left(0.3495\dots \times 0.5227\dots \right)$

OR

uses $\text{P}\left(N_R\ge 1\right)+\text{P}\left(N_S\ge 1\right)-2\times \text{P}\left(N_R\ge 1\right)\times \text{P}\left(N_S\ge 1\right)$          (M1)

$\text{P}\left(\text{one of Rachel or Sophia qualify}\right)=0.4772\dots +0.3495\dots -2\times 0.4772\dots \times 0.3495\dots$

THEN

$=0.4931\dots$

 $=0.493$        A1

Note: M marks are not dependent on the previous A marks.

[7 marks]

In general, well answered with many candidates getting full marks. On the other end of the spectrum, some candidates could only get the first two marks for normal distribution without recognising the binomial distribution and often just stopping there. Some candidates lost the last two marks for not being able to combine the probabilities correctly.

Sketch the probability density function for X, and shade the region representing P(μ − 2σ < X < μ + σ).

Find the value of P(μ − 2σ < X < μ + σ).

Find the value of k for which P(μ − kσ < X < μ + kσ) = 0.5.

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

normal curve centred on 50      A1

vertical lines at $x$ = 42 and $x$ = 54, with shading in between       A1

[2 marks]

P(42 < X < 54) (= P(− 2 < Z < 1))     (M1)

= 0.819       A1

[2 marks]

P(μ − kσ < X < μ + kσ) = 0.5 ⇒ P(X < μ + kσ) = 0.75      (M1)

k = 0.674       A1

Note: Award M1A0 for k = −0.674.

[2 marks]

Calculate the probability that, on a randomly selected day, Timmy makes a profit.

The shop is open for 24 days every month.

Calculate the probability that, in a randomly selected month, Timmy makes a profit on between 5 and 10 days (inclusive).

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

X ~ N(820, 230²)       (M1)

Note: Award M1 for an attempt to use normal distribution. Accept labelled normal graph.

⇒P(X > 1000) = 0.217       A1

[2 marks]

Y ~ B(24,0.217...)       (M1)

Note: Award M1 for recognition of binomial distribution with parameters.

P(Y ≤ 10) − P(Y ≤ 4)         (M1)

Note: Award M1 for an attempt to find P(5 ≤ Y ≤ 10) or P(Y ≤ 10) − P(Y ≤ 4).

= 0.613       A1

[3 marks]

Calculate $k$;

Find $\text{P}(A^{\prime }\cap B)$.

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

use of $\text{P}(A\cup B)=\text{P}(A)+\text{P}(B)-\text{P}(A\cap B)$     M1

$0.5=k+3k-k^2$     A1

$k^2-4k+0.5=0$

$k=0.129$     A1

Note:     Do not award the final A1 if two solutions are given.

[3 marks]

use of $\text{P}(A^{\prime }\cap B)=\text{P}(B)-\text{P}(A\cap B)$ or alternative     (M1)

$\text{P}(A^{\prime }\cap B)=3k-k^2$     (A1)

$=0.371$     A1

[3 marks]

Find the probability that a randomly selected packet has a weight less than $100 \text{g}$.

The probability that a randomly selected packet has a weight greater than $w$ grams is $0.444$. Find the value of $w$.

A packet is randomly selected. Given that the packet has a weight greater than $105 \text{g}$, find the probability that it has a weight greater than $110 \text{g}$.

From a random sample of $500$ packets, determine the number of packets that would be expected to have a weight lying within $1.5$ standard deviations of the mean.

Packets are delivered to supermarkets in batches of $80$. Determine the probability that at least $20$ packets from a randomly selected batch have a weight less than $95 \text{g}$.

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

${\rm X}~\text{N}\left(102, 8^2\right)$

$\text{P}\left({\rm X}<100\right)=0.401$        (M1)A1

[2 marks]

$\text{P}\left({\rm X}>w\right)=0.444$       (M1)

$\Rightarrow w=103 \left(\text{g}\right)$        A1