*Question:* [Maximum mark: 4]

The number of hours spent exercising each week by a group of students is shown in the following table.

Exercising time (in hours) | Number of students |

2 | 5 |

3 | 1 |

4 | 4 |

5 | 3 |

6 | x |

The median is 4.5 hours.

(a) Find the value of x .**(b) Find the standard deviation.**

**▶️Answer/Explanation**

Ans:

(a) **EITHER**

recognising that half the total frequency is 10 (may be seen in an ordered list or indicated on the frequency table)

**OR**

5 + 1 + 4 = 3 + x

**OR**

\(\sum f = 20\)

**THEN**

x = 7

(b) METHOD 1

1.58429…

1.58

**METHOD 2****EITHER**

## Question

Consider the data set \(\{ k – 2,{\text{ }}k,{\text{ }}k + 1,{\text{ }}k + 4\} {\text{ , where }}k \in \mathbb{R}\) .

(a) Find the mean of this data set in terms of *k*.

Each number in the above data set is now **decreased** by 3.

(b) Find the mean of this **new** data set in terms of *k*.

**▶️Answer/Explanation**

## Markscheme

(a) Use of \(\bar x = \frac{{\sum\limits_{i = 1}^4 {{x_i}} }}{n}\) *(M1)*

\(\bar x = \frac{{(k – 2) + k + (k + 1) + (k + 4)}}{4}\) *(A1)*

\(\bar x = \frac{{4k + 3}}{4}\,\,\,\,\,\left( { = k + \frac{3}{4}} \right)\) *A1**N3*

* *

(b) Either attempting to find the new mean or subtracting 3 from **their** \({\bar x}\) *(M1)*

\(\bar x = \frac{{4k + 3}}{4} – 3\,\,\,\,\,\left( { = \frac{{4k – 9}}{4},{\text{ }}k – \frac{9}{4}} \right)\) *A1 N2*

*[5 marks]*

## Examiners report

This was an easy question that was well done by most candidates. Careless arithmetic errors caused some candidates not to earn full marks. Only a few candidates realised that part (b) could be answered correctly by directly subtracting 3 from their answer to part (a). Most successful responses were obtained by redoing the calculation from part (a).

## Question

Consider the data set \(\{ 2,{\text{ }}x,{\text{ }}y,{\text{ }}10,{\text{ }}17\} ,{\text{ }}x,{\text{ }}y \in {\mathbb{Z}^ + }\) and \(x < y\).

The mean of the data set is \(8\) and its variance is \(27.6\).

Find the value of \(x\) and the value of \(y\).

**▶️Answer/Explanation**

## Markscheme

use of \(\mu = \frac{{\sum\limits_{i = 1}^k {{f_i}{x_i}} }}{n}{\text{ }}\) to obtain \(\frac{{2 + x + y + 10 + 17}}{5} = 8\) *(M1)*

\(x + y = 11\) *A1*

**EITHER**

use of \({\sigma ^2} = \frac{{\sum\limits_{i = 1}^k {{f_i}{{({x_i} – \mu )}^2}} }}{n}\) to obtain \(\frac{{{{( – 6)}^2} + {{(x – 8)}^2} + {{(y – 8)}^2} + {2^2} + {9^2}}}{5} = 27.6\) *(M1)*

\({(x – 8)^2} + {(y – 8)^2} = 17\) *A1*

**OR**

use of \({\sigma ^2} = \frac{{\sum\limits_{i = 1}^k {{f_i}x_i^2} }}{n} – {\mu ^2}\) to obtain \(\frac{{{2^2} + {x^2} + {y^2} + {{10}^2} + {{17}^2}}}{5} – {8^2} = 27.6\) *(M1)*

\({x^2} + {y^2} = 65\) *A1*

**THEN**

attempting to solve the two equations *(M1)*

\(x = 4\;\;\;\)and\(\;\;\;y = 7\;\;\;({\text{only as }}x < y)\)*A1 N4*

**Note: **Award ** A0** for \(x = 7\) and \(y = 4\).

**Note: **Award ** (M1)A1(M0)A0(M1)A1** for \(x + y = 11 \Rightarrow x = 4\) and \(y = 7\).

*[6 marks]*

## Examiners report

Reasonably well done. Most candidates were able to obtain \(x + y = 11\). Most manipulation errors occurred when candidates attempted to form the variance equation in terms of \(x\) and \(y\). Some candidates did not apply the condition \(x < y\) when determining their final answer.

## Question

The data of the goals scored by players in a football club during a season are given in the following table.

a.Given that the mean number of goals scored per player is \(1.95\) , find the value of \(k\).[3]

(i) Find the correct mean number of goals scored per player.

(ii) Find the correct standard deviation of the number of goals scored per player.[3]

**▶️Answer/Explanation**

## Markscheme

\(\frac{{0 \bullet 4 + 1 \bullet k + 2 \bullet 3 + 3 \bullet 2 + 4 \bullet 3 + 8 \bullet 1}}{{13 + k}} = 1.95\;\;\;\left( {\frac{{k + 32}}{{k + 13}} = 1.95} \right)\) *(M1)*

attempting to solve for \(k\) *(M1)*

\(k = 7\) *A1*

*[3 marks]*

(i) \(\frac{{7 + 32 + 22}}{{7 + 13 + 1}} = 2.90\;\;\;\left( { = \frac{{61}}{{21}}} \right)\) *(M1)A1*

(ii) standard deviation \( = 4.66\) *A1*

**Note: **Award ** A0 **for \(4.77\).

**[3 marks]**

**Total [6 marks]**

## Examiners report

[N/A]

[N/A]

## Question

Six balls numbered 1, 2, 2, 3, 3, 3 are placed in a bag. Balls are taken one at a time from the bag at random and the number noted. Throughout the question a ball is always replaced before the next ball is taken.

Three balls are taken from the bag. Find the probability that

A single ball is taken from the bag. Let \(X\) denote the value shown on the ball.

a.Find \({\text{E}}(X)\).[2]

b.i.the total of the three numbers is 5;[3]

b.ii.the median of the three numbers is 1.[3]

c.Ten balls are taken from the bag. Find the probability that less than four of the balls are numbered 2.[3]

d.Find the least number of balls that must be taken from the bag for the probability of taking out at least one ball numbered 2 to be greater than 0.95.[3]

e.Another bag also contains balls numbered 1 , 2 or 3.

Eight balls are to be taken from this bag at random. It is calculated that the expected number of balls numbered 1 is 4.8 , and the variance of the number of balls numbered 2 is 1.5.

Find the least possible number of balls numbered 3 in this bag.[8]

**▶️Answer/Explanation**

## Markscheme

\({\text{E}}(X) = 1 \times \frac{1}{6} + 2 \times \frac{2}{6} + 3 \times \frac{3}{6} = \frac{{14}}{6}{\text{ }}\left( { = \frac{7}{3} = 2.33} \right)\) *(M1)A1*

*[2 marks]*

\(3 \times {\text{P}}(113) + 3 \times {\text{P}}(122)\) *(M1)*

\(3 \times \frac{1}{6} \times \frac{1}{6} \times \frac{1}{2} + 3 \times \frac{1}{6} \times \frac{1}{3} \times \frac{1}{3} = \frac{7}{{72}}{\text{ }}( = 0.0972)\) *A1*

**Note: **Award ** M1 **for attempt to find at least four of the cases.

*[3 marks]*

recognising 111 as a possibility \(\left( {{\text{implied by }}\frac{1}{{216}}} \right)\) *(M1)*

recognising 112 and 113 as possibilities \(\left( {{\text{implied by }}\frac{2}{{216}}{\text{ and }}\frac{3}{{216}}} \right)\) *(M1)*

seeing the three arrangements of 112 and 113 *(M1)*

\({\text{P}}(111) + 3 \times {\text{P}}(112) + 3 \times {\text{P}}(113)\)

\( = \frac{1}{{216}} + \frac{6}{{216}} + \frac{9}{{216}} = \frac{{16}}{{216}}{\text{ }}\left( { = \frac{2}{{27}} = 0.0741} \right)\) *A1*

*[3 marks]*

let the number of twos be \(X,{\text{ }}X \sim B\left( {10,{\text{ }}\frac{1}{3}} \right)\) *(M1)*

\({\text{P}}(X < 4) = {\text{P}}(X \leqslant 3) = 0.559\) *(M1)A1*

*[3 marks]*

let \(n\) be the number of balls drawn

\({\text{P}}(X \geqslant 1) = 1 – {\text{P}}(X = 0)\) *M1*

\( = 1 – {\left( {\frac{2}{3}} \right)^n} > 0.95\) *M1*

\({\left( {\frac{2}{3}} \right)^n} > 0.05\)

\(n = 8\) *A1*

*[3 marks]*

\(8{p_1} = 4.8 \Rightarrow {p_1} = \frac{3}{5}\) *(M1)A1*

\(8{p_2}(1 – {p_2}) = 1.5\) *(M1)*

\(p_2^2 – {p_2} – 0.1875 = 0\) *(M1)*

\({p_2} = \frac{1}{4}{\text{ }}\left( {{\text{or }}\frac{3}{4}} \right)\) *A1*

reject \(\frac{3}{4}\) as it gives a total greater than one

\({\text{P}}(1{\text{ or }}2) = \frac{{17}}{{20}}{\text{ or P}}(3) = \frac{3}{{20}}\) *(A1)*

recognising LCM as 20 so min total number is 20 *(M1)*

the least possible number of 3’s is 3 *A1*

*[8 marks]*

## Question

A Chocolate Shop advertises free gifts to customers that collect three vouchers. The vouchers are placed at random into 10% of all chocolate bars sold at this shop. Kati buys some of these bars and she opens them one at a time to see if they contain a voucher. Let \({\text{P}}(X = n)\) be the probability that Kati obtains her third voucher on the \(n{\text{th}}\) bar opened.

(It is assumed that the probability that a chocolate bar contains a voucher stays at 10% throughout the question.)

It is given that \({\text{P}}(X = n) = \frac{{{n^2} + an + b}}{{2000}} \times {0.9^{n – 3}}\) for \(n \geqslant 3,{\text{ }}n \in \mathbb{N}\).

Kati’s mother goes to the shop and buys \(x\) chocolate bars. She takes the bars home for Kati to open.

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

b.Find the values of the constants \(a\) and \(b\).[5]

c.Deduce that \(\frac{{{\text{P}}(X = n)}}{{{\text{P}}(X = n – 1)}} = \frac{{0.9(n – 1)}}{{n – 3}}\) for \(n > 3\).[4]

d.(i) Hence show that \(X\) has two modes \({m_1}\) and \({m_2}\).

(ii) State the values of \({m_1}\) and \({m_2}\).[5]

e.Determine the minimum value of \(x\) such that the probability Kati receives at least one free gift is greater than 0.5.[3]

**▶️Answer/Explanation**

## Markscheme

\({\text{P}}(X = 3) = {(0.1)^3}\) *A1*

\( = 0.001\) *AG*

\({\text{P}}(X = 4) = {\text{P}}(VV\bar VV) + {\text{P}}(V\bar VVV) + {\text{P}}(\bar VVVV)\) *(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*

\(\frac{{9 + 3a + b}}{{2000}} = \frac{1}{{1000}}{\text{ }}(3a + b = – 7)\) *A1*

\(\frac{{16 + 4a + b}}{{2000}} \times \frac{9}{{10}} = \frac{{27}}{{10\,000}}{\text{ }}(4a + b = – 10)\) *A1*

attempting to solve simultaneously *(M1)*

\(a = – 3,{\text{ }}b = 2\) *A1*

**METHOD 2**

\({\text{P}}(X = n) = \left( {\begin{array}{*{20}{c}} {n – 1} \\ 2 \end{array}} \right) \times {0.1^3} \times {0.9^{n – 3}}\) *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**

\({\text{P}}(X = n) = \left( {\begin{array}{*{20}{c}} {n – 1} \\ 2 \end{array}} \right) \times {0.1^3} \times {0.9^{n – 3}}\) *(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

**for a correct denominator.**

*A1*\( = \frac{{0.9(n – 1)(n – 2)}}{{(n – 2)(n – 3)}}\) *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*

\(\frac{{0.9(n – 1)}}{{n – 3}} < 1 \Rightarrow n > 21\) *R1*

\(\frac{{0.9(n – 1)}}{{n – 3}} > 1 \Rightarrow n < 21\) *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**

\(Y \sim {\text{B}}(x,{\text{ }}0.1)\) *(A1)*

attempting to solve \({\text{P}}(Y \geqslant 3) > 0.5\) (or equivalent *eg* \(1 – {\text{P}}(Y \leqslant 2) > 0.5\)) for \(x\) *(M1)*

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

\(x = 27\) *A1*

**METHOD 2**

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

attempting to solve for \(x\) *(M1)*

\(x = 27\) *A1*

*[3 marks]*