18M.2.AHL.TZ1.H_8

pestleMathematicsAIHLPaper 218M· sl-4-6-combined-mutually-exclusive-conditional-independence-prob-diagramssource ↗

Each of the 25 students in a class are asked how many pets they own. Two students own three pets and no students own more than three pets. The mean and standard deviation of the number of pets owned by students in the class are $\frac{{18}}{{25}}$ and $\frac{{24}}{{25}}$ respectively.

Find the number of students in the class who do not own a pet.

Markscheme / solution

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

METHOD 1

let p have no pets, q have one pet and r have two pets (M1)

p + q + r + 2 = 25 (A1)

0p + 1q + 2r + 6 = 18 A1

Note: Accept a statement that there are a total of 12 pets.

attempt to use variance equation, or evidence of trial and error (M1)

$\frac{{0p + 1q + 4r + 18}}{{25}} - {\left( {\frac{{18}}{{25}}} \right)^2} = {\left( {\frac{{24}}{{25}}} \right)^2}$ (A1)

attempt to solve a system of linear equations (M1)

p = 14 A1

METHOD 2

(M1)

$p + q + r + \frac{2}{{25}} = 1$ (A1)

$q + 2r + \frac{6}{{25}} = \frac{{18}}{{25}}\left( { \Rightarrow q + 2r = \frac{{12}}{{25}}} \right)$ A1

$q + 4r + \frac{{18}}{{25}} - {\left( {\frac{{18}}{{25}}} \right)^2} = \frac{{576}}{{625}}\left( { \Rightarrow q + 4r = \frac{{18}}{{25}}} \right)$ (M1)(A1)

$q = \frac{6}{{25}},\,\,r = \frac{3}{{25}}$ (M1)

$p = \frac{{14}}{{25}}$ A1

so 14 have no pets

[7 marks]

Examiners’ report

[N/A]