21M.2.AHL.TZ2.2

pestleMathematicsAIHLPaper 221M· sl-4-9-normal-distribution-and-calculations, sl-4-8-binomial-distribution, sl-4-6-combined-mutually-exclusive-conditional-independence-prob-diagramssource ↗

It is known that the weights of male Persian cats are normally distributed with mean $6.1 kg$ and variance $0 . 5^{2} {kg}^{2}$ .

A group of $80$ male Persian cats are drawn from this population.

The male cats are now joined by $80$ female Persian cats. The female cats are drawn from a population whose weights are normally distributed with mean $4.5 kg$ and standard deviation $0.45 kg$ .

Ten female cats are chosen at random.

Sketch a diagram showing the above information.

[2]

Find the proportion of male Persian cats weighing between $5.5 kg$ and $6.5 kg$ .

[2]

Determine the expected number of cats in this group that have a weight of less than $5.3 kg$ .

[3]

Find the probability that exactly one of them weighs over $4.62 kg$ .

[4]

d.i.

Let $N$ be the number of cats weighing over $4.62 kg$ .

Find the variance of $N$ .

[1]

d.ii.

A cat is selected at random from all $160$ cats.

Find the probability that the cat was female, given that its weight was over $4.7 kg$ .

[4]

Markscheme / solution

A1A1

Note: Award A1 for a normal curve with mean labelled $6.1$ or $μ$ , A1 for indication of SD $(0.5) :$ marks on horizontal axis at $5.6$ and/or $6.6$ OR $μ - 0.5$ and/or $μ + 0.5$ on the correct side and approximately correct position.

[2 marks]

$X ~ N (6.1, 0 . 5^{2})$

$P (5.5 < X < 6.5)$ OR labelled sketch of region (M1)

$= 0.673 (0.673074 \dots)$ A1

[2 marks]

$(P (X < 5.3) =) 0.0547992 \dots$ (A1)

$0.0547992 \dots \times 80$ (M1)

$= 4.38 (4.38393 \dots)$ A1

[3 marks]

$Y ~ N (4.5, 0 . 45^{2})$ ,

$(P (Y > 4.62) =) 0.394862 \dots$ (A1)

use of binomial seen or implied (M1)

using $B (10, 0.394862 \dots)$ (M1)

$0.0430 (0.0429664 \dots)$ A1

[4 marks]

d.i.

$n p (1 - p) = 2.39 (2.38946 \dots)$ A1

[1 mark]

d.ii.

$P (F \cap (W > 4.7)) = 0.5 \times 0.3284 (= 0.1642)$ (A1)

attempt use of tree diagram OR use of $P (F W > 4.7) = \frac{P (F \cap (W > 4.7))}{P (W > 4.7)}$ (M1)

$\frac{0.5 \times 0.3284}{0.5 \times 0.9974 + 0.5 \times 0.3284}$ (A1)

$= 0.248 (0.247669 \dots)$ A1

[4 marks]

Examiners’ report

[N/A]

d.i.

[N/A]

d.ii.

[N/A]