It is known that the weights of male Persian cats are normally distributed with mean 6.1 kgand variance 0.52 kg2.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 drawnfrom a population whose weights are normally distributed with mean 4.5 kg and standarddeviation 0.45 kg.Ten female cats are chosen at random.

Question

HLPaper 2

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 drawnfrom a population whose weights are normally distributed with mean $4.5 kg$ and standarddeviation $0.45 kg$ .

Ten female cats are chosen at random.

Sketch a diagram showing the above information.

[2]

Verified

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]

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

[2]

Verified

Solution

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

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

$= 0.673 (0.673074 \dots)$ A1

[2 marks]

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

[3]

Verified

Solution

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

$0.0547992 \dots \times 80$ (M1)

$= 4.38 (4.38393 \dots)$ A1

[3 marks]

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

[4]

Verified

Solution

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

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

use of binomial seen or implied (M1)

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

$0.0430 (0.0429664 \dots)$ A1

[4 marks]

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

Find the variance of $N$ .

[1]

Verified

Solution

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

[1 mark]

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]

Verified

Solution

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

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