Given that 70 % of the birds weigh more than 2.1 kg and 25 % of the birds weigh more than 2.5 kg, calculate the value of $\mu$ and the value of $\sigma$ .

A random sample of ten of these birds is obtained. Let X denote the number of birds in the sample weighing more than 2.5 kg.

(i)     Calculate ${\text{E}}(X)$ .

(ii)     Calculate the probability that exactly five of these birds weigh more than 2.5 kg.

(iii)     Determine the most likely value of X .

The number of eggs, Y , laid by female birds of this species during the nesting season is modelled by a Poisson distribution with mean $\lambda$ . You are given that ${\text{P}}(Y \geqslant 2) = 0.80085$ , correct to 5 decimal places.

(i)     Determine the value of $\lambda$ .

(ii)     Calculate the probability that two randomly chosen birds lay a total of

two eggs between them.

(iii)     Given that the two birds lay a total of two eggs between them, calculate the probability that they each lay one egg.

we are given that

$2.1 = \mu - 0.5244\sigma$

$2.5 = \mu + 0.6745\sigma$     M1A1

$\mu = 2.27{\text{ , }}\sigma = 0.334$     A1A1

[4 marks]

(i)     let X denote the number of birds weighing more than 2.5 kg

then X is B(10, 0.25)     A1

${\text{E}}(X) = 2.5$     A1

(ii)     0.0584     A1

(iii)     to find the most likely value of X , consider

${p_0} = 0.0563 \ldots ,{\text{ }}{p_1} = 0.1877 \ldots ,{\text{ }}{p_2} = 0.2815 \ldots ,{\text{ }}{p_3} = 0.2502 \ldots$     M1

therefore, most likely value = 2     A1

[5 marks]

(i)     we solve $1 - {\text{P}}(Y \leqslant 1) = 0.80085$ using the GDC     M1

$\lambda = 3.00$     A1

(ii)     let ${X_1}{\text{, }}{X_2}$ denote the number of eggs laid by each bird

${\text{P}}({X_1} + {X_2} = 2) = {\text{P}}({X_1} = 0){\text{P}}({X_2} = 1) + {\text{P}}({X_1} = 1){\text{P}}({X_2} = 1) + {\text{P}}({X_1} = 2){\text{P}}({X_2} = 0)$     M1A1

$= {{\text{e}}^{ - 3}} \times {{\text{e}}^{ - 3}} \times \frac{9}{2} + {({{\text{e}}^{ - 3}} \times 3)^2} + {{\text{e}}^{ - 3}} \times \frac{9}{2} \times {{\text{e}}^{ - 3}} = 0.0446$     A1

(iii)     ${\text{P}}({X_1} = 1,{\text{ }}{X_2} = 1|{X_1} + {X_2} = 2) = \frac{{{\text{P}}({X_1} = 1,{\text{ }}{X_2} = 1)}}{{{\text{P}}({X_1} + {X_2} = 2)}}$     M1A1

$= 0.5$     A1

[8 marks]