S2 June 2016 Q1

EdexcelOld spec14 marksPoisson Distribution

1. A student is investigating the numbers of cherries in a Rays fruit cake. A random sample of Rays fruit cakes is taken and the results are shown in the table below.

Number of cherries	0	1	2	3	4	5	\(\geqslant 6\)
Frequency	24	37	21	12	4	2	0

(a) Calculate the mean and the variance of these data. (3)

(b) Explain why the results in part (a) suggest that a Poisson distribution may be a suitable model for the number of cherries in a Rays fruit cake. (1)

The number of cherries in a Rays fruit cake follows a Poisson distribution with mean 1.5

A Rays fruit cake is to be selected at random.

Find the probability that it contains

(c)

(i) exactly 2 cherries,

(ii) at least 1 cherry. (4)

Rays fruit cakes are sold in packets of 5

(d) Show that the probability that there are more than 10 cherries, in total, in a randomly selected packet of Rays fruit cakes, is 0.1378 correct to 4 decimal places. (3)

Twelve packets of Rays fruit cakes are selected at random.

(e) Find the probability that exactly 3 packets contain more than 10 cherries. (3)

Mark scheme (a)

Note: if a correct answer is given with no incorrect working award full marks unless the markscheme says otherwise.

Scheme	Marks
Mean = 1.41	B1
Variance \(= \dfrac{343}{100} - 1.41^2\)	M1
\(= 1.4419 \qquad (s^2 = 1.456)\)	A1
	(3)

Notes

B1: Cao Allow 141/100

M1: using \(\dfrac{\sum fx^2}{100} - (\text{their mean})^2\) or \(\dfrac{100}{99}\left(\dfrac{\sum fx^2}{100} - (\text{their mean})^2\right)\) oe
NB Allow the square root of this for the M mark.
If no working shown for \(\sum fx^2\) then you must see 343, 3.43 or a correct answer

A1: awrt 1.44 or 1.46 for \(s^2\)

Scheme	Marks
The mean is close to the variance	B1
	(1)

Mark scheme (c)

Scheme	Marks
(i) \(X \sim \mathrm{Po}(1.5)\) \(\mathrm{P}(X = 2) = \dfrac{\mathrm{e}^{-1.5}1.5^2}{2!}\)	M1
\(= 0.2510\)	A1
(ii) \(\mathrm{P}(X \geqslant 1) = 1 - \mathrm{P}(X = 0)\) \(= 1 - \mathrm{e}^{-1.5}\) Or \(1 - 0.2231\)	M1
\(= 0.77686\ldots\)	A1
	(4)

Notes

(i) M1: writing or using \(\dfrac{\mathrm{e}^{-\lambda}\lambda^2}{2!}\) or \(\mathrm{P}(X \leqslant 2) - \mathrm{P}(X \leqslant 1)\)
A1: awrt 0.251

(ii) M1: writing or using \(1 - \mathrm{P}(X = 0)\) oe
A1: awrt 0.777

Mark scheme (d)

Scheme	Marks
\(Y \sim \mathrm{Po}(7.5)\)	B1
\(\mathrm{P}(Y \geqslant 11) = 1 - \mathrm{P}(Y \leqslant 10)\)	M1
\(= 1 - 0.8622\) \(= 0.1378\ *\)	A1cso
	(3)

Notes

B1: Writing Po(7.5)

M1: writing \(\mathrm{P}(Y \geqslant 11)\) or \(1 - \mathrm{P}(Y \leqslant 10)\) oe

A1: Seeing 1 – 0.8622 leading to 0.1378 cso (both B1 and M1 awarded)

Mark scheme (e)

Scheme	Marks
\(A \sim \mathrm{B}(12, 0.1378)\)	M1
\(\mathrm{P}(A = 3) = \dbinom{12}{3}(0.1378)^3(0.8622)^9\)	M1
\(= 0.1516\)	A1
	(3)
	(14 marks)

Notes

M1: using \((p)^n(1 - p)^{12 - n}\) where \(p = 0.1378\) or 0.138 condone missing \(n\mathrm{C}r\)

M1: \(\dbinom{12}{3}(p)^3(1 - p)^9\), with \(0 \lt p \lt 1\) Allow 220 or 12 C 3 instead of \(\dbinom{12}{3}\)

A1: awrt 0.152