June 2019 Paper 3 Q4

EdexcelCurrent spec9 marksBinomial Distribution Large Data Set

4. Magali is studying the mean total cloud cover, in oktas, for Leuchars in 1987 using data from the large data set. The daily mean total cloud cover for all 184 days from the large data set is summarised in the table below.

Daily mean total cloud cover (oktas)	0	1	2	3	4	5	6	7	8
Frequency (number of days)	0	1	4	7	10	30	52	52	28

One of the 184 days is selected at random.

(a) Find the probability that it has a daily mean total cloud cover of 6 or greater. (1)

Magali is investigating whether the daily mean total cloud cover can be modelled using a binomial distribution.

She uses the random variable \(X\) to denote the daily mean total cloud cover and believes that \(X \sim \mathrm{B}(8,\ 0.76)\)

Using Magali’s model,

(b)

(i) find \(\mathrm{P}(X \geqslant 6)\) (2)

(ii) find, to 1 decimal place, the expected number of days in a sample of 184 days with a daily mean total cloud cover of 7 (2)

There were 28 days that had a daily mean total cloud cover of 8
For these 28 days the daily mean total cloud cover for the following day is shown in the table below.

Daily mean total cloud cover (oktas)	0	1	2	3	4	5	6	7	8
Frequency (number of days)	0	0	1	1	2	1	5	9	9

(d) Find the proportion of these days when the daily mean total cloud cover was 6 or greater. (1)

(e) Comment on Magali’s model in light of your answer to part (d). (2)

Mark scheme (a)

Scheme	Marks	AO
\(\dfrac{132}{184} = 0.71739\ldots\) awrt 0.717	B1	1.1b
	(1)

Notes

Allow fractions, decimals or percentages throughout this question.

Allow equivalent fraction, e.g. \(\frac{33}{46}\)

Mark scheme (b)

Scheme	Marks	AO
(i) \(\mathrm{P}(X \geqslant 6) = 1 - \mathrm{P}(X \leqslant 5)\) or \(\mathrm{P}([X =]6) + \mathrm{P}([X =]7) + \mathrm{P}([X =]8)\)	M1	3.4
\(= 1 - 0.296722\ldots\) awrt 0.703	A1	1.1b
	(2)
(ii) \(184 \times \mathrm{P}(X = 7)\) \([= 184 \times 0.2811\ldots]\)	M1	1.1b
\(= 51.7385\ldots\) awrt 51.7	A1	1.1b
	(2)

Notes

(b)(i) M1: for writing or using \(1 - \mathrm{P}(X \leqslant 5)\) or \(\mathrm{P}(X = 6) + \mathrm{P}(X = 7) + \mathrm{P}(X = 8)\)

A1: awrt 0.703 (correct answer scores 2 out of 2)

(b)(ii) M1: for \(184 \times \mathrm{P}(X = 7)\) o.e. e.g., \(184 \times [\mathrm{P}(X \leqslant 7) - \mathrm{P}(X \leqslant 6)]\)

A1: awrt 51.7

Mark scheme (c)

Scheme	Marks	AO
Part (a) and part (b)(i) are similar and the expected number of 7s (51.7 or 0.281) matches with the number of 7s found in the data set (52 or 0.283) so Magali’s model is supported.	B1ft	3.5a
	(1)

Notes

B1ft: comparing ‘0.717’ with ‘0.703’ and ’51.7 or ‘0.281’ with 52 or 0.283 and concluding that Magali’s model is supported (must be comparing prob. with prob. and days with days). Allow not supported or mixed conclusions if consistent with their f.t. answers in (a) and (b)

Scheme	Marks	AO
\(\dfrac{23}{28} = 0.82142\ldots\) awrt 0.821	B1	1.1b
	(1)

Mark scheme (e)

Scheme	Marks	AO
Any one of… Part (d)/‘0.821’ differs from part (a)/(b)(i)/(0.7…) there is a greater/different probability of high cloud cover/more likely to have high cloud cover if the previous day had high cloud cover independence(o.e.) does not hold	B1	2.4
…therefore Magali’s (binomial) model may not be suitable.	dB1	3.5a
	(2)
	(9 marks)

Notes

B1: Any bullet point

dB1: (dep on previous B1) for Magali’s model may not be suitable (o.e.)
Condone not accurate for not suitable

SC: part (d) is similar to part (a)/(b)(i) and a compatible conclusion (i.e. Magali’s model is supported) to score B1B1.