A2 June 2024 Q4

EdexcelCurrent spec12 marksIncludes hypothesis testingCentral Limit Theorem Geometric & Negative Binomial

4. Every morning Geethaka repeatedly rolls a fair, six-sided die until he rolls a 3 and then he stops. The random variable \(X\) represents the number of times he rolls the die each morning.

(a) Suggest a suitable model for the random variable \(X\) (1)

(b) Show that \(\mathrm{P}(X \leqslant 3) = \dfrac{91}{216}\) (2)

After 64 mornings Geethaka will calculate the mean number of times he rolled the die.

Nira wants to check Geethaka’s die to decide whether or not the probability of rolling a 3 with his die is less than \(\dfrac{1}{6}\)

Nira rolls the die repeatedly until she rolls a 3
She obtains \(x = 16\)

(d) By carrying out a suitable test, determine what Nira’s conclusion should be. You should state your hypotheses clearly and use a 5% level of significance. (4)

Mark scheme (a)

Scheme	Marks	AO
\(X \sim \mathrm{Geo}\left(\dfrac{1}{6}\right)\) accept in words: geometric distribution with \(p = \dfrac{1}{6}\)	B1	3.3
	(1)

Notes

B1 for both “geometric” or “Geo” and correct parameter of \(\dfrac{1}{6}\). Probability must be seen in (a)

Mark scheme (b)

Scheme	Marks	AO
\(\mathrm{P}(X \leqslant 3) = 1 - \mathrm{P}(X \gt 3)\) or \(\mathrm{P}(X = 1) + \mathrm{P}(X = 2) + \mathrm{P}(X = 3)\)	M1	1.1b
\(= 1 - \left(\dfrac{5}{6}\right)^3\) or \(\dfrac{1}{6} + \dfrac{5}{6} \times \dfrac{1}{6} + \left(\dfrac{5}{6}\right)^2 \times \dfrac{1}{6} = \dfrac{91}{216}\) (*)	A1cso	1.1b
	(2)

Notes

M1 for a correct method, may be implied by a correct expression

A1cso* for a correct solution leading to printed answer with no incorrect working seen

Mark scheme (c)

Scheme	Marks	AO
\(\mathrm{E}(X) = 6 \qquad \mathrm{Var}(X) = \dfrac{\frac{5}{6}}{\left(\frac{1}{6}\right)^2}\ [= 30]\)	M1 A1	3.4 1.1b
\(\overline{X} \approx\ \sim \mathrm{N}\left(6, \left(\sqrt{\dfrac{30}{64}}\right)^2\right)\)	M1 A1	3.3 1.1b
\(\mathrm{P}(5.6 \lt \overline{X} \lt 7.2) = 0.68064\ldots\)	A1	1.1b
	(5)

Notes

1^st M1 for correct use of formula for \(\mathrm{E}(X)\) or \(\mathrm{Var}(X)\)

1^st A1 for both correct must see value of 6 for \(\mathrm{E}(X)\) and at least correct formula used for \(\mathrm{Var}(X)\) the value of 30 may be implied by 2^nd A1

2^nd M1 for use of CLT to get a normal distribution with correct mean and variance \(\neq\) their 30
Condone use of \(X\) (or their letter) instead of \(\overline{X}\) or seeing \(\mathrm{N}\left(6, \sqrt{\dfrac{30}{64}}\right)\)

2^nd A1 for a correct normal distribution stated or used. May be implied by correct answer.

3^rd A1 for awrt 0.681 but allow 0.68 or 0.680 if correct distribution is seen

Mark scheme (d)

Scheme	Marks	AO
\(\mathrm{H}_0: p = \dfrac{1}{6} \qquad \mathrm{H}_1: p \lt \dfrac{1}{6}\)	B1	2.5
Probability approach: \(\mathrm{P}(X \geqslant 16)\) CR approach: \(\left(\dfrac{5}{6}\right)^{c - 1} \lt 0.05\) or \(\left(\dfrac{5}{6}\right)^d \lt 0.05\) with use of logs	M1	3.4
Probability approach: \(= \mathrm{P}(X \gt 15) = \left(\dfrac{5}{6}\right)^{15} = 0.0649\ldots\) CR approach: CR: \(X \geqslant 18\) or \(X \gt 17\)	A1	1.1b
(Not significant) insufficient evidence that probability is \(\lt \dfrac{1}{6}\) or (Not significant) insufficient evidence that dice is biased	A1	2.2b
	(4)
	(12 marks)

Notes

B1 for both hypotheses correct in terms of \(p\)

M1 for sight or use of \(\mathrm{P}(X \geqslant 16)\) (may be implied)
or correct expression with logs (use of logs may be implied by 17.43…)

1^st A1 for 0.065 or better but accept 0.06 if a correct expression is seen, or correct CR

2^nd A1 for a correct conclusion mentioning probability or biased
Condone ‘evidence suggests probability is equal to \(\dfrac{1}{6}\)’
NB: Must have correct probability or correct CR to gain final A1.
NB: Send responses comparing 0.935 with 0.95 to review