Question
There are 400 students at a school in a certain country. Each student was asked whether they preferred swimming, cycling or running and the results are given in the following table.
Swimming Cycling Running
Female 104 50 66
Male 31 57 92
A student is chosen at random.
(a) (i) Find the probability that the student prefers swimming. [1]
(ii) Determine whether the events ‘the student is male’ and ‘the student prefers swimming’ are independent, justifying your answer. [2]
On average at all the schools in this country 30% of the students do not like any sports.
(b) (i) 10 of the students from this country are chosen at random.
Find the probability that at least 3 of these students do not like any sports. [3]
(ii) 90 students from this country are now chosen at random.
Use an approximation to find the probability that fewer than 32 of them do not like any sports. [5]
Additional Page
If you use the following lined page to complete the answer(s) to any question(s), the question number(s) must be clearly shown.
Mark scheme
7(a)(i) 104 + 31 135 27 , 0∙3375 B1 Evaluated, exact value.
400 = 400 , 80
1
7(a)(ii) Method 1
180 135 31 M1 Their P(M) × their P(S) seen, accept unsimplified.
P(M ) = , 0∙45 P ( S ) = , 0∙3375 P ( M ∩ S ) = , 0∙0775
400 400 400
180 135 243 31
× = , 0.151875 ≠ so NOT independent
400 400 1600 400
A1 P(M), P(S) and P(M ∩ S) notation seen, numerical
comparison and correct conclusion, WWW.
Method 2
P(M ∩ S ) =
31
P(S ) =
135
P(M ) =
180 M1 their P ( M ∩ S )
[P ( M |S ) =] (oe) seen, accept unsimplified.
400 400 400 their P ( S )
31
31 180
P ( M |S ) = 400 = , 0.2296… ≠ so NOT independent
135 135 400
400
A1 P(M), P(S) and P(M ∩ S) notation seen, numerical
comparison and correct conclusion, WWW.
2
7(b)(i) Method 1 [1 – P(0,1,2)]
= 1 – (10C0 0∙30 0∙710 + 10C1 0∙31 0∙79 M1 10
Cx px (1 – p)10-x for 0 < × < 10, 0 < p < 1, any p.
+ 10C2 0∙32 0∙78)
= 1 – (0∙028248 + 0∙121061 + 0∙233474) A1 Correct expression, accept unsimplified, condone omission of
final bracket, condone recovery from poor notation.
= 0∙617 A1 Accept 0∙61715 ⩽ p ⩽ 0∙61722, WWW.
Method 2 [P(3,4,5,6,7,8,9,10) =]
10
C3 0∙33 0∙77 + 10C4 0∙34 0∙76 + 10C5 0∙35 0∙75 M1 10
Cx px (1 – p)10–x for 0 < × < 10, 0 < p < 1, any p.
+ 10C6 0∙36 0∙74 + 10C7 0∙37 0∙73 + 10C8 0∙38 0∙72
+ 10C9 0∙39 0∙71 + 10C10 0∙310 0∙70 A1 Correct unsimplified expression.
= 0∙617 A1 Accept 0∙61715 ⩽ p ⩽ 0∙61722, WWW.
3
7(b)(ii) [p = 0∙3] B1 Correct mean and variance, allow unsimplified.
Mean = 0∙3 × 90 = 27; Condone σ = 4∙347 evaluated.
variance = 0∙3 × 90 × 0∙7 = 18∙9
31.5 − 27 M1 Substituting their μ and σ (not σ2, √σ) into the ±standardising
P(X < 32) = P z < formula with a numerical value for ‘31∙5’.
18.9
M1 Using either 31∙5 or 32∙5 within a ±standardising formula
with numerical values for their μ and σ (condone σ2, √σ).
= Φ(1.035) M1 Appropriate area Φ, from standardisation formula P(z<…) in
final solution, must be probability.
= 0∙850 A1 Allow 0∙8495 < p ⩽ 0∙85(0), final answer WWW.
5