Mathematics · May / June 2022 · Paper 61 · Question 6

V

Z

2.5

= 1 – 0.975

= 0.0250

0.025 or awrt 0.0250 A1

Expected number = 150 "0.025"

×

M1

= 3.75

awrt 3.75 A1

][P(

v

<

V

<

104.9)]

[

P(

>

|

<

104.9)

=

V

v V

= 0.2801

M1

P(

<

104.9)

V

P(

104.9)

(

1 "0.025"

)

P

(

)

v

<

V

<

=

−

−

V

<

v

M1

(

)

= (

)

(

)

P V

<

1 "0.025"

−

−

1 "0.025

−

"

×

0.

280

1

[=0.7019] oe

v

v −

100

=

0.53

M1A1

2.5

v = 101.325 awrt 101.32 or awrt 101.33 A1

Notes

(a)

Standardising with 104.9, 100 and 2.5 (allow ±) implied by 1.96 seen

M1

For use of 1

with 0.9 < p < 1 condone answer of 0.0249 for this mark

M1

−

p

A1

Allow 0.025 or awrt 0.0250 (NB calculator answer is 0.02499….) (answer only scores M1M1A1)

(b)

For 150 "their part ( )"

×

M1

a

awrt 3.75 isw after answer of 3.75 seen. If 3.75 not seen, allow 4 if the method mark is awarded.

A1

(c)

M1

For writing or using a ratio of probabilities with denominator P(

V <

104.9)

oe and equating to 0.2801

p

= 0.2801 implied by awrt 0.273

P(

V <

104.9)

x

×

P(

V

<

104.9)

Use of independence is M0 e.g.

=

0.2801

P(

V

<

104.9)

(

)

(

For writing or using

P(

V

>∩

v

V

<

104.9)

=

P(

V

<

104.9)

−

P(

V

<

v

)[

=

1 "0.025"

−

−

P

V

M1

P(

V

<

v

)

For first two M marks

= 1 – 0.2801 scores M1M1, then follow scheme.

ALT

P(

V

<

104.9)

(dep M1) Dependent on previous M1 for rearranging to find (

)

P V

<

v

(

)1

dM1

Allow equivalent oe

(1 "their( )") 1

−

a

−

0.280

NB: [P(V < v) =] awrt 0.702 implies M1M1M1 or [P(V > v) =] awrt 0.298 implies M1M1M1

v −

100

Standardising with 100, 2.5 and equating to a zvalue,

=z 0.4 < |z| < 0.6

2.5

M1

v −

100

Watch out for

=probability which is M0

2.5

Correct equation with compatible signs

A1

A1

awrt 101.33 (allow awrt 101.32 from use of calculator)

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