Mathematics · Oct / Nov 2025 · Paper 61 · Question 7

P(48.45

<

W

<

54.4)

=

P

Z

<

−

P

Z

<

1.7

1.7

[

]

−

= 0.9104

Cal: 0.91044…

awrt 0.910

0.9772 0.0668

(

)

(

)

P 53.55

<

W

<

54.4

=

"0.9772"

−

1 "0.0668"

−

[= 0.044]

Calc:0.044057….

]"0.044"

[

P(

W

>

53.55| 48.45

<

W

<

54.4)

=

"0.9104"

= 0.04833 awrt 0.0483/0.0484

−

μ

64.58

= −

0.5244

(Calc: – 0.524400…)

69.46

−

μ

=

(Calc: 1.281551)

1.2816

awrt 2.7

awrt 66

4.88 1.806

=

=

2.702...

μ=

65.996...

Notes

Standardising with 54.4, 51 and 1.7 (implied by ±2)

(a)

M1

or standardising with 48.45, 51 and 1.7 (implied by ±1.5)

using 1.72 is M0

Standardising with 54.4, 51 and 1.7 and 48.45, 51 and 1.7

and attempt to find correct probability using p – q oe (where 0.5

M1

< p < 1 and 0 < q < 0.5) or p – (1 – r) oe (where 0.5 < p < 1 and 0.5 <

r < 1)

If 1st M1 scored, then may be implied by correct answer

awrt 0.910 condone 0.91 correct answer with no working

A1

scores 0

for a valid attempt to find P(53.55

<

W

<

54.4)

(b)

M1

(

)

e.g.

"0.9772"

−

1 "0.0668"

−

May be implied by awrt 0.044/0.0441

p

for a conditional probability in the form "0.9104"

where 0 < p <

M1

“0.9104”

p×

"0.9104"

assuming independence is M0 e.g.

"0.9104"

A1

awrt 0.0483/0.0484 No fractions

for standardising 64.58 and setting equal to z value where 0.5 <

(c)

M1

|z| < 0.6 or for standardising 69.46 and setting equal to z value

where 1 < |z| < 1.5

for standardising 64.58 and setting equal to z value where – 0.6 <

M1

z < – 0.5 oe with compatible signs and for standardising 69.46

and setting equal to z value where 1 < z < 1.5 oe with compatible

signs

Dep on 1st M1 for attempting to solve their two linear equations

(using substitution or elimination) leading to an equation in one

dM1

variable

Implied by either

=

or

awrt 2.7

μ=

awrt 66

both

=

and

and both 0.5244 (or better) and

awrt 2.7

μ=

awrt 66

A1

1.2816 (or better) seen

1

1

1

1

1

1

2 *