Mathematics · May / June 2024 · Paper 61 · Question 5
YMA01
As printed in the paper
The question as text
5. A competition consists of two rounds. The time, in minutes, taken by adults to complete round one is modelled by a normal distribution with mean 15 minutes and standard deviation 2 minutes.
Read automatically from the paper; the printed page above is exact.
Show the answer and mark scheme
5(a)[2]
P
X <
18
= P
<±
= ±
1.5
Z
2
= 0.9332
awrt 0.933
x −
15
5(b)[3]
=
0.2533
2
x = 15.506…
awrt 15.5 A1
(
)
5(c)[5]
P
>
−
10
=
0. 7
9
5
T
µ
(
)
μ
±
10
−
μ
10
[
]
= ±
1.96
=
=
5.10...
1.96
μ
−−
5
μ
(
)
[
]
P
T
>
µ
−
5
=
P
Z
>
= −
0.98
[= 0.836…]
"5. 0..."
1
(
)
= "0.836..."
P
T
>
µ
– 5
T
>
µ
−
10
"0.975"
=
0.8579...
awrt 0.858
Notes
(a)
for standardising correctly May be implied by ±1.5
M1
A1
awrt 0.933 (Do not ISW)
(b)
for correct standardisation = to a z value such that 0.25
„
z
„
0.26
M1
for use of awrt ±0.2533
B1
A1
awrt 15.5
for the correct probability of 0.975 – may be seen as the denominator of the
(c)
M1
1.96
fraction. May be implied by use of
z =
or better
μ
+
10
−
μ
μ
−
10
−
μ
For
=
1.96
or
= −
1.96
, leading to a value for
M1
May be implied by ±0.98
for a correct method to find (
P
5
)
T
>
µ
−
using their value for May be implied
M1
by –0.98 If (
)
P
<
+
5
T
µ
is calculated then this may be implied by 0.98
p
for 0.975
where 0.5 < p < 0.975 (must be a probability not their z value) If the
M1
denominator is incorrect only follow through their (
)
P
T
>
µ
−
10
if clearly
labelled and > 0.95
awrt 0.858
A1
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