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2018
P2 Q10
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18 P2 Q10
2018 H2 Mathematics Paper 2 Question 10
The Normal Distribution
Answers
(i)
(ii)
0.605
{0.605}
0.605
(iii)
0.773
{0.773}
0.773
(iv)
0.126
{0.126}
0.126
(v)
137
{137}
137
(vi)
0.163
{0.163}
0.163
Full solutions
(i)
(ii)
Let
X
{X}
X
denote the mass of a randomly selected bulb
X
∼
N
(
50
,
2.25
)
X \sim \textrm{N}( 50, 2.25 )
X
∼
N
(
50
,
2.25
)
P
(
X
<
50.4
)
=
0.605
(3 sf)
■
\begin{align*} & \textrm{P}(X < 50.4) \\ &= 0.605 \textrm{ (3 sf)} \; \blacksquare \end{align*}
P
(
X
<
50.4
)
=
0.605
(3 sf)
■
(iii)
Y
1
+
Y
2
+
…
+
Y
4
∼
N
(
4
×
75
,
4
×
2
2
)
Y
1
+
Y
2
+
…
+
Y
4
∼
N
(
300
,
16
)
\begin{align*} Y_1 + Y_2 + \ldots + Y_4 &\sim \textrm{N}(4 \times 75, 4 \times 2^2) \\ Y_1 + Y_2 + \ldots + Y_{4} &\sim \textrm{N}( 300, 16 ) \end{align*}
Y
1
+
Y
2
+
…
+
Y
4
Y
1
+
Y
2
+
…
+
Y
4
∼
N
(
4
×
75
,
4
×
2
2
)
∼
N
(
300
,
16
)
P
(
Y
1
+
…
+
Y
4
>
297
)
=
0.773
(3 sf)
■
\begin{align*} & \textrm{P}(Y_1 + \ldots + Y_4 > 297) \\ &= 0.773 \textrm{ (3 sf)} \; \blacksquare \end{align*}
P
(
Y
1
+
…
+
Y
4
>
297
)
=
0.773
(3 sf)
■
(iv)
X
+
Y
∼
N
(
50
+
75
,
2.25
+
2
2
)
X
+
Y
∼
N
(
125
,
6.25
)
\begin{align*} X + Y &\sim \textrm{N}(50+75,2.25+2^2) \\ X + Y &\sim \textrm{N}( 125, 6.25 ) \end{align*}
X
+
Y
X
+
Y
∼
N
(
50
+
75
,
2.25
+
2
2
)
∼
N
(
125
,
6.25
)
P
(
124.9
<
X
+
Y
<
125.7
)
=
0.126
(3 sf)
■
\begin{align*} & \textrm{P}(124.9 < X + Y < 125.7) \\ & = 0.126 \textrm{ (3 sf)} \; \blacksquare \end{align*}
P
(
124.9
<
X
+
Y
<
125.7
)
=
0.126
(3 sf)
■
(v)
Let
W
=
0.3
X
{W = 0.3X}
W
=
0.3
X
denote the mass of padding
W
∼
N
(
0.3
×
50
,
0.
3
2
×
2.25
)
W
∼
N
(
15
,
0.2025
)
\begin{align*} W &\sim \textrm{N}(0.3 \times 50, 0.3^2 \times 2.25) \\ W &\sim \textrm{N}(15, 0.2025) \\ \end{align*}
W
W
∼
N
(
0.3
×
50
,
0.
3
2
×
2.25
)
∼
N
(
15
,
0.2025
)
Let
T
=
X
+
Y
+
W
{T=X+Y+W}
T
=
X
+
Y
+
W
denote the mass of a randomly chosen box
T
∼
N
(
50
+
75
+
15
,
2.25
+
2
2
+
0.2025
)
T
∼
N
(
140
,
6.4525
)
\begin{align*} T &\sim \textrm{N}(50+75+15, 2.25+2^2+0.2025) \\ T &\sim \textrm{N}( 140, 6.4525 ) \end{align*}
T
T
∼
N
(
50
+
75
+
15
,
2.25
+
2
2
+
0.2025
)
∼
N
(
140
,
6.4525
)
P
(
T
>
k
)
=
0.9
k
=
137
(3 sf)
■
\begin{gather*} \textrm{P}(T > k) = 0.9 \\ k = 137 \textrm{ (3 sf)} \; \blacksquare \end{gather*}
P
(
T
>
k
)
=
0.9
k
=
137
(3 sf)
■
(vi)
T
1
+
T
2
+
…
+
T
4
∼
N
(
4
×
140
,
4
×
6.4525
)
T
1
+
T
2
+
…
+
T
4
∼
N
(
560
,
25.81
)
\begin{align*} T_1 + T_2 + \ldots + T_4 &\sim \textrm{N}(4\times 140, 4\times 6.4525) \\ T_1 + T_2 + \ldots + T_{4} &\sim \textrm{N}( 560, 25.81 ) \end{align*}
T
1
+
T
2
+
…
+
T
4
T
1
+
T
2
+
…
+
T
4
∼
N
(
4
×
140
,
4
×
6.4525
)
∼
N
(
560
,
25.81
)
P
(
T
1
+
…
+
T
4
>
565
)
=
0.163
(3 sf)
■
\begin{align*} & \textrm{P}(T_1 + \ldots + T_4 > 565) \\ & = 0.163 \textrm{ (3 sf)} \; \blacksquare \end{align*}
P
(
T
1
+
…
+
T
4
>
565
)
=
0.163
(3 sf)
■
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