CAIE S2 2016 June — Question 7 10 marks

Exam BoardCAIE
ModuleS2 (Statistics 2)
Year2016
SessionJune
Marks10
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicLinear combinations of normal random variables
TypeAll components random including container
DifficultyStandard +0.3 This is a straightforward application of linear combinations of normal distributions with clear setup. Part (i) requires finding the distribution of box + 20 bags (sum of independent normals), then a single probability calculation. Part (ii) involves finding the distribution of the difference between two boxes. Both parts are standard textbook exercises requiring only routine application of formulas for sums/differences of normals with no problem-solving insight or tricky manipulation.
Spec5.04a Linear combinations: E(aX+bY), Var(aX+bY)5.04b Linear combinations: of normal distributions

Bags of sugar are packed in boxes, each box containing 20 bags. The masses of the boxes, when empty, are normally distributed with mean 0.4 kg and standard deviation 0.01 kg. The masses of the bags are normally distributed with mean 1.02 kg and standard deviation 0.03 kg.
  1. Find the probability that the total mass of a full box of 20 bags is less than 20.6 kg. [5]
  2. Two full boxes are chosen at random. Find the probability that they differ in mass by less than 0.02 kg. [5]

Question 7:

(ii) ---
7 (i)
AnswerMarks
(ii)E(T) =20.8
2 2
Var(T)= 20 × 0.03 + 0.01 (= 0.0181)
20.6−20.8(= –1.487)
"0.0181"
1 – Φ(“1.487”)
= 0.0684 to 0.686
E(D) = 0
Var(D) = 2 × 0.0181(= 0.0362)
0.02−0 ( = 0.105)
"0.0362)
Φ(“0.105”) = 0.5418 or 1-Φ(0.015)
=0.4582
Φ(“0.105”) – (1 – Φ(“0.105”)
(= 0.5418 – 0.4582)
AnswerMarks
= 0.0836/0.0837B1
B1
M1
M1
A1 [5]
B1
M1
A1
M1
AnswerMarks
A1 [5]2 2
or √(20 × 0.03 + 0.01 ) = 0.135 (3sf)
For standardising (σ must come from
combination)
Area consistent with their working
Any answer within range
Both (Seen or implied)
Allow without √
Allow to 3sf
or 1 – 2(1 – Φ(“0.105”))
(= 1 – 2 × 0.4582)
Question 7:
--- 7 (i)
(ii) ---
7 (i)
(ii) | E(T) =20.8
2 2
Var(T)= 20 × 0.03 + 0.01 (= 0.0181)
20.6−20.8(= –1.487)
"0.0181"
1 – Φ(“1.487”)
= 0.0684 to 0.686
E(D) = 0
Var(D) = 2 × 0.0181(= 0.0362)
0.02−0 ( = 0.105)
"0.0362)
Φ(“0.105”) = 0.5418 or 1-Φ(0.015)
=0.4582
Φ(“0.105”) – (1 – Φ(“0.105”)
(= 0.5418 – 0.4582)
= 0.0836/0.0837 | B1
B1
M1
M1
A1 [5]
B1
M1
A1
M1
A1 [5] | 2 2
or √(20 × 0.03 + 0.01 ) = 0.135 (3sf)
For standardising (σ must come from
combination)
Area consistent with their working
Any answer within range
Both (Seen or implied)
Allow without √
Allow to 3sf
or 1 – 2(1 – Φ(“0.105”))
(= 1 – 2 × 0.4582)
Bags of sugar are packed in boxes, each box containing 20 bags. The masses of the boxes, when empty, are normally distributed with mean 0.4 kg and standard deviation 0.01 kg. The masses of the bags are normally distributed with mean 1.02 kg and standard deviation 0.03 kg.

\begin{enumerate}[label=(\roman*)]
\item Find the probability that the total mass of a full box of 20 bags is less than 20.6 kg. [5]

\item Two full boxes are chosen at random. Find the probability that they differ in mass by less than 0.02 kg. [5]
\end{enumerate}

\hfill \mbox{\textit{CAIE S2 2016 Q7 [10]}}