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Date November 2016 Marks available 2 Reference code 16N.2.sl.TZ0.4
Level SL only Paper 2 Time zone TZ0
Command term Calculate Question number 4 Adapted from N/A

Question

A manufacturer produces 1500 boxes of breakfast cereal every day.

The weights of these boxes are normally distributed with a mean of 502 grams and a standard deviation of 2 grams.

All boxes of cereal with a weight between 497.5 grams and 505 grams are sold. The manufacturer’s income from the sale of each box of cereal is $2.00.

The manufacturer recycles any box of cereal with a weight not between 497.5 grams and 505 grams. The manufacturer’s recycling cost is $0.16 per box.

A different manufacturer produces boxes of cereal with weights that are normally distributed with a mean of 350 grams and a standard deviation of 1.8 grams.

This manufacturer sells all boxes of cereal that are above a minimum weight, \(w\).

They sell 97% of the cereal boxes produced.

Draw a diagram that shows this information.

[2]
a.

(i)     Find the probability that a box of cereal, chosen at random, is sold.

(ii)     Calculate the manufacturer’s expected daily income from these sales.

[4]
b.

Calculate the manufacturer’s expected daily recycling cost.

[2]
c.

Calculate the value of \(w\).

[3]
d.

Markscheme

N16/5/MATSD/SP2/ENG/TZ0/04.a/M

(A1)(A1)

 

Notes:     Award (A1) for bell shape with mean of 502.

Award (A1) for an indication of standard deviation eg 500 and 504.

 

[2 marks]

a.

(i)     \(0.921{\text{ }}(0.920968 \ldots ,{\text{ }}92.0968 \ldots \% )\)     (G2)

 

Note:     Award (M1) for a diagram showing the correct shaded region.

 

(ii)     \(1500 \times 2 \times 0.920968 \ldots \)     (M1)

\( = {\text{ }}(\$ ){\text{ }}2760{\text{ }}(2762.90 \ldots )\)    (A1)(ft)(G2)

 

Note:     Follow through from their answer to part (b)(i).

 

[4 marks]

b.

\(1500 \times 0.16 \times 0.079031 \ldots \)    (M1)

 

Notes:     Award (A1) for \(1500 \times 0.16 \times {\text{ their }}(1 - 0.920968 \ldots )\).

 

OR

\((1500 - 1381.45) \times 0.16\)    (M1)

 

Notes:     Award (M1) for \((1500 - {\text{their }}1381.45) \times 0.16\).

 

\( = (\$ )19.0{\text{ (}}18.9676 \ldots )\)    (A1)(ft)(G2)

[2 marks]

c.

\(347{\text{ }}({\text{grams}}){\text{ }}(346.614 \ldots )\)    (G3)

 

Notes:     Award (G2) for an answer that rounds to 346.

Award (G1) for \(353.385 \ldots \) seen without working (for finding the top 3%).

 

[3 marks]

d.

Examiners report

[N/A]
a.
[N/A]
b.
[N/A]
c.
[N/A]
d.

Syllabus sections

Topic 4 - Statistical applications » 4.1 » The normal distribution.
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