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Date May 2018 Marks available 5 Reference code 18M.2.SL.TZ1.S_9
Level Standard Level Paper Paper 2 Time zone Time zone 1
Command term Find Question number S_9 Adapted from N/A

Question

The weights, in grams, of oranges grown in an orchard, are normally distributed with a mean of 297 g. It is known that 79 % of the oranges weigh more than 289 g and 9.5 % of the oranges weigh more than 310 g.

The weights of the oranges have a standard deviation of σ.

The grocer at a local grocery store will buy the oranges whose weights exceed the 35th percentile.

The orchard packs oranges in boxes of 36.

Find the probability that an orange weighs between 289 g and 310 g.

[2]
a.

Find the standardized value for 289 g.

[2]
b.i.

Hence, find the value of σ.

[3]
b.ii.

To the nearest gram, find the minimum weight of an orange that the grocer will buy.

[3]
c.

Find the probability that the grocer buys more than half the oranges in a box selected at random.

[5]
d.

The grocer selects two boxes at random.

Find the probability that the grocer buys more than half the oranges in each box.

[2]
e.

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

correct approach indicating subtraction      (A1)

eg  0.79 − 0.095, appropriate shading in diagram

P(289 < w < 310) = 0.695 (exact), 69.5 %      A1 N2

[2 marks]

a.

METHOD 1

valid approach      (M1)

eg    1 − p, 21

−0.806421

z = −0.806      A1 N2

 

METHOD 2

(i) & (ii)

correct expression for z (seen anywhere)   (A1)

eg   289 u σ

valid approach      (M1)

eg    1 − p, 21

−0.806421

z = −0.806 (seen anywhere)      A1 N2

 

[2 marks]

b.i.

METHOD 1

attempt to standardize     (M1)

eg    σ = 289 297 z , 289 297 σ

correct substitution with their z (do not accept a probability)     A1

eg   0.806 = 289 297 σ , 289 297 0.806

9.92037

σ = 9.92      A1 N2

 

METHOD 2

(i) & (ii)

correct expression for z (seen anywhere)   (A1)

eg   289 u σ

valid approach      (M1)

eg    1 − p, 21

−0.806421

z = −0.806 (seen anywhere)      A1 N2

valid attempt to set up an equation with their z (do not accept a probability)     (M1)

eg   0.806 = 289 297 σ , 289 297 0.806

9.92037

σ = 9.92      A1 N2

[3 marks]

b.ii.

valid approach      (M1)

eg  P(W < w) = 0.35, −0.338520 (accept 0.385320), diagram showing values in a standard normal distribution

correct score at the 35th percentile      (A1)

eg  293.177

294 (g)       A1 N2

Note: If working shown, award (M1)(A1)A0 for 293.
If no working shown, award N1 for 293.177, N1 for 293.

Exception to the FT rule: If the score is incorrect, and working shown, award A1FT for correctly finding their minimum weight (by rounding up)

[3 marks]

c.

evidence of recognizing binomial (seen anywhere)     (M1)

eg   X B ( 36 , p ) , n C a × p a × q n a

correct probability (seen anywhere) (A1)

eg 0.65

EITHER

finding P(X ≤ 18) from GDC     (A1)

eg 0.045720

evidence of using complement      (M1)

eg 1−P(X ≤ 18)

0.954279

P(X > 18) = 0.954     A1  N2

OR

recognizing P(X > 18) = P(X ≥ 19)     (M1)

summing terms from 19 to 36      (A1)

eg  P(X = 19) + P(X = 20) + … + P(X = 36)

0.954279

P(X > 18) = 0.954     A1  N2

[5 marks]

d.

correct calculation      (A1)

0.954 2 , ( 2 2 ) 0.954 2 ( 1 0.954 ) 0

0.910650

0.911      A1 N2

[2 marks]

e.

Examiners report

[N/A]
a.
[N/A]
b.i.
[N/A]
b.ii.
[N/A]
c.
[N/A]
d.
[N/A]
e.

Syllabus sections

Topic 4—Statistics and probability » SL 4.6—Combined, mutually exclusive, conditional, independence, prob diagrams
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Topic 4—Statistics and probability » SL 4.8—Binomial distribution
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Topic 4—Statistics and probability

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