CAIE S1 2018 June — Question 3 6 marks

Exam BoardCAIE
ModuleS1 (Statistics 1)
Year2018
SessionJune
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicNormal Distribution
TypeExpected frequency with unknown parameter
DifficultyModerate -0.3 Part (i) requires inverse normal calculation using tables/calculator to find μ from a given probability (routine but requires correct setup). Part (ii) is straightforward application of normal probability and expected frequency. Both parts are standard textbook exercises with no novel problem-solving required, making this slightly easier than average.
Spec2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation

3
  1. The volume of soup in Super Soup cartons has a normal distribution with mean \(\mu\) millilitres and standard deviation 9 millilitres. Tests have shown that \(10 \%\) of cartons contain less than 440 millilitres of soup. Find the value of \(\mu\).
  2. A food retailer orders 150 Super Soup cartons. Calculate the number of these cartons for which you would expect the volume of soup to be more than 1.8 standard deviations above the mean.

Question 3(i):
AnswerMarks Guidance
\(z = -1.282\)B1 \(\pm 1.282\) seen
\(-1.282 = \frac{440 - \mu}{9}\)M1 \(\pm\)Standardisation equation with 440, 9 and \(\mu\), equated to a \(z\)-value (not \(1-z\)-value or probability)
\(\mu = 452\)A1 Correct answer rounding to 452, not dependent on B1
Total: 3 marks
Question 3(ii):
AnswerMarks Guidance
\(P(z > 1.8) = 1 - 0.9641 = 0.0359\)B1
Number \(= 0.0359 \times 150 = 5.385\)M1 \(p \times 150\), \(0 < p < 1\)
(Number of cartons \(=\) ) 5A1FT Accept either 5 or 6, not indicated as an approximation; FT *their* \(p \times 150\), answer as an integer
Total: 3 marks
## Question 3(i):

$z = -1.282$ | B1 | $\pm 1.282$ seen

$-1.282 = \frac{440 - \mu}{9}$ | M1 | $\pm$Standardisation equation with 440, 9 and $\mu$, equated to a $z$-value (not $1-z$-value or probability)

$\mu = 452$ | A1 | Correct answer rounding to 452, not dependent on B1

**Total: 3 marks**

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## Question 3(ii):

$P(z > 1.8) = 1 - 0.9641 = 0.0359$ | B1 |

Number $= 0.0359 \times 150 = 5.385$ | M1 | $p \times 150$, $0 < p < 1$

(Number of cartons $=$ ) 5 | A1FT | Accept either 5 or 6, not indicated as an approximation; **FT** *their* $p \times 150$, answer as an integer

**Total: 3 marks**

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3 (i) The volume of soup in Super Soup cartons has a normal distribution with mean $\mu$ millilitres and standard deviation 9 millilitres. Tests have shown that $10 \%$ of cartons contain less than 440 millilitres of soup. Find the value of $\mu$.\\

(ii) A food retailer orders 150 Super Soup cartons. Calculate the number of these cartons for which you would expect the volume of soup to be more than 1.8 standard deviations above the mean.\\

\hfill \mbox{\textit{CAIE S1 2018 Q3 [6]}}