Home / IB Mathematics AHL 4.14 Linear transformation AI HL Paper 1- Exam Style Questions

IB Mathematics AHL 4.14 Linear transformation AI HL Paper 1- Exam Style Questions

IB Mathematics AHL 4.14 Linear transformation AI HL Paper 1- Exam Style Questions- New Syllabus

Question

A manufacturer of chocolates produces them in individual packets, claiming to have an average of 85 chocolates per packet.

Talha bought 30 of these packets in order to check the manufacturer’s claim.

Given that the number of individual chocolates is \( x \), Talha found that, from his 30 packets,

\( \sum x = 2506 \) and \( \sum x^2 = 209738 \).

(a) Find an unbiased estimate for the mean number (\( \mu \)) of chocolates per packet [1]

(b) Use the formula \( s_{n-1}^2 = \frac{\sum x^2 – \frac{(\sum x)^2}{n}}{n-1} \) to determine an unbiased estimate for the variance of the number of chocolates per packet [2]

(c) Find a 95% confidence interval for \( \mu \). You may assume that all conditions for a confidence interval have been met [2]

(d) Suggest, with justification, a valid conclusion that Talha could make [1]

▶️ Answer/Explanation
Markscheme

(a)
83.5

Sample mean: \( \bar{x} = \frac{\sum x}{n} \)

\( \sum x = 2506 \), \( n = 30 \)

\( \bar{x} = \frac{2506}{30} \approx 83.5333 \)

Round to 1 decimal place: 83.5

Result: 83.5 [1]

(b)
13.9

Sample variance: \( s_{n-1}^2 = \frac{\sum x^2 – \frac{(\sum x)^2}{n}}{n-1} \)

\( \sum x^2 = 209738 \), \( \sum x = 2506 \), \( n = 30 \)

\( (\sum x)^2 = 2506^2 = 6280036 \)

\( \frac{(\sum x)^2}{n} = \frac{6280036}{30} \approx 209334.5333 \)

\( \sum x^2 – \frac{(\sum x)^2}{n} = 209738 – 209334.5333 \approx 403.4667 \)

\( s_{n-1}^2 = \frac{403.4667}{29} \approx 13.9126 \)

Round to 1 decimal place: 13.9

Result: 13.9 [2]

(c)
(82.1, 84.9)

95% confidence interval: \( \bar{x} \pm t \times \frac{s}{\sqrt{n}} \)

\( \bar{x} = 83.5333 \), \( s = \sqrt{13.9126} \approx 3.7298 \)

\( \frac{s}{\sqrt{n}} = \frac{3.7298}{\sqrt{30}} \approx 0.6810 \)

\( t \approx 2.045 \) (for 95% confidence, 29 df)

Margin of error: \( 2.045 \times 0.6810 \approx 1.3926 \)

Lower bound: \( 83.5333 – 1.3926 \approx 82.1407 \)

Upper bound: \( 83.5333 + 1.3926 \approx 84.9259 \)

Round to 1 decimal place: (82.1, 84.9)

Result: (82.1, 84.9) [2]

(d)
85 is outside the confidence interval; manufacturer’s claim is incorrect

Confidence interval: (82.1, 84.9)

Claim: \( \mu = 85 \)

85 is not within (82.1, 84.9)

Result: Manufacturer’s claim is incorrect [1]

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