<p>It is required to select a random sample of $30$ pupils from a school with $853$ pupils. A student suggests the following method. ‘Give each pupil sequentially a three-digit number from $001$ to $853$. Use a calculator to generate random three-digit numbers from $0.000$ to $0.999$ inclusive, multiply the answer by $853$, add $1$ and round off to the nearest whole number. Select the corresponding pupil, and repeat as necessary’.</p>
<ol class="i-list"><li><div class="list-item-content"><span class="list-item-text">Determine which pupil would be selected for each of the following calculator outputs: $0.103$,$0.104$, $0.105$, $0.106$, $0.107$.</span></div></li><li><div class="list-item-content"><span class="list-item-text">Use your answers to part (i) to show that this method is biased or non-random, and suggest an improvement.</span></div></li></ol>

It is required to select a random sample of $30$ pupils from a school with $853$ pupils. A student suggests the following method. ‘Give each pupil sequentially a three-digit number from $001$ to $853$. Use a calculator to generate random three-digit numbers from $0.000$ to $0.999$ inclusive, multiply the answer by $853$, add $1$ and round off to the nearest whole number. Select the corresponding pupil, and repeat as necessary’.

Determine which pupil would be selected for each of the following calculator outputs: $0.103$,$0.104$, $0.105$, $0.106$, $0.107$.

Use your answers to part (i) to show that this method is biased or non-random, and suggest an improvement.

Central Limit Theorem

Sampling

H1 Math

2013 June AS/A Level GCE P1 Q1

Let $r$ be the calculator output and $n = 853r + 1$ (rounded to nearest whole number).

Since $0.000 \leq r \leq 0.999$, we get $0 \leq 853r \leq 852.147$, so $1 \leq 853r + 1 \leq 853.147$.

Rounding to nearest whole number: $1 \leq n \leq 853$.

$r = 0.103$: $n = 853(0.103) + 1 = 88.859 \approx 89$. Pupil 89.

$r = 0.104$: $n = 853(0.104) + 1 = 89.712 \approx 90$. Pupil 90.

$r = 0.105$: $n = 853(0.105) + 1 = 90.565 \approx 91$. Pupil 91.

$r = 0.106$: $n = 853(0.106) + 1 = 91.418 \approx 91$. Pupil 91.

$r = 0.107$: $n = 853(0.107) + 1 = 92.271 \approx 92$. Pupil 92.

Since student 91 has a higher chance of being selected compared to the rest in part (i) (two different values of $r$ give pupil 91), this suggests that this method is biased and non-random.

(1) Generate random three-digit numbers from 0.001 to 0.853 inclusive and multiply the answer by 1000. Then, select the corresponding pupil.

(2) Generate random three-digit numbers from 0.001 to 0.999 and multiply the answer by 1000. If a number is larger than 853, ignore this number and generate another random three-digit number. Then, select the corresponding pupil.

2013 June AS/A Level GCE P1 Q1

2013 June AS/A Level GCE P1 Q1

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