2021 H2 Maths Paper 2 Q11 — Worked Solution

Part (a)(i)

2 marks

Select a simple random sample of the $100$ light fittings.

Explanation: number all the light fittings produced that day, then use a random number generator (or random number tables) to choose $100$ of the numbers, and test the fittings with those numbers. This gives every light fitting an equal chance of selection, avoiding bias — for instance, avoiding always testing fittings from the same point in the production run, which might not be representative of the whole day’s output.

$\therefore$ select a simple random sample (e.g. using random numbers), to avoid bias.

Part (a)(ii)

1 mark

Using the frequency table, we find the total number of faulty fittings recorded over the $150$ days.

\Sigma fx = 0(4) + 1(19) + 2(38) + 3(41) + 4(22) + 5(16) + 6(6) + 7(4) + 8(0)

= 450

The sample mean number of faulty fittings per day is

\overline{x} = \frac{450}{150} = 3

Since each day’s sample contains $100$ fittings, we estimate $p$ by

\widehat{p} = \frac{\overline{x}}{100} = \frac{3}{100} = 0.03

To check your answer: the number of days

4 + 19 + 38 + 41 + 22 + 16 + 6 + 4 + 0 = 150

, matching the given total ✓

$\therefore$ $\widehat{p} = 0.03$ .

Part (a)(iii)

2 marks

Let $Y$ be the number of faulty light fittings found on a day, so $Y \sim B(100,\ 0.03)$ , using the estimate of $p$ from part (ii). By GC,

P(Y = 3) = P(Y \leq 3) - P(Y \leq 2)

= 0.6472 - 0.4198

= 0.2275

Over $150$ working days, the expected number of days on which exactly $3$ faulty fittings are found is

150 \times 0.2275 \approx 34.1

To check your answer:

P(Y = 3) \approx 0.2275

is close to (but slightly less than) the probability at the mean, consistent with

Y

being roughly symmetric near its mean of

100(0.03) = 3

✓

$\therefore$ the expected number of days is $34.1$ (3 s.f.).

Part (b)(i)

2 marks

Two assumptions are needed for $X$ to be well modelled by a binomial distribution:

Whether one heating element is faulty is independent of whether any other heating element tested that day is faulty.
The probability that a heating element is faulty is constant for every heating element tested on a given day.

$\therefore$ independence between elements, and a constant probability of a fault, across the elements tested each day.

Part (b)(ii)

2 marks

Assume now $X \sim B(80,\ 0.02)$ . By GC,

P(1 \leq X \leq 4) = P(X \leq 4) - P(X \leq 0)

= 0.9776 - 0.1986

= 0.7790

To check your answer:

P(X \leq 0) \approx 0.1986

is the probability of no faults, plausible for

n = 80

,

p = 0.02

(mean

\text{=}1.6

); the resulting range probability

0.779

is sensibly less than

1

✓

$\therefore$ $P(1 \leq X \leq 4) = 0.779$ (3 s.f.).

Part (b)(iii)

3 marks

Approach

“more than 3 elements found faulty” is a daily event; we first find its probability $q$ for a single day, then treat the number of days (out of 5) on which it happens as a second binomial variable.

We first find $q = P(X > 3)$ , the probability that more than $3$ elements are faulty on a given day. By GC,

P(X \leq 3) = 0.9231

q = P(X > 3) = 1 - P(X \leq 3)

= 1 - 0.9231

= 0.0769

Let $W$ be the number of days, out of $5$ , on which more than $3$ elements are found faulty, so $W \sim B(5,\ 0.0769)$ . By GC,

P(W \leq 1) = 0.9495

P(W \geq 2) = 1 - P(W \leq 1)

= 1 - 0.9495

= 0.0505

To check your answer: since

q \approx 0.0769

is small, “more than 3 faulty” is a rare daily event, so it occurring on

2

or more of

5

days should be a small probability, consistent with

0.0505

✓

$\therefore$ $P(W \geq 2) = 0.0505$ (3 s.f.).

Part (b)(iv)

2 marks

Let $T$ be the total number of faulty elements found over a $5$ -day working week. Since $T$ is the sum of $5$ independent $B(80,\ 0.02)$ variables, $T \sim B(400,\ 0.02)$ . By GC,

P(T \leq 8) = 0.593

To check your answer:

T

has mean

400(0.02) = 8

, so

P(T \leq 8)

being just under

0.6

(slightly more than half) is consistent with

8

being at the mean of a discrete distribution ✓

$\therefore$ $P(T \leq 8) = 0.593$ (3 s.f.).

2021 H2 Maths Paper 2, Question 11

Part (a)(i)

Part (a)(ii)

Part (a)(iii)

Part (b)(i)

Part (b)(ii)

Part (b)(iii)

Part (b)(iv)

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