2021 H2 Maths Paper 2 Q10 — Worked Solution

Part (a)

3 marks

Let $X$ kg be the mass of a seat, where $X \sim N\left( \mu,\ \sigma^{2} \right)$ . We are given $P(X < 2.1) = 0.80$ and $P(X < 1.95) = 0.15$ .

By GC (inverse normal),

\frac{2.1 - \mu}{\sigma} = 0.841621\ldots\quad\quad\text{and}\quad\quad\frac{1.95 - \mu}{\sigma} = - 1.036433\ldots

2.1 = \mu + 0.841621\sigma\quad\quad\text{...(1)}

1.95 = \mu - 1.036433\sigma\quad\quad\text{...(2)}

Subtracting (2) from (1),

2.1 - 1.95 = 0.841621\sigma - ( - 1.036433\sigma)

= 1.878055\sigma

\sigma = \frac{0.15}{1.878055}

= 0.0799\ \left( 3\ \text{s.f.} \right)\ \ \left( \text{shown} \right)

Substituting back into (1),

\mu = 2.1 - 0.841621(0.079870)

= 2.03\ \left( 3\ \text{s.f.} \right)

To check your answer: substituting into (2),

\mu - 1.036433\sigma \approx 2.032780 - 1.036433(0.079870) \approx 1.950

, matching the given value. ✓

$\therefore$ the mean mass of the seats is $\mu \approx 2.03$ kg, and the standard deviation is $\sigma \approx 0.0799$ kg (both 3 s.f.).

Part (b)

2 marks

Let $L$ kg be the mass of a leg, $L \sim N\left( 1.2,\ {0.02}^{2} \right)$ . Standardising,

P(L > 1.21)

= P\left( Z > \frac{1.21 - 1.2}{0.02} \right)

= P(Z > 0.5)

= 0.308538\ \left( \text{by GC} \right)

In a batch of $500$ legs, the expected number with mass more than $1.21$ kg is

500 \times 0.308538 = 154.3\ \left( \text{4 s.f.} \right)

$\therefore$ the expected number of legs with mass more than $1.21$ kg is $154$ (nearest whole number).

Part (c)

3 marks

Let $T = X + L_{1} + L_{2} + L_{3}$ be the total mass of a randomly chosen seat and $3$ randomly chosen legs, where $X \sim N\left( 2.032780,\ {0.079870}^{2} \right)$ (using the unrounded values from (a)) and each $L_{i} \sim N\left( 1.2,\ {0.02}^{2} \right)$ independently. Since a sum of independent normal random variables is normal,

E(T) = \mu + 3(1.2)

= 2.032780 + 3.6

= 5.632780

Var(T) = \sigma^{2} + 3(0.02)^{2}

= 0.006379 + 0.0012

= 0.007579

T \sim N(5.632780,\ 0.007579)

P(5.6 < T < 5.7) = 0.427\ (3\ \text{s.f., by GC)}

To check your answer:

sd(T) \approx 0.0871

kg, and the interval

\lbrack 5.6,\, 5.7\rbrack

spans roughly

\text{±}0.4

standard deviations either side of the mean

5.633

, consistent with a probability in the moderate range around

0.4

–

0.5

. ✓

$\therefore$ $P(5.6 < T < 5.7) \approx 0.427$ (3 s.f.).

Part (d)

3 marks

Approach

drilling reduces every seat’s mass by $9\%$ , so the drilled mass is the linear transformation $0.91X$ , not $X$ minus a fixed amount — scaling a normal random variable by a constant scales its mean by that constant and its variance by the square of that constant.

Since $0.91X \sim N\left( 0.91\mu,\ (0.91\sigma)^{2} \right)$ ,

E(0.91X) = 0.91(2.032780) = 1.849830

Var(0.91X) = (0.91)^{2}(0.079870)^{2} = 0.006483

Let $T'= 0.91X + L_{1} + L_{2} + L_{3}$ be the total mass of a drilled seat and $3$ legs.

E(T') = 1.849830 + 3.6

= 5.449830

Var(T') = 0.006483 + 0.0012

= 0.007683

T'\sim N(5.449830,\ 0.007683)

P(T'< 5.6) = 0.969\ (3\ \text{s.f., by GC)}

To check your answer:

sd(T') \approx 0.0805

kg, so

5.6

kg is about

1.86

standard deviations above the new mean of

5.450

kg, consistent with a high probability close to (but less than)

1

. ✓

$\therefore$ $P(T'< 5.6) \approx 0.969$ (3 s.f.).

Part (e)

4 marks

Approach

the fitting conditions are stated in terms of the difference between a leg’s diameter and its hole’s diameter, so we work with $D = M - H$ rather than $M$ and $H$ separately. “No sanding” and “no padding” each translate into a one-sided bound on $D$ , and together they give a range for $D$ ; since a leg is fitted to only one hole, we can find the probability for one leg-hole pair and then use independence across the $3$ pairs.

Let $H$ mm be the diameter of a hole, $H \sim N\left( 31,\ {0.4}^{2} \right)$ , and let $M$ mm be the diameter of a leg, $M \sim N\left( 30.7,\ {0.3}^{2} \right)$ , with $H$ and $M$ independent. Let $D = M - H$ .

E(D) = 30.7 - 31 = - 0.3

Var(D) = {0.3}^{2} + {0.4}^{2} = 0.09 + 0.16 = 0.25

D \sim N( - 0.3,\ 0.25)

No sanding is needed when the leg is not wider than the hole, i.e. $M \leq H$ , i.e. $D \leq 0$ . No padding is needed when the hole is not more than $0.8$ mm wider than the leg, i.e. $H - M \leq 0.8$ , i.e. $D \geq - 0.8$ . So one leg-hole pair fits without sanding or padding exactly when $- 0.8 \leq D \leq 0$ .

P( - 0.8 \leq D \leq 0) = 0.567\ (3\ \text{s.f., by GC)}

Since the $3$ legs are paired with the $3$ holes at random, and the diameters of different legs and different holes are all independent of one another, whether each of the $3$ leg-hole pairs fits is independent of the other pairs.

P\left( \text{all 3 fit} \right) = (0.567092)^{3}

= 0.182\ \left( 3\ \text{s.f.} \right)

To check your answer: each individual fit has a moderate probability of about

0.567

; cubing this for

3

independent fits gives a noticeably smaller probability, as expected. ✓

$\therefore$ $P\left( \text{all 3 legs fit without sanding or padding} \right) \approx 0.182$ (3 s.f.).

2021 H2 Maths Paper 2, Question 10

Part (a)

Part (b)

Part (c)

Part (d)

Part (e)

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