2021 H2 Mathematics Paper 2 Question 10

The Normal Distribution

Answers

(a)

Mean =2.03 kg{= 2.03 \textrm{ kg}}

(b)

154{154}

(c)

0.427{0.427}

(d)

0.969{0.969}

(e)

0.182{0.182}

Full solutions

(a)

Let S{S} denote the mass of a randomly chosen seat with mean μ{\mu} and standard deviation σ{\sigma}
S∼N(μ,σ2)Z=X−μσ∼N(0,1)\begin{gather*} S \sim \textrm{N}(\mu, \sigma^2) \\ Z = \frac{X-\mu}{\sigma} \sim \textrm{N}(0, 1) \end{gather*}
P(S<2.1)=0.8P(Z<2.1−μσ)=0.8\begin{align*} \textrm{P}(S < 2.1) &= 0.8 \\ \textrm{P}\left(Z < \frac{2.1-\mu}{\sigma}\right) &= 0.8 \\ \end{align*}
2.1−μσ=0.84162\frac{2.1-\mu}{\sigma} = 0.84162
μ+0.84162σ=2.1\begin{equation} \mu + 0.84162 \sigma = 2.1 \end{equation}
P(S<1.95)=0.15P(Z<1.95−μσ)=0.15\begin{align*} \textrm{P}(S < 1.95) &= 0.15 \\ \textrm{P}\left(Z < \frac{1.95-\mu}{\sigma}\right) &= 0.15 \\ \end{align*}
1.95−μσ=0.84162\frac{1.95-\mu}{\sigma} = 0.84162
μ+−1.0364σ=1.95\begin{equation} \mu + -1.0364 \sigma = 1.95 \end{equation}
Solving (1){(1)} and (2){(2)} simultaneously,
Mean =μ=2.0328=2.03 kg (3 sf)  ■\begin{align*} \textrm{Mean } &= \mu \\ &= 2.0328 \\ &= 2.03 \textrm{ kg (3 sf)} \; \blacksquare \end{align*}
Standard deviation =σ=0.079870=0.0799 kg (3 sf)  ■\begin{align*} & \textrm{Standard deviation } \\ &= \sigma \\ &= 0.079870 \\ &= 0.0799 \textrm{ kg (3 sf)} \; \blacksquare \end{align*}

(b)

Let L{L} denote the mass of a randomly selected leg
L∼(1.2,0.022)L \sim(1.2, 0.02^2)
P(L>1.21)=0.30854\begin{align*} & \textrm{P}(L > 1.21) \\ &= 0.30854 \end{align*}
Expected number of legs=500×0.30854=154 (3 sf)  ■\begin{align*} & \textrm{Expected number of legs} \\ &= 500 \times 0.30854 \\ &= 154 \textrm{ (3 sf)} \; \blacksquare \end{align*}

(c)

Let X=S+L1+L2+L3{X=S+L_1+L_2+L_3}
X∼N(2.0328+3(1.2),0.0798702+3(0.022))X∼N(5.6328,0.0075792)\begin{align*} X &\sim \textrm{N}(2.0328+3(1.2), 0.079870^2 + 3 (0.02^2)) \\ X &\sim \textrm{N}( 5.6328, 0.0075792 ) \end{align*}
P(5.6<X<5.7)=0.427 (3 sf)  ■\begin{align*} & \textrm{P}(5.6 < X < 5.7) \\ &= 0.427 \textrm{ (3 sf)} \; \blacksquare \end{align*}

(d)

Let Y=0.91S+L1+L2+L3{Y=0.91S+L_1+L_2+L_3}
Y∼N(0.91(2.0328)+3(1.2),0.912(0.0798702)+3(0.022))Y∼N(5.4498,0.0064826)\begin{align*} Y &\sim \textrm{N}(0.91(2.0328)+3(1.2), 0.91^2(0.079870^2) + 3 (0.02^2)) \\ Y &\sim \textrm{N}( 5.4498, 0.0064826 ) \end{align*}
P(Y<5.6)=0.969 (3 sf)  ■\begin{align*} & \textrm{P}(Y < 5.6) \\ &= 0.969 \textrm{ (3 sf)} \; \blacksquare \end{align*}

(e)

Let H{H} and D{D} denote the diameter of a hole and leg, respectively
D−H∼N(31−30.7,0.42+0.32)D−H∼N(−0.3,0.25)H−D∼N(30.7−31,0.42+0.32)H−D∼N(0.3,0.25)\begin{align*} D - H &\sim \textrm{N}(31-30.7, 0.4^2 + 0.3^2) \\ D - H &\sim \textrm{N}(-0.3, 0.25) \\ H - D &\sim \textrm{N}(30.7-31, 0.4^2 + 0.3^2) \\ H - D &\sim \textrm{N}(0.3, 0.25) \\ \end{align*}
P(sanding needed)=P(D>H)=P(D−H>0)=0.27425\begin{align*} & \textrm{P}(\textrm{sanding needed}) \\ &= \textrm{P}(D > H) \\ &= \textrm{P}(D - H > 0) \\ &= 0.27425 \end{align*}
P(padding needed)=P(H>D+0.8)=P(H−D>0.8)=0.15866\begin{align*} & \textrm{P}(\textrm{padding needed}) \\ &= \textrm{P}(H > D + 0.8) \\ &= \textrm{P}(H - D > 0.8) \\ &= 0.15866 \end{align*}
P(one leg fits)=1−0.27425−0.15866=0.56709\begin{align*} & \textrm{P}(\textrm{one leg fits}) \\ & = 1 - 0.27425 - 0.15866 \\ & = 0.56709 \end{align*}
P(three legs fit)=(0.56709)3=0.182 (3 sf)  ■\begin{align*} & \textrm{P}(\textrm{three legs fit}) \\ & = \left( 0.56709 \right)^3 \\ & = 0.182 \textrm{ (3 sf)} \; \blacksquare \end{align*}