2023 H2 Mathematics Paper 2 Question 9

The Normal Distribution

Answers

(a)

0.440.{{0.440.}}

(b)

0.542.{{0.542.}}

(c)

55.2.{{55.2.}}

(d)

0.237.{{0.237.}}

(e)

0.492.{{0.492.}}

(f)

When there is variable wastage, SL{{S_L}} and SR{{S_R}} will have increased variance, so SL−SR{{S_L - S_R}} will have increased variance as well.

Hence the answer to part (e) will increase.

Full solutions

(a)

Let L{{L}} and R{{R}} denote the length, in metres, of a randomly chosen Long and Regular wooden plank, respectively.

L∼N(1.82,0.22)P(L<1.79)=0.440 (3 sf)  ■\begin{gather*} L \sim N( 1.82,0.2^2 ) \\ \mathrm{P}\left( L < 1.79 \right) = 0.440 \text{ (3 sf)} \; \blacksquare \end{gather*}

(b)

E(L1+⋯+L8)=8  E(L)=14.56\begin{align*} & \mathrm{E}(L_1 + \dotsb + L_8) \\ &= 8 \; \mathrm{E}(L) \\ &= 14.56 \end{align*}
Var(L1+⋯+L8)=8  Var(L)=0.32\begin{align*} & \mathrm{Var}(L_1 + \dotsb + L_8) \\ &= 8 \; \mathrm{Var}(L) \\ &= 0.32 \end{align*}
L1+⋯+L8∼N(14.56,0.32000)P(L1+⋯+L8>14.5)=0.542 (3 sf)  ■\begin{gather*} L_1 + \dotsb + L_8 \sim N( 14.56,0.32000 ) \\ \mathrm{P}\left( L_1 + \dotsb + L_8 > 14.5 \right) = 0.542 \text{ (3 sf)} \; \blacksquare \end{gather*}

(c)

L∼N(1.82,0.22)P(R>1.25)=0.46017\begin{gather*} L \sim N( 1.82,0.2^2 ) \\ \mathrm{P}\left( R > 1.25 \right) = 0.46017 \end{gather*}
Expected number of planks=120×0.46017=55.2 (3 sf)  ■\begin{align*} & \text{Expected number of planks} \\ &= 120 \times 0.46017 \\ &= 55.2 \text{ (3 sf)} \; \blacksquare \end{align*}

(d)

Let

D=L1+⋯+L10−(R1+⋯+R16){D = L_1 + \dotsb + L_{10} - \left( R_1 + \dotsb + R_{16} \right) }
E(D)=10  E(L)−16  E(R)=−1.32\begin{align*} & \mathrm{E}(D) \\ &= 10 \; \mathrm{E}(L) - 16 \; \mathrm{E}(R) \\ &= -1.32 \end{align*}
Var(D)=10  Var(L)+16  Var(R)=1.84\begin{align*} & \mathrm{Var}(D) \\ &= 10 \; \mathrm{Var}(L) + 16 \; \mathrm{Var}(R) \\ &= 1.84 \end{align*}
D∼N(−1.3200,1.84){D \sim N( -1.3200,1.84 )}
P(∣D∣<0.65)=P(−0.65<D<0.65)=0.237 (3 sf)  ■\begin{align*} & \mathrm{P}\left( \left| D \right| < 0.65 \right) \\ &= \mathrm{P}\left( -0.65 < D < 0.65 \right) \\ &= 0.237 \text{ (3 sf)} \; \blacksquare \end{align*}

(e)

Let SL{{S_L}} and SR{{S_R}} denote the length of a Short plank obtained by cutting the Long and Regular planks, respectively.

SL∼N(1.823,0.2232)SL∼N(0.60667,0.0044444)SR∼N(1.222,0.3222)SR∼N(0.61,0.022500)\begin{align*} & S_L \sim \text{N} \left( \frac{1.82}{3}, \frac{0.2^2}{3^2} \right) \\ & S_L \sim N( 0.60667,0.0044444 ) \\ & S_R \sim \text{N} \left( \frac{1.22}{2}, \frac{0.3^2}{2^2} \right) \\ & S_R \sim N( 0.61,0.022500 ) \end{align*}
SL−SR∼N(−0.0033333,0.026944){S_L - S_R \sim N( -0.0033333,0.026944 )}
P(SL>SR)=P(SL−SR>0)=0.492 (3 sf)  ■\begin{align*} & \mathrm{P}\left( S_L > S_R \right) \\ &= \mathrm{P}\left( S_L - S_R > 0 \right) \\ &= 0.492 \text{ (3 sf)} \; \blacksquare \end{align*}

(f)

When there is variable wastage, SL{{S_L}} and SR{{S_R}} will have increased variance, so SL−SR{{S_L - S_R}} will have increased variance as well.

Hence Hence the answer to part (e) will increase   ■{{\; \blacksquare}}

Question Commentary

Parts (a) to (e) are relatively straightforward normal distribution questions, testing the use of our GC along with calculating the mean and variance of new distributions like X1+X2{{X_1 + X_2}} and 2X.{{2X.}}

Part (f) links to theoretical knowledge, where I suspect any reasonable explanation will suffice (eg if the wastage of each cut is a fixed amount that is the same for all planks, the answer to part (e) will not change. However, I believe this scenario is less likely than the increased variance answer we have suggested above).

To see why an increased variance will cause the answer to part (e) to increase, we recommend sketching out a normal distribution and locating the region representing part (e). Increasing the variance means that the normal curve will be more spread out, leading to a larger area under the curve.