2017 H2 Mathematics Paper 2 Question 10

The Normal Distribution

Answers

{0.345}

{0.612}

Mean

{= 12.3 \textrm{ g}}

Standard deviation

{= 0.475 \textrm{ g}}

{55.7}

Let

{X}

denote the mass of a randomly selected sphere

X \sim \textrm{N}( 20, 0.25 )

\begin{align*} & \textrm{P}(X > 20.2) \\ &= 0.345 \textrm{ (3 sf)} \; \blacksquare \end{align*}

\begin{align*} 1.1X &\sim \textrm{N}(1.1 \times 20, 1.1^2 \times 0.5^2) \\ 1.1X &\sim \textrm{N}(22,0.3025) \end{align*}

\begin{align*} & \textrm{P}(21.5 < 1.1X < 22.45) \\ &= 0.612 \textrm{ (3 sf)} \; \blacksquare \end{align*}

Let

{Y}

denote the mass of a randomly chosen metal bar with mean

{\mu}

and standard deviation

{\sigma}

\begin{gather*} Y \sim \textrm{N}(\mu, \sigma^2) \\ Z = \frac{X-\mu}{\sigma} \sim \textrm{N}(0, 1) \end{gather*}

\begin{align*} \textrm{P}(Y > 12.2) &= 0.6 \\ \textrm{P}\left(Z > \frac{12.2-\mu}{\sigma}\right) &= 0.6 \\ \end{align*}

\frac{12.2-\mu}{\sigma} = -0.25335

\begin{equation} \mu -0.25335 \sigma = 12.2 \end{equation}

\begin{align*} \textrm{P}(Y < 12) &= 0.25 \\ \textrm{P}\left(Z < \frac{12-\mu}{\sigma}\right) &= 0.25 \\ \end{align*}

\frac{12-\mu}{\sigma} = -0.25335

\begin{equation} \mu -0.67449 \sigma = 12 \end{equation}

Solving

{(1)}

and

{(2)}

simultaneously,

\begin{align*} \textrm{Mean } &= \mu \\ &= 12.320 \\ &= 12.3 \textrm{ g (3 sf)} \; \blacksquare \end{align*}

\begin{align*} & \textrm{Standard deviation } \\ &= \sigma \\ &= 0.47490 \\ &= 0.475 \textrm{ g (3 sf)} \; \blacksquare \end{align*}

Y \sim \textrm{N}(12.320, 0.47490^2)

Let

{W = 1.1X_1 + 1.1X_2 + Y}

denote the total mass of a randomly chosen component

\begin{align*} W &\sim \textrm{N}(2 \times 22 + 12.320, 2 \times 0.3025 + 0.47490^2) \\ W &\sim \textrm{N}( 56.32, 0.83053 ) \end{align*}

\begin{align*} \textrm{P}(W > k) &= 0.75 \\ k &= 55.7 \textrm{ (3 sf)} \; \blacksquare \end{align*}