Mathematical Methods — Question 19

ANSWER

$0.57735$ (equivalently $1/\sqrt3$)

Worked solution

**Method 1.** The density is $f(x)=1/80$ for $100\le x\le180$. Using the given rules, $E(X)=140$ and $\operatorname{Var}(X)=1600/3$, so the standard deviation is $40/\sqrt3$. The required probability is $$P\left(140-\frac{40}{\sqrt3}<X<140+\frac{40}{\sqrt3}\right)=\int_{140-40/\sqrt3}^{140+40/\sqrt3}\frac1{80}\,dx=0.57735.$$

**Alternative Method 2.** Graph the uniform density and identify the interval of width $2\sigma$ centred at 140. Its rectangular area is $$\frac1{80}\times2\times\frac{40}{\sqrt3}=\frac1{\sqrt3}.$$

correctly determines the probability density function

[1 mark]

correctly determines the mean and the standard deviation

[1 mark]

establishes definite integral to represent probability

[1 mark]

determines probability

[1 mark]

One QCAA sample method typeset for web; criterion wording is adapted from the official marking guide.