Question 5
The continuous random variable $X$ has the probability density function
$$f(x)=\begin{cases}\dfrac{\cos(x)}{2}, & \frac{-\pi}{2}\le x\le\frac{\pi}{2}\\[6pt] 0, & \text{otherwise}\end{cases}$$
The standard deviation of $X$ is
Not quite. 0.467 is the variance, $\operatorname{Var}(X)=\dfrac{\pi^2-8}{4}\approx0.467$. The standard deviation is its square root. In 2022, 37.9% of Queensland students chose (A).
That’s it. Step through the hints below to see why. In 2022, 37.7% of Queensland students chose (B).
Not quite. Check it against the graph. A standard deviation of 1.211 would need much of the probability out near $\pm\tfrac{\pi}{2}$, but this pdf is highest at $x=0$ and falls to 0 at the ends. Find $\operatorname{Var}(X)$ first, then take the square root. In 2022, 13.1% of Queensland students chose (C).
Not quite. 1.467 is almost $\tfrac{\pi}{2}\approx1.571$, the furthest $X$ can ever be from its mean of 0. That would need nearly all the probability at the ends, but this pdf is highest in the middle. Find $\operatorname{Var}(X)$ first, then take the square root. In 2022, 9.2% of Queensland students chose (D).
Too narrow. Here $s^2$ is less than $\operatorname{Var}(X)\approx0.467$, so increase $s$.
Too wide. Here $s^2$ is more than $\operatorname{Var}(X)\approx0.467$, so decrease $s$.
Very close, but not equal. Here $s^2$ is near 0.467 but not equal to it. Use the button for the exact value.
This is option (A). Here $s=0.467$, but that is the variance itself. Its square is only about 0.218. The standard deviation is the square root of the variance.
This is option (C). Here $s^2\approx1.467$, more than three times the variance. The bracket reaches out to where the pdf is low.
This is option (D). Here $s^2\approx2.153$, more than four times the variance. The bracket covers almost the whole domain.
This is it. $s^2=\operatorname{Var}(X)$, so $s\approx0.684$ is the standard deviation, option (B).
Start with the mean. The pdf is symmetric about $x=0$, so the mean is 0. Your calculator agrees:
$$\mu=\int_{-\pi/2}^{\pi/2}x\cdot\frac{\cos(x)}{2}\,dx=0$$
The variance is $E(X^2)-\mu^2$. Integrate $x^2f(x)$ over the domain on your calculator.
$\displaystyle \operatorname{Var}(X)$$\displaystyle =\int_{-\pi/2}^{\pi/2}x^2\cdot\frac{\cos(x)}{2}\,dx-0^2$$\displaystyle \approx0.4674$
The standard deviation is the square root of the variance.
$$\sigma=\sqrt{0.4674}\approx0.684$$
Stopping at the variance gives 0.467, which is option (A). In the figure, a bracket of half-width 0.684 is exactly the one where $s^2=\operatorname{Var}(X)$.
$$\mu=\int_{-\pi/2}^{\pi/2}\frac{x\cos(x)}{2}\,dx=0$$
$\displaystyle \operatorname{Var}(X)$$\displaystyle =\int_{-\pi/2}^{\pi/2}\frac{x^2\cos(x)}{2}\,dx-0^2$$\displaystyle \approx0.4674$
$$\sigma=\sqrt{0.4674}\approx0.684$$
The exact variance is $\dfrac{\pi^2-8}{4}$, but on Paper 2 your calculator does the integral.
How Queensland students answered in 2022
More students chose (A) than the answer. From the QCAA Mathematical Methods subject report for 2022. The percentages can add to less than 100 because some students didn’t answer.
Only mark this done when you could do it without help. Reading the solution does not count.
Question wording is from the 2022 QCAA Mathematical Methods external assessment, © State of Queensland (QCAA) 2022, licensed under CC BY 4.0, and has been adapted. Tangent Tuition is not affiliated with the QCAA.
Spot a mistake on this page? Report it