Question 7
The life expectancy (in years) of an electronic component can be represented by the probability density function
$$p(x)=\begin{cases}\dfrac{1}{x^2}, & x\ge1\\[6pt] 0, & \text{otherwise}\end{cases}$$
The probability that the component lasts between 1 and 10 years is
Not quite. 0.010 is $p(10)=\tfrac{1}{10^2}$, the height of the pdf at $x=10$. A probability is an area under the pdf, not a height.
Not quite. 0.100 is the probability that the component lasts more than 10 years, $\int_{10}^{\infty}\tfrac{1}{x^2}\,dx=\tfrac{1}{10}$. You also get 0.1 if you substitute only the upper limit, $-\tfrac{1}{10}$, and drop the sign. You want the area from 1 to 10, which is the rest: $1-0.1$.
That’s it. Step through the hints below to see why.
Not quite. 0.990 is what you get if the power stays at $-2$ when you integrate: $\Bigl[-x^{-2}\Bigr]_1^{10}=-0.01+1$. That is the same as subtracting the heights, $p(1)-p(10)$. The power goes up by one, so $\int x^{-2}\,dx=-x^{-1}+c$.
The shaded area is $P(1<X<b)$. The bar along the bottom runs from probability 0 to 1. Its round marker shows the shaded area, and the grey markers are the four options.
The shaded area is $1-\tfrac{1}{10}=0.9$ exactly. That’s option (C). The tail past 10 holds the other 0.1, so the two add to the total area of 1.
Both limits are 1, so there is no area yet. Drag the upper limit to the right.
Very close to option (C), but not equal: the area is because the upper limit isn’t 10. Use the button below.
The shaded area is . The rest of the probability, , lies past the upper limit.
The marker happens to sit near option () here, but only because the upper limit stops so early.
Past $x=5$ the curve runs very close to the axis, but it never reaches it. That thin tail still holds probability.
For a continuous random variable, a probability is an area under the pdf. Lasting between 1 and 10 years means $1<X<10$.
$P(1<X<10)=\displaystyle\int_1^{10}\frac{1}{x^2}\,dx$
Write $\tfrac{1}{x^2}$ as $x^{-2}$ and raise the power by one.
$\displaystyle\int x^{-2}\,dx=\frac{x^{-1}}{-1}$$=-\dfrac{1}{x}+c$
$\Bigl[-\dfrac{1}{x}\Bigr]_1^{10}$$=-\dfrac{1}{10}-(-1)$$=0.9$
Sense check: the total area from 1 onwards is 1, and the tail past 10 is $\int_{10}^{\infty}\tfrac{1}{x^2}\,dx=\tfrac{1}{10}$. So the area from 1 to 10 is what’s left, 0.9.
0.010 is the height $p(10)$, and 0.990 comes from leaving the power at $-2$, which is the same as $p(1)-p(10)$.
$$\begin{aligned}P(1<X<10)&=\int_1^{10}\frac{1}{x^2}\,dx\\&=\Bigl[-\frac{1}{x}\Bigr]_1^{10}\\&=-\frac{1}{10}+1\\&=0.9\end{aligned}$$
A probability is an area, so integrate the pdf between the two values. The answer is 0.900.
Only mark this done when you could do it without help. Reading the solution does not count.
Question wording is from the 2020 QCAA Mathematical Methods external assessment, © State of Queensland (QCAA) 2020, licensed under CC BY 4.0, and has been adapted. Tangent Tuition is not affiliated with the QCAA.
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