Question 4
The random variable $X$ has the probability density function
$$f(x)=\left\{\begin{array}{lcl}0.25e^{-0.25x} & , & x\geq0\\ 0 & , & \text{otherwise}\end{array}\right.$$
Rounded to four decimal places, $P(X>10)$ is
That’s it. Step through the hints below to see why.
Not quite. You gave the pdf value $f(10)=0.25e^{-2.5}=0.0205$ as the probability.
Not quite. You treated $X$ as normal with mean and standard deviation $4$, so you found $P(Z>1.5)=0.0668$.
Not quite. You found the wrong tail, $P(X<10)$${}=1-e^{-2.5}=0.9179$.
Integrate the pdf from $10$ to $\infty$.
$$\begin{aligned}P(X>10)&=\int_{10}^{\infty}0.25e^{-0.25x}\,dx\\&\approx\int_{10}^{1000}0.25e^{-0.25x}\,dx\\&=0.0820849\ldots\end{aligned}$$
Use a large upper limit (e.g. 1000) in the calculator’s $\int dx$ template in place of $\infty$. Output: $0.0820849\ldots$
$$\begin{aligned}\int_{10}^{\infty}0.25e^{-0.25x}\,dx&\approx\int_{10}^{1000}0.25e^{-0.25x}\,dx\\&=0.0820849\ldots\end{aligned}$$
$P(X>10)=0.0821$ (to four decimal places).
Only mark this done when you could do it without help. Reading the solution does not count.
A Tangent Tuition mock exam question. It is not a past QCAA question, and Tangent Tuition is not affiliated with the QCAA.
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