Continuous Random Variables - IB Mathematics Applications and Interpretation

1. Given that a customer has already waited for 2 minutes, find the probability that they will wait for at least one more minute.
   Answer: 0.607 We need to find the conditional probability $P(T \ge 3 | T \ge 2)$. The exponential distribution is 'memoryless', which means $P(T > s+t | T > s) = P(T > t)$. Therefore, the probability of waiting at least one more minute, given they have already waited 2 minutes, is the same as the initial probability of waiting at least one minute. We calculate $P(T \ge 1)$. $$P(T \ge 1) = \int_{1}^{\infty} 0.5e^{-0.5t} \,dt = \left[-e^{-0.5t}\right]_{1}^{\infty} = 0 - (-e^{-0.5}) = e^{-0.5} \approx 0.607$$ Alternatively, using the conditional probability formula: $P(T \ge 3 | T \ge 2) = \frac{P(T \ge 3)}{P(T \ge 2)}$. $P(T \ge 3) = e^{-0.5 \times 3} = e^{-1.5}$. $P(T \ge 2) = e^{-0.5 \times 2} = e^{-1}$. The ratio is $e^{-1.5} / e^{-1} = e^{-0.5} \approx 0.607$. Distractor A is $P(2 \le T \le 3)$. Distractor B is $P(T \ge 3)$. Distractor C is $P(T \le 1)$.
2. The waiting time $X$ in minutes for a particular service is modelled by the exponential probability density function $$f(x) = 3e^{-3x}, \quad x \ge 0$$ Find the median waiting time.
   Answer: 0.231 The median $m$ is the value such that $P(X \le m) = 0.5$. $$\int_{0}^{m} 3e^{-3x} \,dx = 0.5$$ $$\left[-e^{-3x}\right]_{0}^{m} = 0.5$$ $$(-e^{-3m}) - (-e^0) = 0.5$$ $$1 - e^{-3m} = 0.5 \implies e^{-3m} = 0.5$$ $$-3m = \ln(0.5) = -\ln(2)$$ $$m = \frac{\ln(2)}{3} \approx 0.231$$ Distractor B is the mean of the distribution ($1/\lambda = 1/3$). Distractor C is the probability value itself. Distractor D is the value of $\ln(2)$, forgetting to divide by 3.
3. The height $H$ of a plant, in cm, after one month is modelled by the probability density function $f(h) = \frac{1}{\sqrt{8\pi}} e^{-(h-15)^2/8}$ for $h \ge 0$. Find the probability that a randomly selected plant has a height between 12 cm and 16 cm.
   Answer: 0.625 The PDF is of the form for a normal distribution, $f(x) = \frac{1}{\sigma\sqrt{2\pi}} e^{-(x-\mu)^2/(2\sigma^2)}$. By comparing the given function with the general form, we can identify the parameters. The mean is $\mu = 15$. We have $2\sigma^2 = 8$, which means $\sigma^2 = 4$ and the standard deviation is $\sigma = 2$. We need to calculate $P(12 \le H \le 16)$ for $H \sim N(15, 2^2)$. Using a graphing calculator's normal cumulative distribution function (normalcdf), with lower bound=12, upper bound=16, $\mu=15$, and $\sigma=2$, we get a probability of approximately 0.6247. \newline Distractor A results from incorrectly identifying $\sigma = 4$. Distractor B results from incorrectly identifying $\sigma = \sqrt{8}$. Distractor D is the probability of being within one standard deviation (13 to 17), which is approximately 68.3%.
4. The distribution of daily rainfall, $X$, in a certain region has a probability density function that is skewed to the right. Which of the following statements is most likely to be true for this distribution?
   Answer: The mean daily rainfall is greater than the median daily rainfall. For a distribution that is skewed to the right (positively skewed), the tail of the distribution is on the right side. This long tail contains high values that pull the mean to the right. The median is less affected by these extreme values, and the mode is the peak of the distribution. The general relationship for a right-skewed distribution is: Mode < Median < Mean. Therefore, the mean is greater than the median.
5. Find the expected cost of a repair.
   Answer: 236.67 For a uniform distribution on $[a,b]$, the PDF is $f(t) = \frac{1}{b-a}$. Here, $f(t) = \frac{1}{4-2} = \frac{1}{2}$ for $2 \le t \le 4$. The expected cost is $E(C(T)) = E(50 + 20T^2) = 50 + 20E(T^2)$. We must first calculate $E(T^2)$. $$E(T^2) = \int_{2}^{4} t^2 f(t) \,dt = \int_{2}^{4} t^2 \cdot \frac{1}{2} \,dt$$ $$= \frac{1}{2} \left[\frac{t^3}{3}\right]_{2}^{4} = \frac{1}{6} (4^3 - 2^3) = \frac{1}{6} (64 - 8) = \frac{56}{6} = \frac{28}{3}$$ Now, find the expected cost: $$E(C(T)) = 50 + 20 \left(\frac{28}{3}\right) = 50 + \frac{560}{3} = \frac{150+560}{3} = \frac{710}{3} \approx 236.67$$ Distractor A is $C(E(T))$, a common error where the function is evaluated at the mean time $E(T)=3$. Distractors C and D are from other calculation errors.
6. Find the median lifetime of these components.
   Answer: 3.466 years The exponential distribution with mean $\mu = 1/\lambda$ has the PDF $f(x) = \lambda e^{-\lambda x}$. Given the mean is 5, we have $1/\lambda = 5$, so $\lambda = 1/5 = 0.2$. The PDF is $f(x) = 0.2e^{-0.2x}$ for $x \ge 0$. The median $m$ is the value for which $P(X \le m) = 0.5$. $$\int_{0}^{m} 0.2e^{-0.2x} \,dx = 0.5$$ $$\left[-e^{-0.2x}\right]_{0}^{m} = 0.5$$ $$-e^{-0.2m} - (-e^0) = 0.5$$ $$1 - e^{-0.2m} = 0.5 \implies e^{-0.2m} = 0.5$$ $$-0.2m = \ln(0.5) = -\ln(2)$$ $$m = \frac{\ln(2)}{0.2} = 5\ln(2) \approx 3.466$$ Distractor A is the value of $\ln(2)$ without multiplying by 5. Distractor B is half the mean. Distractor D is the mean, which is confused with the median.
7. What is the probability that a component will fail after the warranty expires but before 4 years of use?
   Answer: 0.667 We need to find the probability $P(1 < L < 4)$. This is calculated by the integral of the PDF from 1 to 4. $$P(1 < L < 4) = \int_{1}^{4} \frac{1}{36}l(6-l) \,dl = \frac{1}{36} \int_{1}^{4} (6l - l^2) \,dl$$ $$= \frac{1}{36} \left[3l^2 - \frac{l^3}{3}\right]_{1}^{4}$$ $$= \frac{1}{36} \left[ \left(3(4^2) - \frac{4^3}{3}\right) - \left(3(1^2) - \frac{1^3}{3}\right) \right]$$ $$= \frac{1}{36} \left[ \left(48 - \frac{64}{3}\right) - \left(3 - \frac{1}{3}\right) \right] = \frac{1}{36} \left[ \frac{80}{3} - \frac{8}{3} \right] = \frac{1}{36} \left( \frac{72}{3} \right) = \frac{24}{36} = \frac{2}{3} \approx 0.667$$ Distractor A is $P(L < 1)$, the probability it fails during warranty. Distractor C is $P(L < 4)$. Distractor D is $P(L > 1)$.

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