IB Mathematics: Applications and Interpretation Quiz: Continuous Random Variables
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Continuous Random VariablesQuestion 1 of 14

The lifespan LL in years of an electronic component is modelled by the probability density function f(l)=136l(6−l)f(l) = \frac{1}{36}l(6-l) for 0≤l≤60 \le l \le 6. The component comes with a 1-year warranty.

What is the probability that a component will fail after the warranty expires but before 4 years of use?

0.074
0.667
0.741
0.926
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IB Mathematics: Applications and Interpretation Quiz

IB Mathematics: Applications and Interpretation Quiz: Continuous Random Variables

Practice Continuous Random Variables in IB Mathematics: Applications and Interpretation with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Continuous Random Variables, giving you a quick way to practice the rules, question types, and explanations that matter most for IB Mathematics: Applications and Interpretation.

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Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

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Question 1

The lifespan LL in years of an electronic component is modelled by the probability density function f(l)=136l(6−l)f(l) = \frac{1}{36}l(6-l) for 0≤l≤60 \le l \le 6. The component comes with a 1-year warranty.

What is the probability that a component will fail after the warranty expires but before 4 years of use?

  1. 0.074
  2. 0.667 (correct answer)
  3. 0.741
  4. 0.926
Explanation: We need to find the probability P(1<L<4)P(1 < L < 4). This is calculated by the integral of the PDF from 1 to 4. P(1<L<4)=∫14136l(6−l) dl=136∫14(6l−l2) dl P(1 < L < 4) = \int_{1}^{4} \frac{1}{36}l(6-l) \,dl = \frac{1}{36} \int_{1}^{4} (6l - l^2) \,dl =136[3l2−l33]14= \frac{1}{36} \left[3l^2 - \frac{l^3}{3}\right]_{1}^{4} =136[(3(42)−433)−(3(12)−133)]= \frac{1}{36} \left[ \left(3(4^2) - \frac{4^3}{3}\right) - \left(3(1^2) - \frac{1^3}{3}\right) \right] =136[(48−643)−(3−13)]=136[803−83]=136(723)=2436=23≈0.667= \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)P(L < 1), the probability it fails during warranty. Distractor C is P(L<4)P(L < 4). Distractor D is P(L>1)P(L > 1).

Question 2

The height HH of a plant, in cm, after one month is modelled by the probability density function f(h)=18πe−(h−15)2/8f(h) = \frac{1}{\sqrt{8\pi}} e^{-(h-15)^2/8} for h≥0h \ge 0. Find the probability that a randomly selected plant has a height between 12 cm and 16 cm.

  1. 0.372
  2. 0.511
  3. 0.625 (correct answer)
  4. 0.683
Explanation: The PDF is of the form for a normal distribution, f(x)=1σ2πe−(x−μ)2/(2σ2)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 μ=15\mu = 15. We have 2σ2=82\sigma^2 = 8, which means σ2=4\sigma^2 = 4 and the standard deviation is σ=2\sigma = 2. We need to calculate P(12≤H≤16)P(12 \le H \le 16) for H∼N(15,22)H \sim N(15, 2^2). Using a graphing calculator's normal cumulative distribution function (normalcdf), with lower bound=12, upper bound=16, μ=15\mu=15, and σ=2\sigma=2, we get a probability of approximately 0.6247. \newline Distractor A results from incorrectly identifying σ=4\sigma = 4. Distractor B results from incorrectly identifying σ=8\sigma = \sqrt{8}. Distractor D is the probability of being within one standard deviation (13 to 17), which is approximately 68.3%.

Question 3

The distribution of daily rainfall, XX, 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?

  1. The mean daily rainfall is equal to the median daily rainfall.
  2. The mean daily rainfall is less than the median daily rainfall.
  3. The mode of the daily rainfall is greater than the mean daily rainfall.
  4. The mean daily rainfall is greater than the median daily rainfall. (correct answer)
Explanation: 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.

Question 4

The time TT in hours to repair a machine follows a uniform distribution on the interval [2,4][2, 4]. The cost of the repair, in dollars, is given by the function C(T)=50+20T2C(T) = 50 + 20T^2.

Find the expected cost of a repair.

  1. 230.00
  2. 236.67 (correct answer)
  3. 240.00
  4. 263.33
Explanation: For a uniform distribution on [a,b][a,b], the PDF is f(t)=1b−af(t) = \frac{1}{b-a}. Here, f(t)=14−2=12f(t) = \frac{1}{4-2} = \frac{1}{2} for 2≤t≤42 \le t \le 4. The expected cost is E(C(T))=E(50+20T2)=50+20E(T2)E(C(T)) = E(50 + 20T^2) = 50 + 20E(T^2). We must first calculate E(T2)E(T^2). E(T2)=∫24t2f(t) dt=∫24t2⋅12 dtE(T^2) = \int_{2}^{4} t^2 f(t) \,dt = \int_{2}^{4} t^2 \cdot \frac{1}{2} \,dt =12[t33]24=16(43−23)=16(64−8)=566=283= \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(283)=50+5603=150+5603=7103≈236.67E(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))C(E(T)), a common error where the function is evaluated at the mean time E(T)=3E(T)=3. Distractors C and D are from other calculation errors.

Question 5

The lifetime of a particular electronic component is modelled by a continuous random variable XX with an exponential distribution. The mean lifetime is known to be 5 years.

Find the median lifetime of these components.

  1. 0.693 years
  2. 2.500 years
  3. 3.466 years (correct answer)
  4. 5.000 years
Explanation: The exponential distribution with mean μ=1/λ\mu = 1/\lambda has the PDF f(x)=λe−λxf(x) = \lambda e^{-\lambda x}. Given the mean is 5, we have 1/λ=51/\lambda = 5, so λ=1/5=0.2\lambda = 1/5 = 0.2. The PDF is f(x)=0.2e−0.2xf(x) = 0.2e^{-0.2x} for x≥0x \ge 0. The median mm is the value for which P(X≤m)=0.5P(X \le m) = 0.5. ∫0m0.2e−0.2x dx=0.5\int_{0}^{m} 0.2e^{-0.2x} \,dx = 0.5 [−e−0.2x]0m=0.5\left[-e^{-0.2x}\right]_{0}^{m} = 0.5 −e−0.2m−(−e0)=0.5-e^{-0.2m} - (-e^0) = 0.5 1−e−0.2m=0.5  ⟹  e−0.2m=0.51 - e^{-0.2m} = 0.5 \implies e^{-0.2m} = 0.5 −0.2m=ln⁡(0.5)=−ln⁡(2)-0.2m = \ln(0.5) = -\ln(2) m=ln⁡(2)0.2=5ln⁡(2)≈3.466m = \frac{\ln(2)}{0.2} = 5\ln(2) \approx 3.466 Distractor A is the value of ln⁡(2)\ln(2) without multiplying by 5. Distractor B is half the mean. Distractor D is the mean, which is confused with the median.

Question 6

The waiting time XX in minutes for a particular service is modelled by the exponential probability density function f(x)=3e−3x,x≥0f(x) = 3e^{-3x}, \quad x \ge 0 Find the median waiting time.

  1. 0.231 (correct answer)
  2. 0.333
  3. 0.500
  4. 0.693
Explanation: The median mm is the value such that P(X≤m)=0.5P(X \le m) = 0.5. ∫0m3e−3x dx=0.5\int_{0}^{m} 3e^{-3x} \,dx = 0.5 [−e−3x]0m=0.5\left[-e^{-3x}\right]_{0}^{m} = 0.5 (−e−3m)−(−e0)=0.5(-e^{-3m}) - (-e^0) = 0.5 1−e−3m=0.5  ⟹  e−3m=0.51 - e^{-3m} = 0.5 \implies e^{-3m} = 0.5 −3m=ln⁡(0.5)=−ln⁡(2)-3m = \ln(0.5) = -\ln(2) m=ln⁡(2)3≈0.231m = \frac{\ln(2)}{3} \approx 0.231 Distractor B is the mean of the distribution (1/λ=1/31/\lambda = 1/3). Distractor C is the probability value itself. Distractor D is the value of ln⁡(2)\ln(2), forgetting to divide by 3.

Question 7

The height HH of a plant, in cm, after one month is modelled by the probability density function f(h)=18πe−(h−15)2/8f(h) = \frac{1}{\sqrt{8\pi}} e^{-(h-15)^2/8} for h≥0h \ge 0. Find the probability that a randomly selected plant has a height between 12 cm and 16 cm.

  1. 0.372
  2. 0.511
  3. 0.625 (correct answer)
  4. 0.683
Explanation: The PDF is of the form for a normal distribution, f(x)=1σ2πe−(x−μ)2/(2σ2)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 μ=15\mu = 15. We have 2σ2=82\sigma^2 = 8, which means σ2=4\sigma^2 = 4 and the standard deviation is σ=2\sigma = 2. We need to calculate P(12≤H≤16)P(12 \le H \le 16) for H∼N(15,22)H \sim N(15, 2^2). Using a graphing calculator's normal cumulative distribution function (normalcdf), with lower bound=12, upper bound=16, μ=15\mu=15, and σ=2\sigma=2, we get a probability of approximately 0.6247. \newline Distractor A results from incorrectly identifying σ=4\sigma = 4. Distractor B results from incorrectly identifying σ=8\sigma = \sqrt{8}. Distractor D is the probability of being within one standard deviation (13 to 17), which is approximately 68.3%.

Question 8

The time TT in hours to repair a machine follows a uniform distribution on the interval [2,4][2, 4]. The cost of the repair, in dollars, is given by the function C(T)=50+20T2C(T) = 50 + 20T^2.

Find the expected cost of a repair.

  1. 230.00
  2. 236.67 (correct answer)
  3. 240.00
  4. 263.33
Explanation: For a uniform distribution on [a,b][a,b], the PDF is f(t)=1b−af(t) = \frac{1}{b-a}. Here, f(t)=14−2=12f(t) = \frac{1}{4-2} = \frac{1}{2} for 2≤t≤42 \le t \le 4. The expected cost is E(C(T))=E(50+20T2)=50+20E(T2)E(C(T)) = E(50 + 20T^2) = 50 + 20E(T^2). We must first calculate E(T2)E(T^2). E(T2)=∫24t2f(t) dt=∫24t2⋅12 dtE(T^2) = \int_{2}^{4} t^2 f(t) \,dt = \int_{2}^{4} t^2 \cdot \frac{1}{2} \,dt =12[t33]24=16(43−23)=16(64−8)=566=283= \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(283)=50+5603=150+5603=7103≈236.67E(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))C(E(T)), a common error where the function is evaluated at the mean time E(T)=3E(T)=3. Distractors C and D are from other calculation errors.

Question 9

The waiting time XX in minutes for a particular service is modelled by the exponential probability density function f(x)=3e−3x,x≥0f(x) = 3e^{-3x}, \quad x \ge 0 Find the median waiting time.

  1. 0.231 (correct answer)
  2. 0.333
  3. 0.500
  4. 0.693
Explanation: The median mm is the value such that P(X≤m)=0.5P(X \le m) = 0.5. ∫0m3e−3x dx=0.5\int_{0}^{m} 3e^{-3x} \,dx = 0.5 [−e−3x]0m=0.5\left[-e^{-3x}\right]_{0}^{m} = 0.5 (−e−3m)−(−e0)=0.5(-e^{-3m}) - (-e^0) = 0.5 1−e−3m=0.5  ⟹  e−3m=0.51 - e^{-3m} = 0.5 \implies e^{-3m} = 0.5 −3m=ln⁡(0.5)=−ln⁡(2)-3m = \ln(0.5) = -\ln(2) m=ln⁡(2)3≈0.231m = \frac{\ln(2)}{3} \approx 0.231 Distractor B is the mean of the distribution (1/λ=1/31/\lambda = 1/3). Distractor C is the probability value itself. Distractor D is the value of ln⁡(2)\ln(2), forgetting to divide by 3.

Question 10

The time TT, in minutes, a customer waits for a coffee is modelled by the probability density function f(t)=0.5e−0.5tf(t) = 0.5e^{-0.5t} for t≥0t \ge 0.

Given that a customer has already waited for 2 minutes, find the probability that they will wait for at least one more minute.

  1. 0.145
  2. 0.223
  3. 0.393
  4. 0.607 (correct answer)
Explanation: We need to find the conditional probability P(T≥3∣T≥2)P(T \ge 3 | T \ge 2). The exponential distribution is 'memoryless', which means P(T>s+t∣T>s)=P(T>t)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≥1)P(T \ge 1). P(T≥1)=∫1∞0.5e−0.5t dt=[−e−0.5t]1∞=0−(−e−0.5)=e−0.5≈0.607 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≥3∣T≥2)=P(T≥3)P(T≥2)P(T \ge 3 | T \ge 2) = \frac{P(T \ge 3)}{P(T \ge 2)}. P(T≥3)=e−0.5×3=e−1.5P(T \ge 3) = e^{-0.5 \times 3} = e^{-1.5}. P(T≥2)=e−0.5×2=e−1P(T \ge 2) = e^{-0.5 \times 2} = e^{-1}. The ratio is e−1.5/e−1=e−0.5≈0.607e^{-1.5} / e^{-1} = e^{-0.5} \approx 0.607. Distractor A is P(2≤T≤3)P(2 \le T \le 3). Distractor B is P(T≥3)P(T \ge 3). Distractor C is P(T≤1)P(T \le 1).

Question 11

The lifespan LL in years of an electronic component is modelled by the probability density function f(l)=136l(6−l)f(l) = \frac{1}{36}l(6-l) for 0≤l≤60 \le l \le 6. The component comes with a 1-year warranty.

What is the probability that a component will fail after the warranty expires but before 4 years of use?

  1. 0.074
  2. 0.667 (correct answer)
  3. 0.741
  4. 0.926
Explanation: We need to find the probability P(1<L<4)P(1 < L < 4). This is calculated by the integral of the PDF from 1 to 4. P(1<L<4)=∫14136l(6−l) dl=136∫14(6l−l2) dl P(1 < L < 4) = \int_{1}^{4} \frac{1}{36}l(6-l) \,dl = \frac{1}{36} \int_{1}^{4} (6l - l^2) \,dl =136[3l2−l33]14= \frac{1}{36} \left[3l^2 - \frac{l^3}{3}\right]_{1}^{4} =136[(3(42)−433)−(3(12)−133)]= \frac{1}{36} \left[ \left(3(4^2) - \frac{4^3}{3}\right) - \left(3(1^2) - \frac{1^3}{3}\right) \right] =136[(48−643)−(3−13)]=136[803−83]=136(723)=2436=23≈0.667= \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)P(L < 1), the probability it fails during warranty. Distractor C is P(L<4)P(L < 4). Distractor D is P(L>1)P(L > 1).

Question 12

The distribution of daily rainfall, XX, 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?

  1. The mean daily rainfall is equal to the median daily rainfall.
  2. The mean daily rainfall is less than the median daily rainfall.
  3. The mode of the daily rainfall is greater than the mean daily rainfall.
  4. The mean daily rainfall is greater than the median daily rainfall. (correct answer)
Explanation: 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.

Question 13

The lifetime of a particular electronic component is modelled by a continuous random variable XX with an exponential distribution. The mean lifetime is known to be 5 years.

Find the median lifetime of these components.

  1. 0.693 years
  2. 2.500 years
  3. 3.466 years (correct answer)
  4. 5.000 years
Explanation: The exponential distribution with mean μ=1/λ\mu = 1/\lambda has the PDF f(x)=λe−λxf(x) = \lambda e^{-\lambda x}. Given the mean is 5, we have 1/λ=51/\lambda = 5, so λ=1/5=0.2\lambda = 1/5 = 0.2. The PDF is f(x)=0.2e−0.2xf(x) = 0.2e^{-0.2x} for x≥0x \ge 0. The median mm is the value for which P(X≤m)=0.5P(X \le m) = 0.5. ∫0m0.2e−0.2x dx=0.5\int_{0}^{m} 0.2e^{-0.2x} \,dx = 0.5 [−e−0.2x]0m=0.5\left[-e^{-0.2x}\right]_{0}^{m} = 0.5 −e−0.2m−(−e0)=0.5-e^{-0.2m} - (-e^0) = 0.5 1−e−0.2m=0.5  ⟹  e−0.2m=0.51 - e^{-0.2m} = 0.5 \implies e^{-0.2m} = 0.5 −0.2m=ln⁡(0.5)=−ln⁡(2)-0.2m = \ln(0.5) = -\ln(2) m=ln⁡(2)0.2=5ln⁡(2)≈3.466m = \frac{\ln(2)}{0.2} = 5\ln(2) \approx 3.466 Distractor A is the value of ln⁡(2)\ln(2) without multiplying by 5. Distractor B is half the mean. Distractor D is the mean, which is confused with the median.

Question 14

The time TT, in minutes, a customer waits for a coffee is modelled by the probability density function f(t)=0.5e−0.5tf(t) = 0.5e^{-0.5t} for t≥0t \ge 0.

Given that a customer has already waited for 2 minutes, find the probability that they will wait for at least one more minute.

  1. 0.145
  2. 0.223
  3. 0.393
  4. 0.607 (correct answer)
Explanation: We need to find the conditional probability P(T≥3∣T≥2)P(T \ge 3 | T \ge 2). The exponential distribution is 'memoryless', which means P(T>s+t∣T>s)=P(T>t)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≥1)P(T \ge 1). P(T≥1)=∫1∞0.5e−0.5t dt=[−e−0.5t]1∞=0−(−e−0.5)=e−0.5≈0.607 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≥3∣T≥2)=P(T≥3)P(T≥2)P(T \ge 3 | T \ge 2) = \frac{P(T \ge 3)}{P(T \ge 2)}. P(T≥3)=e−0.5×3=e−1.5P(T \ge 3) = e^{-0.5 \times 3} = e^{-1.5}. P(T≥2)=e−0.5×2=e−1P(T \ge 2) = e^{-0.5 \times 2} = e^{-1}. The ratio is e−1.5/e−1=e−0.5≈0.607e^{-1.5} / e^{-1} = e^{-0.5} \approx 0.607. Distractor A is P(2≤T≤3)P(2 \le T \le 3). Distractor B is P(T≥3)P(T \ge 3). Distractor C is P(T≤1)P(T \le 1).