IB Mathematics: Applications and Interpretation Quiz: Further Distributions
20 questions · exam conditions
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Further DistributionsQuestion 1 of 20

A call centre models the number of incoming calls using a Poisson distribution. On average, there are 5 calls every 10 minutes. What is the probability that there are exactly 2 calls in a 5-minute period and exactly 3 calls in the next 5-minute period?

0.055
0.175
0.235
0.470
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IB Mathematics: Applications and Interpretation Quiz

IB Mathematics: Applications and Interpretation Quiz: Further Distributions

Practice Further Distributions 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.

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This quiz focuses on Further Distributions, 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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Question 1

A call centre models the number of incoming calls using a Poisson distribution. On average, there are 5 calls every 10 minutes. What is the probability that there are exactly 2 calls in a 5-minute period and exactly 3 calls in the next 5-minute period?

  1. 0.055 (correct answer)
  2. 0.175
  3. 0.235
  4. 0.470
Explanation: The average rate is 5 calls per 10 minutes, so the rate for a 5-minute period is λ=2.5\lambda = 2.5. Let X1X_1 be the number of calls in the first 5-minute period and X2X_2 be the number of calls in the second. We assume X1,X2∼Po(2.5)X_1, X_2 \sim Po(2.5) and that they are independent. We want to find P(X1=2 and X2=3)P(X_1=2 \text{ and } X_2=3). Because of independence, this is P(X1=2)×P(X2=3)P(X_1=2) \times P(X_2=3). Using a GDC: P(X1=2)=poissonpdf(2.5,2)≈0.2565P(X_1=2) = \text{poissonpdf}(2.5, 2) \approx 0.2565 and P(X2=3)=poissonpdf(2.5,3)≈0.2138P(X_2=3) = \text{poissonpdf}(2.5, 3) \approx 0.2138. The combined probability is 0.2565×0.2138≈0.05480.2565 \times 0.2138 \approx 0.0548, which is closest to 0.055.

Question 2

The number of errors on a page of a manuscript follows a Poisson distribution with a mean of 0.75 errors per page. The manuscript has 400 pages. Using a suitable approximation, what is the probability that the entire manuscript contains at least 310 errors?

  1. 0.198
  2. 0.229 (correct answer)
  3. 0.261
  4. 0.771
Explanation: First, find the mean number of errors for the entire manuscript. The rate is 0.75 errors/page for 400 pages, so the total mean is λ=0.75×400=300\lambda = 0.75 \times 400 = 300. Let XX be the number of errors, so X∼Po(300)X \sim Po(300). Since λ=300\lambda = 300 is large, we can use a normal approximation, Y∼N(300,300)Y \sim N(300, 300). The mean is μ=300\mu = 300 and the standard deviation is σ=300\sigma = \sqrt{300}. We need to find P(X≥310)P(X \ge 310). With continuity correction, this becomes P(Y>309.5)P(Y > 309.5). Using a GDC, P(Y>309.5)=normalcdf(309.5,∞,300,300)≈0.229P(Y > 309.5) = \text{normalcdf}(309.5, \infty, 300, \sqrt{300}) \approx 0.229.

Question 3

The number of accidents on a stretch of highway follows a Poisson distribution with an average of 144 accidents per year. Using a normal approximation, what is the probability of there being fewer than 130 accidents in a year?

  1. 0.117
  2. 0.125
  3. 0.136 (correct answer)
  4. 0.148
Explanation: Let XX be the number of accidents. We are given X∼Po(144)X \sim Po(144). Since λ=144\lambda=144 is large, we can approximate this with a normal distribution Y∼N(μ,σ2)Y \sim N(\mu, \sigma^2), where μ=λ=144\mu = \lambda = 144 and σ2=λ=144\sigma^2 = \lambda = 144. The standard deviation is σ=144=12\sigma = \sqrt{144} = 12. We need to find P(X<130)P(X < 130), which is equivalent to P(X≤129)P(X \le 129). Applying the continuity correction, we calculate P(Y<129.5)P(Y < 129.5). Using a GDC, P(Y<129.5)=normalcdf(−∞,129.5,144,12)≈0.136P(Y < 129.5) = \text{normalcdf}(-\infty, 129.5, 144, 12) \approx 0.136.

Question 4

A manufacturer claims that their batteries have a 99.5% success rate. A quality control check involves testing 2000 batteries. Using a Poisson approximation, what is the probability of finding exactly 10 faulty batteries?

  1. 0.0053
  2. 0.1048
  3. 0.1251 (correct answer)
  4. 0.5830
Explanation: Let XX be the number of faulty batteries. The probability of a battery being faulty is p=1−0.995=0.005p = 1 - 0.995 = 0.005. The number of trials is n=2000n = 2000. So, X∼B(2000,0.005)X \sim B(2000, 0.005). Since nn is large and pp is small, we can use a Poisson approximation. The mean is λ=np=2000×0.005=10\lambda = np = 2000 \times 0.005 = 10. So we use Y∼Po(10)Y \sim Po(10). We need to find the probability of finding exactly 10 faulty batteries, which is P(Y=10)P(Y = 10). Using a GDC, P(Y=10)=poissonpdf(10,10)≈0.1251P(Y=10) = \text{poissonpdf}(10, 10) \approx 0.1251.

Question 5

The number of flaws in a 100-metre roll of fabric follows a Poisson distribution with a standard deviation of 5\sqrt{5}. The company produces a special order of 20 of these rolls. What is the probability that the total number of flaws in the special order is less than 90?

  1. 0.125
  2. 0.149 (correct answer)
  3. 0.174
  4. 0.851
Explanation: For a Poisson distribution, the variance σ2\sigma^2 equals the mean λ\lambda. Given the standard deviation is 5\sqrt{5}, the variance is (5)2=5(\sqrt{5})^2 = 5. So, the mean number of flaws per roll is λroll=5\lambda_{roll} = 5. For a special order of 20 rolls, the total number of flaws, TT, will follow a Poisson distribution with a mean λT=20×λroll=20×5=100\lambda_T = 20 \times \lambda_{roll} = 20 \times 5 = 100. So, T∼Po(100)T \sim Po(100). We need to find P(T<90)P(T < 90), which is P(T≤89)P(T \le 89). Since λ=100\lambda=100 is large, we can use a normal approximation Y∼N(100,100)Y \sim N(100, 100). Using continuity correction, P(T≤89)P(T \le 89) becomes P(Y<89.5)P(Y < 89.5). Using a GDC, P(Y<89.5)=normalcdf(−∞,89.5,100,100)=normalcdf(−∞,89.5,100,10)≈0.1469P(Y < 89.5) = \text{normalcdf}(-\infty, 89.5, 100, \sqrt{100}) = \text{normalcdf}(-\infty, 89.5, 100, 10) \approx 0.1469.

Question 6

The number of radioactive particles detected by a Geiger counter in a 1-minute interval is a Poisson variable with mean λ\lambda. If the probability of detecting zero particles in a 1-minute interval is 0.05, what is the approximate mean number of particles detected in a 30-second interval?

  1. 1.50 (correct answer)
  2. 1.61
  3. 2.99
  4. 3.00
Explanation: Let XX be the number of particles in a 1-minute interval, so X∼Po(λ)X \sim Po(\lambda). The probability of detecting zero particles is given by P(X=0)=e−λλ00!=e−λP(X=0) = \frac{e^{-\lambda}\lambda^0}{0!} = e^{-\lambda}. We are given that P(X=0)=0.05P(X=0) = 0.05, so e−λ=0.05e^{-\lambda} = 0.05. Solving for λ\lambda, we get −λ=ln⁡(0.05)-\lambda = \ln(0.05), which gives λ=−ln⁡(0.05)≈2.9957\lambda = -\ln(0.05) \approx 2.9957. This is the mean for a 1-minute interval. The question asks for the mean number of particles in a 30-second interval. Since the rate is constant, the mean for a 30-second interval will be half of the mean for a 1-minute interval. λ30s=λ2=2.99572≈1.4978\lambda_{30s} = \frac{\lambda}{2} = \frac{2.9957}{2} \approx 1.4978. This is approximately 1.50.

Question 7

A binomial distribution is given by X∼B(n,0.08)X \sim B(n, 0.08). What is the smallest integer value of nn for which the number of successes can be reasonably approximated by a normal distribution, using the condition that both npnp and n(1−p)n(1-p) must be greater than 10?

  1. 11
  2. 12
  3. 125
  4. 126 (correct answer)
Explanation: For a normal distribution to be a reasonable approximation of a binomial distribution, we require both np>10np > 10 and n(1−p)>10n(1-p) > 10. First condition: n(0.08)>10  ⟹  n>100.08  ⟹  n>125n(0.08) > 10 \implies n > \frac{10}{0.08} \implies n > 125. Second condition: n(1−0.08)>10  ⟹  n(0.92)>10  ⟹  n>100.92  ⟹  n>10.869...n(1-0.08) > 10 \implies n(0.92) > 10 \implies n > \frac{10}{0.92} \implies n > 10.869.... Both conditions must be satisfied, so we must choose the value of nn that satisfies the stricter condition, which is n>125n > 125. The smallest integer value for nn is 126.

Question 8

In a batch of 1500 seeds, the probability of any given seed not germinating is 0.02. Let XX be the number of seeds that do not germinate. Which calculation would provide the most accurate estimate for P(X=35)P(X=35)?

  1. The exact value from the distribution X∼B(1500,0.02)X \sim B(1500, 0.02). (correct answer)
  2. Using a Poisson approximation with λ=30\lambda = 30.
  3. Using a normal approximation with μ=30\mu = 30 and σ2=29.4\sigma^2 = 29.4, finding P(34.5<Y<35.5)P(34.5 < Y < 35.5).
  4. Using a normal approximation with μ=30\mu = 30 and σ2=30\sigma^2 = 30, finding P(34.5<Y<35.5)P(34.5 < Y < 35.5).
Explanation: While approximations are useful, the most accurate value is always the exact probability calculated from the original distribution. Here, XX follows a binomial distribution B(1500,0.02)B(1500, 0.02). A GDC can calculate P(X=35)=binompdf(1500,0.02,35)P(X=35) = \text{binompdf}(1500, 0.02, 35) directly. Both Poisson (B) and Normal (C) approximations are valid and would give good estimates, but neither is as accurate as the exact binomial calculation. Choice D uses the wrong variance for the normal approximation of a binomial distribution (it uses σ2=np\sigma^2 = np instead of σ2=np(1−p)\sigma^2 = np(1-p)). Therefore, the exact binomial calculation is the most accurate.

Question 9

A factory produces a large number of computer chips, of which 2% are defective. A sample of 600 chips is selected for testing. Using a normal approximation with continuity correction, what is the probability that fewer than 10 chips in the sample are defective?

  1. 0.183
  2. 0.208
  3. 0.235 (correct answer)
  4. 0.264
Explanation: Let XX be the number of defective chips. X∼B(600,0.02)X \sim B(600, 0.02). The conditions for a normal approximation are met: np=600×0.02=12>5np = 600 \times 0.02 = 12 > 5 and n(1−p)=600×0.98=588>5n(1-p) = 600 \times 0.98 = 588 > 5. The parameters for the normal approximation YY are mean μ=np=12\mu = np = 12 and variance σ2=np(1−p)=12×0.98=11.76\sigma^2 = np(1-p) = 12 \times 0.98 = 11.76. The standard deviation is σ=11.76≈3.429\sigma = \sqrt{11.76} \approx 3.429. The probability required is P(X<10)P(X < 10), which is equivalent to P(X≤9)P(X \le 9). Applying the continuity correction, we calculate P(Y<9.5)P(Y < 9.5). Using a GDC, P(Y<9.5)=normalcdf(−∞,9.5,12,11.76)≈0.235P(Y < 9.5) = \text{normalcdf}(-\infty, 9.5, 12, \sqrt{11.76}) \approx 0.235.

Question 10

The number of typos on a website's homepage, XX, follows a Poisson distribution. The standard deviation of the number of typos is known to be 2.5. What is the probability of finding exactly 4 typos on the homepage?

  1. 0.134 (correct answer)
  2. 0.158
  3. 0.175
  4. 0.195
Explanation: For a Poisson distribution, the variance σ2\sigma^2 is equal to the mean λ\lambda. Given the standard deviation σ=2.5\sigma = 2.5, the variance is σ2=(2.5)2=6.25\sigma^2 = (2.5)^2 = 6.25. Therefore, λ=6.25\lambda = 6.25. The distribution is X∼Po(6.25)X \sim Po(6.25). We need to find the probability of exactly 4 typos, which is P(X=4)P(X=4). Using a GDC, P(X=4)=poissonpdf(6.25,4)≈0.1338P(X=4) = \text{poissonpdf}(6.25, 4) \approx 0.1338.

Question 11

A biased coin with P(Heads)=0.6P(\text{Heads}) = 0.6 is flipped 400 times. Let HH be the number of heads. Using a normal approximation, find the probability that the number of heads is between 230 and 250, inclusive.

  1. 0.789
  2. 0.811
  3. 0.832 (correct answer)
  4. 0.854
Explanation: The number of heads HH follows a binomial distribution B(400,0.6)B(400, 0.6). The conditions for a normal approximation are met since np=400×0.6=240>5np = 400 \times 0.6 = 240 > 5 and n(1−p)=400×0.4=160>5n(1-p) = 400 \times 0.4 = 160 > 5. The parameters for the normal approximation YY are μ=240\mu = 240 and σ2=400×0.6×0.4=96\sigma^2 = 400 \times 0.6 \times 0.4 = 96. So, σ=96\sigma = \sqrt{96}. We need to find P(230≤H≤250)P(230 \le H \le 250). Applying the continuity correction, we calculate P(229.5<Y<250.5)P(229.5 < Y < 250.5). Using a GDC, P(229.5<Y<250.5)=normalcdf(229.5,250.5,240,96)≈0.832P(229.5 < Y < 250.5) = \text{normalcdf}(229.5, 250.5, 240, \sqrt{96}) \approx 0.832.

Question 12

The number of accidents on a stretch of highway follows a Poisson distribution with an average of 144 accidents per year. Using a normal approximation, what is the probability of there being fewer than 130 accidents in a year?

  1. 0.117
  2. 0.125
  3. 0.136 (correct answer)
  4. 0.148
Explanation: Let XX be the number of accidents. We are given X∼Po(144)X \sim Po(144). Since λ=144\lambda=144 is large, we can approximate this with a normal distribution Y∼N(μ,σ2)Y \sim N(\mu, \sigma^2), where μ=λ=144\mu = \lambda = 144 and σ2=λ=144\sigma^2 = \lambda = 144. The standard deviation is σ=144=12\sigma = \sqrt{144} = 12. We need to find P(X<130)P(X < 130), which is equivalent to P(X≤129)P(X \le 129). Applying the continuity correction, we calculate P(Y<129.5)P(Y < 129.5). Using a GDC, P(Y<129.5)=normalcdf(−∞,129.5,144,12)≈0.136P(Y < 129.5) = \text{normalcdf}(-\infty, 129.5, 144, 12) \approx 0.136.

Question 13

A manufacturer claims that their batteries have a 99.5% success rate. A quality control check involves testing 2000 batteries. Using a Poisson approximation, what is the probability of finding exactly 10 faulty batteries?

  1. 0.0053
  2. 0.1048
  3. 0.1251 (correct answer)
  4. 0.5830
Explanation: Let XX be the number of faulty batteries. The probability of a battery being faulty is p=1−0.995=0.005p = 1 - 0.995 = 0.005. The number of trials is n=2000n = 2000. So, X∼B(2000,0.005)X \sim B(2000, 0.005). Since nn is large and pp is small, we can use a Poisson approximation. The mean is λ=np=2000×0.005=10\lambda = np = 2000 \times 0.005 = 10. So we use Y∼Po(10)Y \sim Po(10). We need to find the probability of finding exactly 10 faulty batteries, which is P(Y=10)P(Y = 10). Using a GDC, P(Y=10)=poissonpdf(10,10)≈0.1251P(Y=10) = \text{poissonpdf}(10, 10) \approx 0.1251.

Question 14

The number of radioactive particles detected by a Geiger counter in a 1-minute interval is a Poisson variable with mean λ\lambda. If the probability of detecting zero particles in a 1-minute interval is 0.05, what is the approximate mean number of particles detected in a 30-second interval?

  1. 1.50 (correct answer)
  2. 1.61
  3. 2.99
  4. 3.00
Explanation: Let XX be the number of particles in a 1-minute interval, so X∼Po(λ)X \sim Po(\lambda). The probability of detecting zero particles is given by P(X=0)=e−λλ00!=e−λP(X=0) = \frac{e^{-\lambda}\lambda^0}{0!} = e^{-\lambda}. We are given that P(X=0)=0.05P(X=0) = 0.05, so e−λ=0.05e^{-\lambda} = 0.05. Solving for λ\lambda, we get −λ=ln⁡(0.05)-\lambda = \ln(0.05), which gives λ=−ln⁡(0.05)≈2.9957\lambda = -\ln(0.05) \approx 2.9957. This is the mean for a 1-minute interval. The question asks for the mean number of particles in a 30-second interval. Since the rate is constant, the mean for a 30-second interval will be half of the mean for a 1-minute interval. λ30s=λ2=2.99572≈1.4978\lambda_{30s} = \frac{\lambda}{2} = \frac{2.9957}{2} \approx 1.4978. This is approximately 1.50.

Question 15

The number of typos on a website's homepage, XX, follows a Poisson distribution. The standard deviation of the number of typos is known to be 2.5. What is the probability of finding exactly 4 typos on the homepage?

  1. 0.134 (correct answer)
  2. 0.158
  3. 0.175
  4. 0.195
Explanation: For a Poisson distribution, the variance σ2\sigma^2 is equal to the mean λ\lambda. Given the standard deviation σ=2.5\sigma = 2.5, the variance is σ2=(2.5)2=6.25\sigma^2 = (2.5)^2 = 6.25. Therefore, λ=6.25\lambda = 6.25. The distribution is X∼Po(6.25)X \sim Po(6.25). We need to find the probability of exactly 4 typos, which is P(X=4)P(X=4). Using a GDC, P(X=4)=poissonpdf(6.25,4)≈0.1338P(X=4) = \text{poissonpdf}(6.25, 4) \approx 0.1338.

Question 16

A rare disease affects 0.1% of a large population. A researcher takes a random sample of 5000 individuals. Which of the following best justifies the use of a Poisson distribution to model the number of individuals with the disease and gives the correct parameter λ\lambda?

  1. The sample size nn is large and the probability pp is small; λ=5\lambda = 5. (correct answer)
  2. The conditions np>5np > 5 and n(1−p)>5n(1-p) > 5 are met; λ=5\lambda = 5.
  3. The sample size nn is large and the probability pp is small; λ=0.001\lambda = 0.001.
  4. The distribution is discrete and events are independent; λ=50\lambda = 50.
Explanation: The number of individuals with the disease follows a binomial distribution B(5000,0.001)B(5000, 0.001). The Poisson distribution can be used to approximate the binomial distribution when the number of trials nn is large and the probability of success pp is small. Here, n=5000n=5000 is large and p=0.001p=0.001 is small, so the approximation is appropriate. The parameter λ\lambda for the Poisson distribution is given by λ=np=5000×0.001=5\lambda = np = 5000 \times 0.001 = 5. Choice B gives the conditions for a normal approximation. Choice C uses the wrong parameter. Choice D gives general conditions but calculates an incorrect parameter.

Question 17

A biased coin with P(Heads)=0.6P(\text{Heads}) = 0.6 is flipped 400 times. Let HH be the number of heads. Using a normal approximation, find the probability that the number of heads is between 230 and 250, inclusive.

  1. 0.789
  2. 0.811
  3. 0.832 (correct answer)
  4. 0.854
Explanation: The number of heads HH follows a binomial distribution B(400,0.6)B(400, 0.6). The conditions for a normal approximation are met since np=400×0.6=240>5np = 400 \times 0.6 = 240 > 5 and n(1−p)=400×0.4=160>5n(1-p) = 400 \times 0.4 = 160 > 5. The parameters for the normal approximation YY are μ=240\mu = 240 and σ2=400×0.6×0.4=96\sigma^2 = 400 \times 0.6 \times 0.4 = 96. So, σ=96\sigma = \sqrt{96}. We need to find P(230≤H≤250)P(230 \le H \le 250). Applying the continuity correction, we calculate P(229.5<Y<250.5)P(229.5 < Y < 250.5). Using a GDC, P(229.5<Y<250.5)=normalcdf(229.5,250.5,240,96)≈0.832P(229.5 < Y < 250.5) = \text{normalcdf}(229.5, 250.5, 240, \sqrt{96}) \approx 0.832.

Question 18

A machine produces bolts with a defect rate of 0.4%. The bolts are sold in boxes of 1000. A box is rejected if it contains 8 or more defective bolts. Using a Poisson approximation, what is the probability that a randomly selected box is rejected?

  1. 0.021
  2. 0.051 (correct answer)
  3. 0.092
  4. 0.949
Explanation: Let XX be the number of defective bolts. This can be modeled by X∼B(1000,0.004)X \sim B(1000, 0.004). Since nn is large and pp is small, we use a Poisson approximation with λ=np=1000×0.004=4\lambda = np = 1000 \times 0.004 = 4. So we use Y∼Po(4)Y \sim Po(4). A box is rejected if it contains 8 or more defective bolts, which corresponds to P(Y≥8)P(Y \ge 8). This is calculated as 1−P(Y≤7)1 - P(Y \le 7). Using a GDC, P(Y≤7)=poissoncdf(4,7)≈0.9489P(Y \le 7) = \text{poissoncdf}(4, 7) \approx 0.9489. Therefore, the probability of rejection is P(Y≥8)=1−0.9489=0.0511P(Y \ge 8) = 1 - 0.9489 = 0.0511.

Question 19

The number of goals scored by a football team in a match follows a Poisson distribution with mean 2.1. The number of goals conceded by the team in a match follows an independent Poisson distribution with mean 1.1. What is the probability that a total of exactly 3 goals are scored in a match (by either team)?

  1. 0.073
  2. 0.189
  3. 0.223 (correct answer)
  4. 0.412
Explanation: Let XX be the number of goals scored, so X∼Po(2.1)X \sim Po(2.1). Let YY be the number of goals conceded, so Y∼Po(1.1)Y \sim Po(1.1). Since the variables are independent, the total number of goals in the match, T=X+YT = X+Y, follows a Poisson distribution with a mean that is the sum of the individual means: λT=λX+λY=2.1+1.1=3.2\lambda_T = \lambda_X + \lambda_Y = 2.1 + 1.1 = 3.2. We need to find the probability of exactly 3 goals in total, which is P(T=3)P(T=3). Using a GDC, P(T=3)=poissonpdf(3.2,3)≈0.2226P(T=3) = \text{poissonpdf}(3.2, 3) \approx 0.2226, which is closest to 0.223.

Question 20

The number of calls to a technical support line follows a Poisson distribution with a mean of 150 calls per day. Using a suitable approximation, find the value kk such that the probability of receiving more than kk calls in a day is approximately 0.025.

  1. 173
  2. 174 (correct answer)
  3. 175
  4. 176
Explanation: Let XX be the number of calls. X∼Po(150)X \sim Po(150). Since λ=150\lambda=150 is large, we can use a normal approximation Y∼N(150,150)Y \sim N(150, 150), with μ=150\mu=150 and σ=150\sigma = \sqrt{150}. We are looking for kk such that P(X>k)≈0.025P(X > k) \approx 0.025. This is equivalent to P(X≥k+1)P(X \ge k+1). With continuity correction, this becomes P(Y>k+0.5)P(Y > k+0.5). We need to solve P(Y>k+0.5)=0.025P(Y > k+0.5) = 0.025. This means P(Y<k+0.5)=1−0.025=0.975P(Y < k+0.5) = 1 - 0.025 = 0.975. Using the inverse normal function on a GDC, we find the value yy for which P(Y<y)=0.975P(Y < y) = 0.975: y=invNorm(0.975,150,150)≈174.00y = \text{invNorm}(0.975, 150, \sqrt{150}) \approx 174.00. So, we have k+0.5=174.00k+0.5 = 174.00, which gives k=173.5k = 173.5. Since kk must be an integer, we consider the boundary. For k=173k=173, P(X>173)≈P(Y>173.5)≈0.030P(X>173) \approx P(Y>173.5) \approx 0.030. For k=174k=174, P(X>174)≈P(Y>174.5)≈0.023P(X>174) \approx P(Y>174.5) \approx 0.023. The value closest to 0.025 is for k=174k=174.