Further Distributions - IB Mathematics Applications and Interpretation

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?
   Answer: 0.055 The average rate is 5 calls per 10 minutes, so the rate for a 5-minute period is $\lambda = 2.5$. Let $X_1$ be the number of calls in the first 5-minute period and $X_2$ be the number of calls in the second. We assume $X_1, X_2 \sim Po(2.5)$ and that they are independent. We want to find $P(X_1=2 \text{ and } X_2=3)$. Because of independence, this is $P(X_1=2) \times P(X_2=3)$. Using a GDC: $P(X_1=2) = \text{poissonpdf}(2.5, 2) \approx 0.2565$ and $P(X_2=3) = \text{poissonpdf}(2.5, 3) \approx 0.2138$. The combined probability is $0.2565 \times 0.2138 \approx 0.0548$, which is closest to 0.055.
2. 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)?
   Answer: 0.223 Let $X$ be the number of goals scored, so $X \sim Po(2.1)$. Let $Y$ be the number of goals conceded, so $Y \sim Po(1.1)$. Since the variables are independent, the total number of goals in the match, $T = X+Y$, follows a Poisson distribution with a mean that is the sum of the individual means: $\lambda_T = \lambda_X + \lambda_Y = 2.1 + 1.1 = 3.2$. We need to find the probability of exactly 3 goals in total, which is $P(T=3)$. Using a GDC, $P(T=3) = \text{poissonpdf}(3.2, 3) \approx 0.2226$, which is closest to 0.223.
3. 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 $\lambda$?
   Answer: The sample size $n$ is large and the probability $p$ is small; $\lambda = 5$. The number of individuals with the disease follows a binomial distribution $B(5000, 0.001)$. The Poisson distribution can be used to approximate the binomial distribution when the number of trials $n$ is large and the probability of success $p$ is small. Here, $n=5000$ is large and $p=0.001$ is small, so the approximation is appropriate. The parameter $\lambda$ for the Poisson distribution is given by $\lambda = np = 5000 \times 0.001 = 5$. 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.
4. 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?
   Answer: 0.235 Let $X$ be the number of defective chips. $X \sim B(600, 0.02)$. The conditions for a normal approximation are met: $np = 600 \times 0.02 = 12 > 5$ and $n(1-p) = 600 \times 0.98 = 588 > 5$. The parameters for the normal approximation $Y$ are mean $\mu = np = 12$ and variance $\sigma^2 = np(1-p) = 12 \times 0.98 = 11.76$. The standard deviation is $\sigma = \sqrt{11.76} \approx 3.429$. The probability required is $P(X < 10)$, which is equivalent to $P(X \le 9)$. Applying the continuity correction, we calculate $P(Y < 9.5)$. Using a GDC, $P(Y < 9.5) = \text{normalcdf}(-\infty, 9.5, 12, \sqrt{11.76}) \approx 0.235$.
5. A binomial distribution is given by $X \sim B(n, 0.08)$. What is the smallest integer value of $n$ for which the number of successes can be reasonably approximated by a normal distribution, using the condition that both $np$ and $n(1-p)$ must be greater than 10?
   Answer: 126 For a normal distribution to be a reasonable approximation of a binomial distribution, we require both $np > 10$ and $n(1-p) > 10$. First condition: $n(0.08) > 10 \implies n > \frac{10}{0.08} \implies n > 125$. Second condition: $n(1-0.08) > 10 \implies n(0.92) > 10 \implies n > \frac{10}{0.92} \implies n > 10.869...$. Both conditions must be satisfied, so we must choose the value of $n$ that satisfies the stricter condition, which is $n > 125$. The smallest integer value for $n$ is 126.
6. A student is taking a multiple-choice test with 200 questions, where each question has 5 possible answers. The student guesses randomly on every question. Using a normal approximation, what is the probability that the student gets at least 50 questions correct?
   Answer: 0.057 Let $X$ be the number of correct answers. The probability of guessing a question correctly is $p = 1/5 = 0.2$. The number of trials is $n = 200$. Thus, $X \sim B(200, 0.2)$. We check the conditions for normal approximation: $np = 200 \times 0.2 = 40 > 5$ and $n(1-p) = 200 \times 0.8 = 160 > 5$. The conditions are met. The parameters for the normal approximation $Y$ are $\mu = np = 40$ and $\sigma^2 = np(1-p) = 40 \times 0.8 = 32$. The standard deviation is $\sigma = \sqrt{32}$. We need to find $P(X \ge 50)$. With continuity correction, this becomes $P(Y > 49.5)$. Using a GDC, $P(Y > 49.5) = \text{normalcdf}(49.5, \infty, 40, \sqrt{32}) \approx 0.057$.
7. The number of flaws in a 100-metre roll of fabric follows a Poisson distribution with a standard deviation of $\sqrt{5}$. 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?
   Answer: 0.149 For a Poisson distribution, the variance $\sigma^2$ equals the mean $\lambda$. Given the standard deviation is $\sqrt{5}$, the variance is $(\sqrt{5})^2 = 5$. So, the mean number of flaws per roll is $\lambda_{roll} = 5$. For a special order of 20 rolls, the total number of flaws, $T$, will follow a Poisson distribution with a mean $\lambda_T = 20 \times \lambda_{roll} = 20 \times 5 = 100$. So, $T \sim Po(100)$. We need to find $P(T < 90)$, which is $P(T \le 89)$. Since $\lambda=100$ is large, we can use a normal approximation $Y \sim N(100, 100)$. Using continuity correction, $P(T \le 89)$ becomes $P(Y < 89.5)$. Using a GDC, $P(Y < 89.5) = \text{normalcdf}(-\infty, 89.5, 100, \sqrt{100}) = \text{normalcdf}(-\infty, 89.5, 100, 10) \approx 0.1469$.
8. 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?
   Answer: 0.051 Let $X$ be the number of defective bolts. This can be modeled by $X \sim B(1000, 0.004)$. Since $n$ is large and $p$ is small, we use a Poisson approximation with $\lambda = np = 1000 \times 0.004 = 4$. So we use $Y \sim Po(4)$. A box is rejected if it contains 8 or more defective bolts, which corresponds to $P(Y \ge 8)$. This is calculated as $1 - P(Y \le 7)$. Using a GDC, $P(Y \le 7) = \text{poissoncdf}(4, 7) \approx 0.9489$. Therefore, the probability of rejection is $P(Y \ge 8) = 1 - 0.9489 = 0.0511$.
9. For a random variable $X \sim B(n, p)$, a normal approximation is considered valid if $np > 5$ and $n(1-p) > 5$. For which of the following distributions is this approximation not valid?
   Answer: $X \sim B(90, 0.05)$ We check the conditions $np > 5$ and $n(1-p) > 5$ for each option. A: For $X \sim B(60, 0.1)$, $np = 60 \times 0.1 = 6 > 5$ and $n(1-p) = 60 \times 0.9 = 54 > 5$. The approximation is valid. B: For $X \sim B(100, 0.92)$, $np = 100 \times 0.92 = 92 > 5$ and $n(1-p) = 100 \times 0.08 = 8 > 5$. The approximation is valid. C: For $X \sim B(30, 0.7)$, $np = 30 \times 0.7 = 21 > 5$ and $n(1-p) = 30 \times 0.3 = 9 > 5$. The approximation is valid. D: For $X \sim B(90, 0.05)$, $np = 90 \times 0.05 = 4.5$. Since $np$ is not greater than 5, the normal approximation is not valid for this distribution.
10. In a batch of 1500 seeds, the probability of any given seed not germinating is 0.02. Let $X$ be the number of seeds that do not germinate. Which calculation would provide the most accurate estimate for $P(X=35)$?
    Answer: The exact value from the distribution $X \sim B(1500, 0.02)$. While approximations are useful, the most accurate value is always the exact probability calculated from the original distribution. Here, $X$ follows a binomial distribution $B(1500, 0.02)$. A GDC can calculate $P(X=35) = \text{binompdf}(1500, 0.02, 35)$ 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 $\sigma^2 = np$ instead of $\sigma^2 = np(1-p)$). Therefore, the exact binomial calculation is the most accurate.
11. A biased coin with $P(\text{Heads}) = 0.6$ is flipped 400 times. Let $H$ be the number of heads. Using a normal approximation, find the probability that the number of heads is between 230 and 250, inclusive.
    Answer: 0.832 The number of heads $H$ follows a binomial distribution $B(400, 0.6)$. The conditions for a normal approximation are met since $np = 400 \times 0.6 = 240 > 5$ and $n(1-p) = 400 \times 0.4 = 160 > 5$. The parameters for the normal approximation $Y$ are $\mu = 240$ and $\sigma^2 = 400 \times 0.6 \times 0.4 = 96$. So, $\sigma = \sqrt{96}$. We need to find $P(230 \le H \le 250)$. Applying the continuity correction, we calculate $P(229.5 < Y < 250.5)$. Using a GDC, $P(229.5 < Y < 250.5) = \text{normalcdf}(229.5, 250.5, 240, \sqrt{96}) \approx 0.832$.
12. 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 $k$ such that the probability of receiving more than $k$ calls in a day is approximately 0.025.
    Answer: 174 Let $X$ be the number of calls. $X \sim Po(150)$. Since $\lambda=150$ is large, we can use a normal approximation $Y \sim N(150, 150)$, with $\mu=150$ and $\sigma = \sqrt{150}$. We are looking for $k$ such that $P(X > k) \approx 0.025$. This is equivalent to $P(X \ge k+1)$. With continuity correction, this becomes $P(Y > k+0.5)$. We need to solve $P(Y > k+0.5) = 0.025$. This means $P(Y < k+0.5) = 1 - 0.025 = 0.975$. Using the inverse normal function on a GDC, we find the value $y$ for which $P(Y < y) = 0.975$: $y = \text{invNorm}(0.975, 150, \sqrt{150}) \approx 174.00$. So, we have $k+0.5 = 174.00$, which gives $k = 173.5$. Since $k$ must be an integer, we consider the boundary. For $k=173$, $P(X>173) \approx P(Y>173.5) \approx 0.030$. For $k=174$, $P(X>174) \approx P(Y>174.5) \approx 0.023$. The value closest to 0.025 is for $k=174$.
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?
    Answer: 0.1251 Let $X$ be the number of faulty batteries. The probability of a battery being faulty is $p = 1 - 0.995 = 0.005$. The number of trials is $n = 2000$. So, $X \sim B(2000, 0.005)$. Since $n$ is large and $p$ is small, we can use a Poisson approximation. The mean is $\lambda = np = 2000 \times 0.005 = 10$. So we use $Y \sim Po(10)$. We need to find the probability of finding exactly 10 faulty batteries, which is $P(Y = 10)$. Using a GDC, $P(Y=10) = \text{poissonpdf}(10, 10) \approx 0.1251$.
14. 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?
    Answer: 0.229 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 $\lambda = 0.75 \times 400 = 300$. Let $X$ be the number of errors, so $X \sim Po(300)$. Since $\lambda = 300$ is large, we can use a normal approximation, $Y \sim N(300, 300)$. The mean is $\mu = 300$ and the standard deviation is $\sigma = \sqrt{300}$. We need to find $P(X \ge 310)$. With continuity correction, this becomes $P(Y > 309.5)$. Using a GDC, $P(Y > 309.5) = \text{normalcdf}(309.5, \infty, 300, \sqrt{300}) \approx 0.229$.
15. Let $X \sim Po(100)$. Which of the following calculations gives the best approximation for $P(X > 115)$?
    Answer: $P(Y > 115.5)$ where $Y \sim N(100, 100)$ We are approximating a Poisson distribution $X \sim Po(100)$ with a normal distribution $Y$. The parameters for the normal distribution are $\mu = \lambda = 100$ and $\sigma^2 = \lambda = 100$. Note that the normal distribution notation is $N(\mu, \sigma^2)$, not $N(\mu, \sigma)$. So, $Y \sim N(100, 100)$. We need to approximate $P(X > 115)$, which is equivalent to $P(X \ge 116)$. When applying the continuity correction for a discrete distribution being approximated by a continuous one, the interval for $X \ge 116$ corresponds to $Y > 115.5$. Therefore, the best approximation is $P(Y > 115.5)$ where $Y \sim N(100, 100)$.
16. The number of radioactive particles detected by a Geiger counter in a 1-minute interval is a Poisson variable with mean $\lambda$. 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?
    Answer: 1.50 Let $X$ be the number of particles in a 1-minute interval, so $X \sim Po(\lambda)$. The probability of detecting zero particles is given by $P(X=0) = \frac{e^{-\lambda}\lambda^0}{0!} = e^{-\lambda}$. We are given that $P(X=0) = 0.05$, so $e^{-\lambda} = 0.05$. Solving for $\lambda$, we get $-\lambda = \ln(0.05)$, which gives $\lambda = -\ln(0.05) \approx 2.9957$. 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. $\lambda_{30s} = \frac{\lambda}{2} = \frac{2.9957}{2} \approx 1.4978$. This is approximately 1.50.
17. 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?
    Answer: 0.136 Let $X$ be the number of accidents. We are given $X \sim Po(144)$. Since $\lambda=144$ is large, we can approximate this with a normal distribution $Y \sim N(\mu, \sigma^2)$, where $\mu = \lambda = 144$ and $\sigma^2 = \lambda = 144$. The standard deviation is $\sigma = \sqrt{144} = 12$. We need to find $P(X < 130)$, which is equivalent to $P(X \le 129)$. Applying the continuity correction, we calculate $P(Y < 129.5)$. Using a GDC, $P(Y < 129.5) = \text{normalcdf}(-\infty, 129.5, 144, 12) \approx 0.136$.
18. The number of typos on a website's homepage, $X$, 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?
    Answer: 0.134 For a Poisson distribution, the variance $\sigma^2$ is equal to the mean $\lambda$. Given the standard deviation $\sigma = 2.5$, the variance is $\sigma^2 = (2.5)^2 = 6.25$. Therefore, $\lambda = 6.25$. The distribution is $X \sim Po(6.25)$. We need to find the probability of exactly 4 typos, which is $P(X=4)$. Using a GDC, $P(X=4) = \text{poissonpdf}(6.25, 4) \approx 0.1338$.

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