Normal Distribution - IB Mathematics Analysis and Approaches

1. Let $X$ be a random variable with distribution $N(\mu, \sigma^2)$ and let $Z = \frac{X - \mu}{\sigma}$ be the standardized variable. Which of the following statements is not always true?
   Answer: $P(X < k) = P(Z < k)$ This question tests the understanding of the standardization process. A: The mean of a standard normal distribution is 0. This is true by definition. $E(Z) = E(\frac{X - \mu}{\sigma}) = \frac{1}{\sigma}(E(X) - \mu) = \frac{1}{\sigma}(\mu - \mu) = 0$. B: This is the definition of standardization for calculating probabilities. To find the probability for $X$, we must convert the value $k$ to its corresponding z-score. This is true. C: The normal distribution is symmetric about its mean $\mu$. Therefore, half the probability lies below the mean and half lies above. This is true. D: This statement claims that the probability of $X$ being less than $k$ is the same as the probability of $Z$ being less than $k$. This is incorrect unless $\mu=0$ and $\sigma=1$. One must use the standardized value of $k$, not $k$ itself. This is a common conceptual error.
2. A machine fills bags of sugar. The weight $W$ of sugar in a bag is normally distributed with mean 505 g and standard deviation 3 g. A bag is selected at random. What is the probability that its weight is between 500 g and 510 g?
   Answer: 0.9044 We need to find $P(500 < W < 510)$. We can use a calculator's normal cumulative distribution function (normalCdf). normalCdf(lower=500, upper=510, $\mu$=505, $\sigma$=3). Alternatively, we can use z-scores: For $W = 500$: $z_1 = \frac{500 - 505}{3} = -\frac{5}{3} \approx -1.667$. For $W = 510$: $z_2 = \frac{510 - 505}{3} = \frac{5}{3} \approx 1.667$. We need to find $P(-1.667 < Z < 1.667)$. This is $P(Z < 1.667) - P(Z < -1.667)$. Using a standard normal table or calculator: $P(Z < 1.667) \approx 0.9522$. $P(Z < -1.667) \approx 0.0478$. The probability is $0.9522 - 0.0478 = 0.9044$.
3. The weight of a product, $W$, is normally distributed with a mean of 500g and a standard deviation of 10g. A product is rejected if its weight is less than 485g. If 10 products are chosen at random, what is the probability that exactly two are rejected?
   Answer: 0.113 This is a two-step problem. First, find the probability (p) that a single product is rejected. Then, use the binomial distribution to find the probability of exactly two rejections in a sample of 10. Step 1: Find p. $p = P(W < 485)$. Standardize the value 485: $z = \frac{485 - 500}{10} = \frac{-15}{10} = -1.5$. $p = P(Z < -1.5) \approx 0.0668$. Step 2: Use the binomial distribution $X \sim B(n=10, p=0.0668)$. We want to find $P(X = 2)$. $P(X = 2) = \binom{10}{2} (0.0668)^2 (1 - 0.0668)^8$. $\binom{10}{2} = \frac{10 \times 9}{2} = 45$. $P(X = 2) = 45 \times (0.0668)^2 \times (0.9332)^8 \approx 45 \times 0.004462 \times 0.5768 \approx 0.1156$. The closest answer is 0.113.
4. Let $X \sim N(25, \sigma^2)$. The interval $(19, 31)$ is symmetric about the mean. Given that $P(19 < X < 31) = 0.86$, find the value of $\sigma$.
   Answer: 4.1 The interval $(19, 31)$ is symmetric about the mean $\mu = \frac{19+31}{2} = 25$. The total probability is 1. The area in the two tails outside this interval is $1 - 0.86 = 0.14$. Due to symmetry, the area in each tail is equal: $P(X < 19) = P(X > 31) = \frac{0.14}{2} = 0.07$. We can use $P(X < 19) = 0.07$ to find $\sigma$. First, find the corresponding z-score: $z = \text{invNorm}(0.07, 0, 1) \approx -1.476$. Using the formula $z = \frac{x - \mu}{\sigma}$: $-1.476 = \frac{19 - 25}{\sigma}$. $-1.476 = \frac{-6}{\sigma}$. $\sigma = \frac{-6}{-1.476} \approx 4.065$.
5. The random variable $X$ is normally distributed with mean 40. Given that $P(X > 50) = 0.1$, find $P(30 < X < 50)$.
   Answer: 0.8 When you encounter a normal distribution problem with symmetry properties, look for ways to use the bell curve's balanced nature to your advantage. Since $X$ is normally distributed with mean 40, the distribution is perfectly symmetric around this center point. You're given that $P(X > 50) = 0.1$, which means the area under the curve to the right of 50 is 0.1. Here's the key insight: because 50 is exactly 10 units above the mean (40), and the normal distribution is symmetric, the probability $P(X < 30)$ must also equal 0.1, since 30 is exactly 10 units below the mean. The areas in both tails are equal. Now you can find $P(30 < X < 50)$. Since the total probability under any distribution equals 1, and you know both tail areas: $P(30 < X < 50) = 1 - P(X < 30) - P(X > 50) = 1 - 0.1 - 0.1 = 0.8$ Looking at the wrong answers: A) 0.4 incorrectly assumes this covers only half the central area. B) "Cannot be determined" ignores the powerful symmetry property that makes this solvable without knowing the standard deviation. C) 0.9 mistakenly subtracts only one tail probability instead of both. The answer is D) 0.8. Strategy tip: In normal distribution problems, always check if you can exploit symmetry around the mean. When given information about one tail, you often get the opposite tail "for free" due to the bell curve's balanced shape.
6. Let $X_1 \sim N(100, 10^2)$ and $X_2 \sim N(100, 15^2)$. Which statement correctly compares the probabilities of these two distributions?
   Answer: $P(90 < X_1 < 110) > P(90 < X_2 < 110)$ Both distributions have the same mean $\mu = 100$. The interval $(90, 110)$ is symmetric about this mean. $X_1$ has a smaller standard deviation ($\sigma_1 = 10$) than $X_2$ ($\sigma_2 = 15$). A smaller standard deviation means the data is more concentrated around the mean. Therefore, the probability of finding a value within a certain range around the mean is higher for the distribution with the smaller standard deviation. Thus, $P(90 < X_1 < 110) > P(90 < X_2 < 110)$. We can also verify this by calculating z-scores. For $X_1$, the interval corresponds to $-1 < Z < 1$. For $X_2$, the interval corresponds to $-10/15 < Z < 10/15$, i.e., $-0.67 < Z < 0.67$. The area from -1 to 1 is greater than the area from -0.67 to 0.67.
7. The scores on a test, $S$, are normally distributed with a mean of 75 and a standard deviation of 8. A student's score of 65 has a z-score of $z_1$, and a score of 90 has a z-score of $z_2$. Find the value of $z_2 - z_1$.
   Answer: 3.125 First, calculate the z-score for each score using the formula $z = \frac{s - \mu}{\sigma}$. For a score of 65: $z_1 = \frac{65 - 75}{8} = \frac{-10}{8} = -1.25$. For a score of 90: $z_2 = \frac{90 - 75}{8} = \frac{15}{8} = 1.875$. The question asks for the value of $z_2 - z_1$. $z_2 - z_1 = 1.875 - (-1.25) = 1.875 + 1.25 = 3.125$.
8. Let $X \sim N(50, 4^2)$. Find the value $k$ such that $P(X > k) = 0.3$.
   Answer: 52.1 We are given $P(X > k) = 0.3$. The inverse normal function on most calculators works with the area to the left, so we need $P(X < k) = 1 - 0.3 = 0.7$. We need to find the value $k$ corresponding to a cumulative probability of 0.7. Using a calculator: $k = \text{invNorm}(\text{area}=0.7, \mu=50, \sigma=4) \approx 52.098$. Alternatively, using z-scores: Find the z-score for a cumulative probability of 0.7: $z = \text{invNorm}(0.7, 0, 1) \approx 0.5244$. Then use $k = \mu + z\sigma$: $k = 50 + (0.5244)(4) = 50 + 2.0976 = 52.0976 \approx 52.1$.
9. The scores for two exams, Mathematics and Physics, are normally distributed. For Mathematics, scores $M$ follow $N(70, 12^2)$. For Physics, scores $P$ follow $N(72, 8^2)$. To receive a distinction, a student must score above 85. In which subject is it more likely for a student to receive a distinction?
   Answer: Mathematics, because its corresponding z-score for a score of 85 is smaller. When comparing probabilities across different normal distributions, you need to standardize the values using z-scores to make meaningful comparisons. The key insight is that smaller z-scores (closer to the mean) correspond to higher probabilities in the tail beyond that point. To find which subject makes distinction more likely, calculate the z-score for a score of 85 in each subject. For Mathematics: $z_M = \frac{85 - 70}{12} = \frac{15}{12} = 1.25$. For Physics: $z_P = \frac{85 - 72}{8} = \frac{13}{8} = 1.625$. Since Mathematics has the smaller z-score (1.25 < 1.625), a score of 85 is closer to the mean in Mathematics, meaning more area lies above this point. Therefore, distinction is more likely in Mathematics. Option A incorrectly suggests that larger standard deviation alone makes distinction more likely, but you must consider how far 85 is from each mean relative to the spread. Option B states the opposite of what's true - Physics actually has the larger z-score, not smaller. Option C focuses only on the higher mean in Physics (72 vs 70) but ignores that Physics also has much smaller variability (σ = 8 vs 12), making scores above 85 relatively rarer despite the higher mean. Remember: when comparing probabilities across different normal distributions, always standardize using z-scores. The distribution with the smaller z-score for your threshold value will have the higher probability of exceeding that threshold.
10. The random variable $X$ is normally distributed with a mean $\mu$ and standard deviation $\sigma$. If $P(X < 88) = P(X > 112)$, what is the value of $\mu$?
    Answer: 100 The condition $P(X < a) = P(X > b)$ for a normal distribution implies that the values $a$ and $b$ are symmetric with respect to the mean $\mu$. The mean must therefore be the midpoint of $a$ and $b$. In this case, $a = 88$ and $b = 112$. $\mu = \frac{88 + 112}{2} = \frac{200}{2} = 100$.
11. The heights of students in a school are normally distributed. Let $H$ be the height of a randomly chosen student. It is known that $P(H > 180) = 0.25$ and $P(H < 165) = 0.35$. Find the conditional probability $P(H > 180 | H > 165)$.
    Answer: 0.385 We need to find $P(H > 180 | H > 165)$. Using the formula for conditional probability, $P(A|B) = \frac{P(A \cap B)}{P(B)}$. Here, event A is $H > 180$ and event B is $H > 165$. The intersection $A \cap B$ is the event that $H > 180$ AND $H > 165$, which simplifies to just $H > 180$. So, we need to calculate $\frac{P(H > 180)}{P(H > 165)}$. We are given $P(H > 180) = 0.25$. We are given $P(H < 165) = 0.35$, so $P(H > 165) = 1 - P(H < 165) = 1 - 0.35 = 0.65$. Therefore, the conditional probability is $\frac{0.25}{0.65} \approx 0.3846$.
12. The random variable $X$ follows a normal distribution $X \sim N(120, \sigma^2)$. Given that $P(X < 135) = 0.92$, find the standard deviation $\sigma$.
    Answer: 10.6 First, we find the z-score corresponding to a cumulative probability of 0.92. Using an inverse normal function on a calculator, $z = \text{invNorm}(0.92, 0, 1) \approx 1.405$. The formula for the z-score is $z = \frac{x - \mu}{\sigma}$. Substitute the known values: $1.405 = \frac{135 - 120}{\sigma}$. This simplifies to $1.405 = \frac{15}{\sigma}$. Solving for $\sigma$ gives $\sigma = \frac{15}{1.405} \approx 10.67$.
13. The lifetimes of a brand of battery are normally distributed with a standard deviation of 30 hours. It is found that 18% of batteries have a lifetime of less than 450 hours. Find the mean lifetime, $\mu$, of the batteries.
    Answer: 477 hours We are given $P(X < 450) = 0.18$. First, find the z-score corresponding to this probability: $z = \text{invNorm}(0.18, 0, 1) \approx -0.915$. Using the standardization formula $z = \frac{x - \mu}{\sigma}$, we have $-0.915 = \frac{450 - \mu}{30}$. Multiply by 30: $-0.915 \times 30 = 450 - \mu$, which gives $-27.45 = 450 - \mu$. Solving for $\mu$: $\mu = 450 + 27.45 = 477.45$. Thus, the mean lifetime is approximately 477 hours.
14. Let $X \sim N(\mu, \sigma^2)$. Given that $P(X < \mu - a) = 0.3$, find $P(\mu - a < X < \mu + a)$.
    Answer: 0.4 The normal distribution is symmetric about the mean $\mu$. Given $P(X < \mu - a) = 0.3$. By symmetry, the probability in the upper tail is the same: $P(X > \mu + a) = P(X < \mu - a) = 0.3$. The total probability under the curve is 1. The probability we want is the area between the two tails. So, $P(\mu - a < X < \mu + a) = 1 - P(X < \mu - a) - P(X > \mu + a)$. $P(\mu - a < X < \mu + a) = 1 - 0.3 - 0.3 = 0.4$.
15. The random variable $X \sim N(\mu, \sigma^2)$. The value $x_1 = 80$ has a z-score of $-0.5$. The value $x_2 = 110$ has a z-score of $2.0$. Find the mean, $\mu$.
    Answer: 86 We can set up a system of two linear equations using the z-score formula $z = (x - \mu) / \sigma$. 1) $-0.5 = \frac{80 - \mu}{\sigma} \implies -0.5\sigma = 80 - \mu$ 2) $2.0 = \frac{110 - \mu}{\sigma} \implies 2.0\sigma = 110 - \mu$ From equation (1), we can write $\mu = 80 + 0.5\sigma$. From equation (2), we can write $\mu = 110 - 2.0\sigma$. Set the expressions for $\mu$ equal to each other: $80 + 0.5\sigma = 110 - 2.0\sigma$ $2.5\sigma = 30$ $\sigma = \frac{30}{2.5} = 12$. Now substitute the value of $\sigma$ back into one of the equations for $\mu$: $\mu = 80 + 0.5(12) = 80 + 6 = 86$.
16. The masses of apples from a certain orchard are normally distributed with a mean of 150 g and a standard deviation of 18 g. What is the interquartile range of the masses?
    Answer: 24.3 g The interquartile range (IQR) is the difference between the third quartile ($Q_3$) and the first quartile ($Q_1$). $Q_1$ is the value for which $P(X < Q_1) = 0.25$. The corresponding z-score is $z_1 = \text{invNorm}(0.25) \approx -0.6745$. $Q_3$ is the value for which $P(X < Q_3) = 0.75$. The corresponding z-score is $z_3 = \text{invNorm}(0.75) \approx 0.6745$. We calculate $Q_1$ and $Q_3$ using $x = \mu + z\sigma$: $Q_1 = 150 + (-0.6745)(18) \approx 150 - 12.141 = 137.859$ $Q_3 = 150 + (0.6745)(18) \approx 150 + 12.141 = 162.141$ $IQR = Q_3 - Q_1 = 162.141 - 137.859 = 24.282$. Alternatively, $IQR = (z_3 - z_1)\sigma = (0.6745 - (-0.6745)) \times 18 = 2 \times 0.6745 \times 18 \approx 24.3$ g.
17. The length of a fish, $L$, is normally distributed with a mean of 30 cm. It is known that the 95th percentile for the length is 38 cm. Find the standard deviation of the lengths.
    Answer: 4.86 cm The 95th percentile means that $P(L < 38) = 0.95$. We need to find the z-score that corresponds to a cumulative probability of 0.95. Using an inverse normal function, $z = \text{invNorm}(0.95, 0, 1) \approx 1.645$. Now we use the standardization formula $z = \frac{L - \mu}{\sigma}$. $1.645 = \frac{38 - 30}{\sigma}$ $1.645 = \frac{8}{\sigma}$ $\sigma = \frac{8}{1.645} \approx 4.86$ cm.
18. For a standard normal variable $Z \sim N(0, 1)$, the statement $P(Z > c) = P(Z < -c)$ illustrates its symmetry. Which of the following statements correctly expresses this same property of symmetry for a general normal variable $X \sim N(\mu, \sigma^2)$?
    Answer: $P(X > \mu + c) = P(X < \mu - c)$ for any constant c > 0. When you encounter questions about normal distribution symmetry, focus on how the center of symmetry changes from the standard normal to a general normal distribution. For the standard normal $Z \sim N(0, 1)$, the distribution is symmetric around 0, so $P(Z > c) = P(Z < -c)$ means equal areas exist at equal distances from the center. For a general normal variable $X \sim N(\mu, \sigma^2)$, the center of symmetry shifts to the mean $\mu$, not zero. The correct symmetry property becomes: $P(X > \mu + c) = P(X < \mu - c)$ for any constant $c > 0$. This states that the probability of being $c$ units above the mean equals the probability of being $c$ units below the mean, which perfectly captures the symmetric nature around $\mu$. This makes answer D correct. Let's examine why the other options fail: Option A incorrectly assumes the distribution is still symmetric around zero, ignoring that the mean has shifted to $\mu$. Option B has a logical error - it states "for any constant c" but doesn't actually involve $c$ in the probability statement, and the statement $P(X > \mu) = P(X < \mu)$ is always true regardless of $c$. Option C simply expresses the complement rule ($P(A) = 1 - P(A^c)$), which applies to any random variable and has nothing to do with symmetry. Study tip: For normal distribution symmetry questions, always identify the center first - it's 0 for standard normal, but $\mu$ for general normal distributions. The symmetry property must reflect equal distances from this center.

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