Normal Distribution Extensions - IB Mathematics Analysis and Approaches

1. The lifetime of a particular type of battery is a random variable with a mean of 400 hours and a standard deviation of 50 hours. The distribution of the lifetime is unknown. A random sample of 100 batteries is selected. Which of the following allows for the calculation of the approximate probability that the sample mean lifetime is greater than 410 hours?
   Answer: The Central Limit Theorem. The problem asks for the probability related to the sample mean $\bar{X}$. The distribution of the original population is explicitly stated as unknown. Therefore, we cannot assume it is normal.\ However, the sample size $n=100$ is large (typically $n \ge 30$ is considered sufficient).\ The Central Limit Theorem (CLT) states that for a sufficiently large sample size, the sampling distribution of the sample mean $\bar{X}$ will be approximately normal, regardless of the shape of the population distribution. This is precisely the theorem that justifies approximating the distribution of $\bar{X}$ with a normal distribution to calculate the required probability.\ * A: The assumption that the population distribution is normal. If this were true, the distribution of $\bar{X}$ would be exactly normal, not approximately. The CLT allows us to proceed without this assumption. * C: The Law of Large Numbers. This law states that as the sample size grows, the sample mean $\bar{X}$ converges to the population mean $\mu$. It describes the long-term behavior of the mean, but not its sampling distribution for a finite sample. * D: The use of a continuity correction. This is used when approximating a discrete probability distribution (like Binomial or Poisson) with a continuous one (like Normal). Battery lifetime is a continuous variable, so this correction is not applicable.
2. The time taken for a student to complete a puzzle is normally distributed with a mean of 15 minutes and a standard deviation of 4 minutes. A teacher takes a random sample of $n$ students. The probability that the mean time for the sample is less than 14 minutes is found to be 0.0228. Find the sample size $n$.
   Answer: 64 Let $X$ be the time for a single student. $X \sim N(15, 4^2)$.\ The sample mean $\bar{X}$ follows the distribution $\bar{X} \sim N(15, \frac{4^2}{n})$. The standard error is $\sigma_{\bar{X}} = \frac{4}{\sqrt{n}}$.\ We are given $P(\bar{X} < 14) = 0.0228$.\ We first find the z-score corresponding to a cumulative probability of 0.0228. This is a standard value, $z = -2$.\ Now we use the standardization formula: $z = \frac{\bar{x} - \mu}{\sigma_{\bar{X}}}$.\ $-2 = \frac{14 - 15}{\frac{4}{\sqrt{n}}}$.\ $-2 = \frac{-1}{\frac{4}{\sqrt{n}}}$.\ $2 = \frac{\sqrt{n}}{4}$.\ $\sqrt{n} = 2 \times 4 = 8$.\ $n = 8^2 = 64$.\ * A: 4 This can result from the algebraic error $\frac{\sqrt{n}}{4} = \frac{1}{2} \implies \sqrt{n}=2 \implies n=4$ or from $\frac{16}{n} = z^2 = 4 \implies n=4$. * B: 8 This is the value of $\sqrt{n}$, obtained by forgetting the final step of squaring. * C: 16 This would give a standard error of $4/\sqrt{16} = 1$, leading to a Z-score of $(14-15)/1 = -1$, which corresponds to a probability of 0.1587, not 0.0228.
3. A biased coin with a probability of landing heads of 0.6 is tossed 100 times. Using a normal approximation, what is the probability of obtaining at least 70 heads?
   Answer: 0.026 Let $X$ be the number of heads. $X$ follows a binomial distribution, $X \sim B(100, 0.6)$.\ We can approximate this with a normal distribution $Y \sim N(\mu, \sigma^2)$.\ The mean is $\mu = np = 100 \times 0.6 = 60$.\ The variance is $\sigma^2 = np(1-p) = 100 \times 0.6 \times 0.4 = 24$.\ The standard deviation is $\sigma = \sqrt{24} \approx 4.899$.\ We want to find $P(X \ge 70)$. Since we are approximating a discrete distribution with a continuous one, we must use a continuity correction.\ $P(X \ge 70)$ becomes $P(Y > 69.5)$.\ Standardizing the value: $Z = \frac{69.5 - 60}{\sqrt{24}} = \frac{9.5}{4.899} \approx 1.939$.\ We need to find $P(Z > 1.939)$, which is $1 - P(Z < 1.939)$. Using a calculator, this is approximately $1 - 0.9737 = 0.0263$.\ * A: 0.016 is the result of using an incorrect continuity correction, $P(Y > 70.5)$. * B: 0.021 is the result of not using a continuity correction, i.e., calculating $P(Y > 70)$. * D: 0.033 is the result of misinterpreting 'at least 70' as 'more than 69' and using an incorrect continuity correction of $P(Y > 69)$ or mixing up the boundary.
4. A machine produces bolts with lengths that are normally distributed with a standard deviation of 0.1 mm. The mean length, $\mu$, is adjustable. For a batch to be accepted, the probability that the mean length of a sample of 25 bolts is within 0.05 mm of $\mu$ must be at least 0.99. What is the maximum possible value of the standard deviation of the population for this condition to be met?
   Answer: 0.0970 mm Let $X$ be the length of a bolt, $X \sim N(\mu, \sigma^2)$. We are given $n=25$. The sample mean $\bar{X}$ is distributed as $\bar{X} \sim N(\mu, \sigma^2/25)$. The standard error is $\sigma_{\bar{X}} = \sigma/\sqrt{25} = \sigma/5$.\ We require the probability that the sample mean is 'within 0.05 mm of $\mu$' to be at least 0.99. This is written as $P(|\bar{X} - \mu| \le 0.05) \ge 0.99$.\ This is equivalent to $P(-0.05 \le \bar{X} - \mu \le 0.05) \ge 0.99$.\ To standardize, we divide by the standard error $\sigma/5$:\ $P(\frac{-0.05}{\sigma/5} \le Z \le \frac{0.05}{\sigma/5}) \ge 0.99$.\ Let $z_c = \frac{0.05}{\sigma/5} = \frac{0.25}{\sigma}$. We need $P(-z_c \le Z \le z_c) \ge 0.99$.\ This means the area in the two tails is at most $1 - 0.99 = 0.01$. The area in the upper tail is at most 0.005. So $P(Z > z_c) \le 0.005$.\ This implies $P(Z < z_c) \ge 0.995$.\ Using the inverse normal function, the critical z-score for a cumulative probability of 0.995 is $z_{crit} \approx 2.5758$.\ So we must have $z_c \ge 2.5758$.\ $\frac{0.25}{\sigma} \ge 2.5758$.\ $\sigma \le \frac{0.25}{2.5758} \approx 0.09705$.\ The maximum possible value for the standard deviation $\sigma$ is approximately 0.0970 mm.\ * A: 0.0194 is $0.05 / 2.5758$, from forgetting to use the standard error $\sigma/5$. * C: 0.4852 might come from an algebraic error, e.g., $\sigma \le 0.25 \times 1.96$ using the wrong Z-score for 95% confidence. * D: 2.5758 is the Z-score itself, not the standard deviation.
5. Let $X_1, X_2, \dots, X_n$ be $n$ independent random variables, each with distribution $N(\mu, \sigma^2)$. Let $S = \sum_{i=1}^n X_i$ be the sum and $\bar{X} = \frac{1}{n}S$ be the sample mean. Which of the following statements is incorrect?
   Answer: $Var(S) = n^2\sigma^2$ Let's analyze each statement based on the properties of expectation and variance.\ * **A: $E(S) = E(\sum_{i=1}^n X_i) = \sum_{i=1}^n E(X_i) = \sum_{i=1}^n \mu = n\mu$.** This statement is correct.\ * **B: $Var(S) = Var(\sum_{i=1}^n X_i)$.** Because the variables are independent, $Var(\sum_{i=1}^n X_i) = \sum_{i=1}^n Var(X_i) = \sum_{i=1}^n \sigma^2 = n\sigma^2$.** This statement is correct.\ * **C: $Var(\bar{X}) = Var(\frac{1}{n}S) = (\frac{1}{n})^2 Var(S) = \frac{1}{n^2} (n\sigma^2) = \frac{\sigma^2}{n}$.** This statement is correct.\ * **D: $Var(S) = n^2\sigma^2$.** This statement is incorrect. It confuses the variance of a sum of independent variables with the variance of a multiple of a single variable, i.e., $Var(nX_1) = n^2 Var(X_1) = n^2\sigma^2$. The correct formula for the sum is $n\sigma^2$ as shown in B.
6. Two independent random variables $X$ and $Y$ are normally distributed, such that $X \sim N(100, 12^2)$ and $Y \sim N(95, 5^2)$. Find the probability $P(X > Y)$, correct to three significant figures.
   Answer: 0.650 To find $P(X > Y)$, we consider the difference $D = X - Y$. The distribution of $D$ is also normal.\ $E(D) = E(X) - E(Y) = 100 - 95 = 5$. Since $X$ and $Y$ are independent, the variances add: $Var(D) = Var(X) + Var(Y) = 12^2 + 5^2 = 144 + 25 = 169$. The standard deviation of $D$ is $\sigma_D = \sqrt{169} = 13$.\ So, $D \sim N(5, 13^2)$.\ The probability $P(X > Y)$ is equivalent to $P(D > 0)$.\ We standardize this value: $Z = \frac{0 - 5}{13} = -\frac{5}{13} \approx -0.3846$.\ We need to find $P(Z > -0.3846)$. Using a calculator's normal cumulative distribution function, this is approximately 0.6498.\ Rounded to three significant figures, the probability is 0.650.\ * A: 0.350 is $1 - 0.650$, which corresponds to $P(X < Y)$. * B: 0.616 is the result of incorrectly adding standard deviations ($\sigma_D = 12 + 5 = 17$) instead of variances. * D: 0.677 is the result of incorrectly subtracting variances ($Var(D) = 144 - 25 = 119$) because the variable is a difference.
7. In a large population, 40% of people have blood type A. A random sample of 200 people is selected. Using a normal approximation with continuity correction, which expression calculates the probability that between 75 and 90 people (inclusive) in the sample have blood type A?
   Answer: $P(74.5 < Y < 90.5)$, where $Y \sim N(80, 48)$ Let $X$ be the number of people with blood type A in the sample. This follows a binomial distribution $X \sim B(n, p)$ with $n=200$ and $p=0.4$.\ To use a normal approximation, we find the mean and variance:\ Mean: $\mu = np = 200 \times 0.4 = 80$.\ Variance: $\sigma^2 = np(1-p) = 200 \times 0.4 \times 0.6 = 48$.\ So we approximate $X$ with a normal distribution $Y \sim N(80, 48)$.\ The question asks for the probability that the number of people is between 75 and 90 inclusive, which is $P(75 \le X \le 90)$.\ When applying the continuity correction for an inclusive range $[a, b]$, the interval becomes $(a-0.5, b+0.5)$.\ Therefore, $P(75 \le X \le 90)$ is approximated by $P(74.5 < Y < 90.5)$.\ Combining these, the correct expression is $P(74.5 < Y < 90.5)$, where $Y \sim N(80, 48)$.\ * A: This expression does not use a continuity correction. * C: This expression uses an incorrect continuity correction, which would be appropriate for a strict inequality $P(76 < X < 89)$. * D: This expression calculates the variance incorrectly, possibly by using $np(p)$ instead of $np(1-p)$.
8. The number of calls received by a call centre in an hour follows a Poisson distribution with a mean of 40. Using a suitable approximation, find the probability that the call centre receives between 35 and 45 calls (inclusive) in a given hour.
   Answer: 0.615 Let $X$ be the number of calls. We are given $X \sim Po(40)$. Since the mean $\lambda = 40$ is large (typically $\lambda > 10$), we can use a normal approximation $Y \sim N(\lambda, \lambda)$, so $Y \sim N(40, 40)$.\ We want to find $P(35 \le X \le 45)$. We must use a continuity correction.\ $P(35 \le X \le 45)$ is approximated by $P(34.5 < Y < 45.5)$.\ The standard deviation is $\sigma = \sqrt{40} \approx 6.325$.\ We standardize the endpoints:\ $Z_1 = \frac{34.5 - 40}{\sqrt{40}} = \frac{-5.5}{6.325} \approx -0.8696$\ $Z_2 = \frac{45.5 - 40}{\sqrt{40}} = \frac{5.5}{6.325} \approx 0.8696$\ We need $P(-0.8696 < Z < 0.8696) = P(Z < 0.8696) - P(Z < -0.8696) \approx 0.8077 - 0.1923 = 0.6154$.\ * A: 0.109 is the result of incorrectly using $\sigma = 40$ instead of $\sigma = \sqrt{40}$. * B: 0.523 is the result of using an incorrect continuity correction, such as $P(35.5 < Y < 44.5)$. * C: 0.571 is the result of not applying the continuity correction, calculating $P(35 < Y < 45)$.
9. Let $X \sim N(\mu, \sigma^2)$. A random sample of size $n$ is taken. Consider the probabilities $P_1 = P(X > \mu + \sigma)$ and $P_2 = P(\bar{X} > \mu + \sigma)$, where $\bar{X}$ is the sample mean. For $n > 1$, which of the following is true?
   Answer: $P_1 > P_2$ Let's analyze both probabilities by standardizing them.\ For $P_1 = P(X > \mu + \sigma)$, the random variable is $X$, which has mean $\mu$ and standard deviation $\sigma$. The Z-score is $Z = \frac{(\mu + \sigma) - \mu}{\sigma} = \frac{\sigma}{\sigma} = 1$. So, $P_1 = P(Z > 1)$.\ For $P_2 = P(\bar{X} > \mu + \sigma)$, the random variable is $\bar{X}$, which has mean $\mu$ and standard deviation $\sigma_{\bar{X}} = \frac{\sigma}{\sqrt{n}}$. The Z-score is $Z = \frac{(\mu + \sigma) - \mu}{\sigma/\sqrt{n}} = \frac{\sigma}{\sigma/\sqrt{n}} = \sqrt{n}$. So, $P_2 = P(Z > \sqrt{n})$.\ Since we are given $n > 1$, we have $\sqrt{n} > 1$.\ The normal distribution's cumulative probability function is strictly increasing. Therefore, for $z_2 > z_1$, $P(Z > z_2) < P(Z > z_1)$.\ Since $\sqrt{n} > 1$, it follows that $P(Z > \sqrt{n}) < P(Z > 1)$.\ Therefore, $P_2 < P_1$, which is the same as $P_1 > P_2$.\ * **A: $P_1 = P_2$** would only be true if $\sqrt{n}=1$, i.e., $n=1$, but the question specifies $n>1$. * **C: $P_1 < P_2$** would imply $\sqrt{n}<1$, which contradicts $n>1$. * **D: The relationship depends on the value of $n$.** The relationship $P_1 > P_2$ holds for any $n>1$, so it is not dependent on the specific value.
10. The weights of two breeds of dog, A and B, are independent and normally distributed. For breed A, the mean weight is 30 kg and 10% of dogs weigh more than 35 kg. For breed B, the mean weight is 25 kg and the standard deviation is 3 kg. What is the probability that the total weight of two randomly selected dogs, one from each breed, exceeds 60 kg?
    Answer: 0.155 Let $X$ be the weight of a dog of breed A and $Y$ be the weight of a dog of breed B.\ Step 1: Find the standard deviation of breed A, $\sigma_A$. We are given $X \sim N(30, \sigma_A^2)$ and $P(X > 35) = 0.10$. Standardizing, $P(Z > \frac{35-30}{\sigma_A}) = 0.10$. This means $P(Z < \frac{5}{\sigma_A}) = 0.90$. Using the inverse normal function, the z-score corresponding to a cumulative probability of 0.90 is approximately 1.2816. So, $\frac{5}{\sigma_A} \approx 1.2816 \implies \sigma_A \approx \frac{5}{1.2816} \approx 3.901$ kg.\ Step 2: Define the total weight $T = X + Y$. We have $Y \sim N(25, 3^2)$. The distribution of $T$ is also normal.\ $E(T) = E(X) + E(Y) = 30 + 25 = 55$ kg.\ $Var(T) = Var(X) + Var(Y) = (3.901)^2 + 3^2 \approx 15.218 + 9 = 24.218$.\ $\sigma_T = \sqrt{24.218} \approx 4.921$ kg.\ Step 3: Calculate $P(T > 60)$. Standardizing, $Z = \frac{60 - 55}{4.921} \approx 1.016$.\ $P(Z > 1.016) = 1 - P(Z < 1.016) \approx 1 - 0.8452 = 0.1548$. This is approximately 0.155.\ * A: 0.121 results from using an incorrect z-score for 10% tail, such as 1.645 (for a 5% tail). * C: 0.196 results from incorrectly assuming $\sigma_A = 35 - 30 = 5$. * D: 0.234 results from incorrectly adding standard deviations ($\sigma_T = 3.901 + 3$) instead of variances.
11. Let $X_1, X_2, X_3$ be independent random variables with $X_i \sim N(i, 3)$ for $i=1, 2, 3$. Let $Y = X_1 - 2X_2 + X_3$. Find the variance of $Y$.
    Answer: 18 We are given that $X_i \sim N(i, 3)$ for $i=1, 2, 3$, so each variable has variance 3.\ $X_1 \sim N(1, 3)$, $X_2 \sim N(2, 3)$, $X_3 \sim N(3, 3)$\ So $Var(X_1) = Var(X_2) = Var(X_3) = 3$.\ The random variable $Y$ is a linear combination $Y = X_1 - 2X_2 + X_3$.\ For independent variables, the variance of a linear combination is:\ $Var(Y) = Var(X_1 - 2X_2 + X_3) = (1)^2Var(X_1) + (-2)^2Var(X_2) + (1)^2Var(X_3)$\ $Var(Y) = 1 \cdot 3 + 4 \cdot 3 + 1 \cdot 3 = 3 + 12 + 3 = 18$\ * A: -1: Variance cannot be negative. * B: 6: This might result from incorrectly using the coefficients without squaring: $(1 + 2 + 1) \times 3 = 12$ or from assuming variance 1 for each variable. * C: 14: This could come from summing $1^2 + 2^2 + 3^2 = 14$ if the variances were $i^2$ and coefficients were all 1.
12. The weights of apples from a certain orchard are normally distributed with a mean of 150g and a standard deviation of 20g. A random sample of 16 apples is taken. What is the probability that the mean weight of the apples in the sample is less than 145g?
    Answer: 0.1587 Let $X$ be the weight of a single apple. We have $X \sim N(150, 20^2)$.\ For a sample of size $n=16$, the distribution of the sample mean $\bar{X}$ is given by $\bar{X} \sim N(\mu, \frac{\sigma^2}{n})$.\ So, $\bar{X} \sim N(150, \frac{20^2}{16})$.\ The standard deviation of the sample mean (the standard error) is $\sigma_{\bar{X}} = \frac{\sigma}{\sqrt{n}} = \frac{20}{\sqrt{16}} = \frac{20}{4} = 5$.\ Thus, $\bar{X} \sim N(150, 5^2)$.\ We need to find $P(\bar{X} < 145)$.\ Standardizing the value: $Z = \frac{145 - 150}{5} = \frac{-5}{5} = -1$.\ We need $P(Z < -1)$. From standard normal tables or a calculator, this probability is approximately 0.1587.\ * A: 0.0228 corresponds to a Z-score of -2, which might result from a calculation error for the standard error, such as $\sigma_{\bar{X}} = 2.5$. * C: 0.4013 is the result of forgetting to divide the standard deviation by $\sqrt{n}$ and using $\sigma = 20$ instead of $\sigma_{\bar{X}}=5$. This gives $Z = \frac{145-150}{20} = -0.25$, and $P(Z < -0.25) \approx 0.4013$. * D: 0.8413 is $P(Z > -1)$, which would answer the question 'what is the probability the mean weight is more than 145g?'.
13. The random variable $X$ is the sum of $n$ independent and identically distributed random variables, each with mean $\mu$ and variance $\sigma^2$. The random variable $Y$ is the average of these $n$ variables. What are $E(X)$ and $Var(Y)$?
    Answer: $E(X)=n\mu$, $Var(Y)=\sigma^2/n$ Let the $n$ i.i.d. random variables be $Z_1, Z_2, \dots, Z_n$, where $E(Z_i)=\mu$ and $Var(Z_i)=\sigma^2$.\ The variable $X$ is the sum: $X = \sum_{i=1}^n Z_i$.\ The expectation of the sum is the sum of the expectations: $E(X) = E(\sum Z_i) = \sum E(Z_i) = \sum \mu = n\mu$.\ The variable $Y$ is the average (sample mean): $Y = \bar{Z} = \frac{1}{n} \sum Z_i = \frac{1}{n}X$.\ The variance of the average is $Var(Y) = Var(\frac{1}{n}X) = (\frac{1}{n})^2 Var(X)$.\ First, we need $Var(X)$. Since the variables are independent, the variance of the sum is the sum of the variances: $Var(X) = Var(\sum Z_i) = \sum Var(Z_i) = \sum \sigma^2 = n\sigma^2$.\ Now, substitute this into the expression for $Var(Y)$: $Var(Y) = (\frac{1}{n})^2 (n\sigma^2) = \frac{n\sigma^2}{n^2} = \frac{\sigma^2}{n}$.\ Therefore, $E(X)=n\mu$ and $Var(Y)=\sigma^2/n$.\ * B: Incorrectly calculates $Var(Y)$ as $\sigma^2/n^2$, perhaps from squaring the $1/n$ but forgetting the $n$ from $Var(X)$. * C: Incorrectly assumes the expectation of the sum is just $\mu$ and the variance of the average is just $\sigma^2$. * D: Correctly finds $E(X)$ but incorrectly states that $Var(Y) = n\sigma^2$. This is the variance of the sum $X$, not the average $Y$.
14. A manufacturer claims that their light bulbs have a mean lifetime of 1000 hours. The lifetimes are not normally distributed, but have a known standard deviation of 120 hours. A consumer group tests a sample of 36 bulbs. Using the Central Limit Theorem, what is the probability that the sample mean lifetime is less than 950 hours?
    Answer: 0.0062 We are given $\mu = 1000$, $\sigma = 120$, and $n=36$. The population is not normal.\ However, since the sample size $n=36$ is large enough ($n \ge 30$), we can apply the Central Limit Theorem. The CLT states that the sampling distribution of the sample mean $\bar{X}$ is approximately normal.\ The mean of this distribution is $\mu_{\bar{X}} = \mu = 1000$.\ The standard deviation of this distribution (standard error) is $\sigma_{\bar{X}} = \frac{\sigma}{\sqrt{n}} = \frac{120}{\sqrt{36}} = \frac{120}{6} = 20$.\ So, $\bar{X} \approx N(1000, 20^2)$.\ We want to find $P(\bar{X} < 950)$.\ Standardize the value: $Z = \frac{950 - 1000}{20} = \frac{-50}{20} = -2.5$.\ We need $P(Z < -2.5)$. Using a calculator or standard normal table, this probability is approximately 0.0062.\ * B: 0.0228 corresponds to a Z-score of -2. This would occur if the standard error was miscalculated as 25 (e.g. $1000-950=50, 50/2 = 25$). * C: 0.3385 is the result of forgetting to use the standard error and using the population standard deviation instead: $Z = \frac{950 - 1000}{120} \approx -0.4167$, for which $P(Z < -0.4167) \approx 0.3385$. * D: It cannot be determined... This is incorrect because the Central Limit Theorem allows us to approximate the probability due to the large sample size.

Start this quiz