College Statistics Quiz: Normal Distribution Applications
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Normal Distribution ApplicationsQuestion 1 of 20

The heights of adult males in a country are normally distributed with a mean of 70 inches and a standard deviation of 3 inches. The heights of adult females are normally distributed with a mean of 65 inches and a standard deviation of 2.65 inches. If one male and one female are chosen at random, what is the approximate probability that the male is at least 3 inches taller than the female?

0.3085
0.6554
0.6915
0.8413
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College Statistics Quiz

College Statistics Quiz: Normal Distribution Applications

Practice Normal Distribution Applications in College Statistics 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 Normal Distribution Applications, giving you a quick way to practice the rules, question types, and explanations that matter most for College Statistics.

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Question 1

The heights of adult males in a country are normally distributed with a mean of 70 inches and a standard deviation of 3 inches. The heights of adult females are normally distributed with a mean of 65 inches and a standard deviation of 2.65 inches. If one male and one female are chosen at random, what is the approximate probability that the male is at least 3 inches taller than the female?

  1. 0.3085
  2. 0.6554
  3. 0.6915 (correct answer)
  4. 0.8413
Explanation: Let M be the height of the male and F be the height of the female. We want to find P(M ≥ F + 3), which can be rewritten as P(M - F ≥ 3). Let the difference in heights be D = M - F. Since M and F are independent normal variables, D is also normally distributed. The mean of the difference is E[D] = E[M] - E[F] = 70 - 65 = 5 inches. The variance of the difference is the sum of the variances: Var(D) = Var(M) + Var(F) = 3² + (2.65)² = 9 + 7.0225 = 16.0225. The standard deviation of the difference is σD=16.0225≈4\sigma_D = \sqrt{16.0225} \approx 4. We need to find P(D ≥ 3). The z-score is z = (3 - E[D]) / σD\sigma_D = (3 - 5) / 4 = -2 / 4 = -0.5. The probability is P(Z ≥ -0.5) = 1 - P(Z < -0.5) = 1 - 0.3085 = 0.6915.

Question 2

Let X be a normally distributed random variable with mean μ\mu and variance σ2\sigma^2. Consider the transformed random variable Y=10−2XY = 10 - 2X. Which of the following correctly describes the distribution of Y?

  1. Normally distributed with mean 10−2μ10 - 2\mu and variance 4σ24\sigma^2 (correct answer)
  2. Normally distributed with mean 10−2μ10 - 2\mu and variance −2σ2-2\sigma^2
  3. Normally distributed with mean 10−2μ10 - 2\mu and variance 2σ22\sigma^2
  4. The distribution of Y is not necessarily normal.
Explanation: A linear transformation of a normally distributed random variable results in another normally distributed random variable. If X∼N(μ,σ2)X \sim N(\mu, \sigma^2) and Y=aX+bY = aX + b, then Y∼N(aμ+b,a2σ2)Y \sim N(a\mu + b, a^2\sigma^2). In this case, a = -2 and b = 10. The mean of Y is E[Y] = E[10 - 2X] = 10 - 2E[X] = 10 - 2μ\mu. The variance of Y is Var(Y) = Var(10 - 2X) = (-2)²Var(X) = 4σ2\sigma^2. Therefore, Y is normally distributed with mean 10−2μ10 - 2\mu and variance 4σ24\sigma^2.

Question 3

A manufacturer produces bolts with a mean length of 50.0 mm and a standard deviation of 0.15 mm. The lengths are normally distributed. A nut is produced with a mean inner diameter of 50.4 mm and a standard deviation of 0.20 mm, also normally distributed. For a secure fit, the clearance (nut diameter minus bolt length) must be between 0.1 mm and 0.6 mm. What is the approximate probability that a randomly paired bolt and nut will NOT have a secure fit?

  1. 0.115
  2. 0.212
  3. 0.327 (correct answer)
  4. 0.673
Explanation: Let N be the nut diameter and B be the bolt length. The clearance is C = N - B. Since N and B are independent and normal, C is also normal. The mean of C is E[C] = E[N] - E[B] = 50.4 - 50.0 = 0.4 mm. The variance of C is Var(C) = Var(N) + Var(B) = (0.20)² + (0.15)² = 0.04 + 0.0225 = 0.0625. The standard deviation of C is σC=0.0625=0.25\sigma_C = \sqrt{0.0625} = 0.25 mm. A secure fit occurs if 0.1 < C < 0.6. We first find the probability of a secure fit, P(0.1 < C < 0.6). The z-scores are z₁ = (0.1 - 0.4) / 0.25 = -1.2 and z₂ = (0.6 - 0.4) / 0.25 = 0.8. The probability of a secure fit is P(-1.2 < Z < 0.8) = P(Z < 0.8) - P(Z < -1.2) = 0.7881 - 0.1151 = 0.673. The question asks for the probability of NOT a secure fit, which is 1 - 0.673 = 0.327.

Question 4

The time to complete a project is the sum of the times for two independent stages, Stage A and Stage B. The time for Stage A is normally distributed with a mean of 40 hours and a standard deviation of 4 hours. The time for Stage B is normally distributed with a mean of 30 hours and a standard deviation of 3 hours. The project has a deadline of 75 hours. What is the probability that the project is completed on time?

  1. 0.1587
  2. 0.8413 (correct answer)
  3. 0.8944
  4. 0.9772
Explanation: Let T_A be the time for Stage A and T_B be the time for Stage B. The total time T = T_A + T_B. Since they are independent normal variables, T is also normal. The mean of T is E[T] = E[T_A] + E[T_B] = 40 + 30 = 70 hours. The variance of T is Var(T) = Var(T_A) + Var(T_B) = 4² + 3² = 16 + 9 = 25. The standard deviation of T is σT=25=5\sigma_T = \sqrt{25} = 5 hours. We need to find the probability that the project is completed on time, which is P(T ≤ 75). The z-score is z = (75 - 70) / 5 = 5 / 5 = 1. The probability P(Z ≤ 1) is 0.8413.

Question 5

IQ scores are normally distributed with a mean of 100 and a standard deviation of 15. A person is selected at random and found to have an IQ score of at least 115. What is the probability that this person's IQ score is actually greater than 130?

  1. 0.0228
  2. 0.1359
  3. 0.1437 (correct answer)
  4. 0.1587
Explanation: This is a conditional probability problem: P(X > 130 | X > 115). The formula for conditional probability is P(A|B) = P(A and B) / P(B). Here, A is X > 130 and B is X > 115. The event 'A and B' is just X > 130, since if a score is greater than 130, it is automatically greater than 115. So, we need to calculate P(X > 130) / P(X > 115). First, find P(X > 130): z = (130 - 100) / 15 = 2. P(Z > 2) = 1 - 0.9772 = 0.0228. Next, find P(X > 115): z = (115 - 100) / 15 = 1. P(Z > 1) = 1 - 0.8413 = 0.1587. Finally, divide the probabilities: 0.0228 / 0.1587 ≈ 0.1437.

Question 6

The fill volume of a soda machine is normally distributed. It is known that 2.5% of bottles are underfilled and have a volume less than 490 mL, while 16% of bottles are overfilled with a volume greater than 505 mL. Based on this information, what are the approximate mean and standard deviation of the fill volume?

  1. μ=498.4\mu = 498.4 mL, σ=4.3\sigma = 4.3 mL
  2. μ=497.5\mu = 497.5 mL, σ=3.8\sigma = 3.8 mL
  3. μ=499.9\mu = 499.9 mL, σ=5.0\sigma = 5.0 mL (correct answer)
  4. μ=500.2\mu = 500.2 mL, σ=4.7\sigma = 4.7 mL
Explanation: We have two pieces of information we can translate into z-scores.
  1. P(X < 490) = 0.025. The z-score for a cumulative probability of 0.025 is z = -1.96. So, (490 - μ) / σ = -1.96, which gives the equation: 490 = μ - 1.96σ.
  2. P(X > 505) = 0.16. This means P(X < 505) = 1 - 0.16 = 0.84. The z-score for a cumulative probability of 0.84 is approximately z = 1.0. (A more precise value is 0.9945, but 1.0 is standard). So, (505 - μ) / σ = 1.0, which gives the equation: 505 = μ + 1.0σ. We now have a system of two linear equations: (I) μ - 1.96σ = 490 (II) μ + 1.0σ = 505 Subtracting (I) from (II): (μ + σ) - (μ - 1.96σ) = 505 - 490 => 2.96σ = 15 => σ = 15 / 2.96 ≈ 5.07. Substitute σ back into (II): μ + 5.07 = 505 => μ = 499.93. The closest answer is μ = 499.9 mL and σ = 5.0 mL.

Question 7

A population of measurements is normally distributed with mean μ\mu and standard deviation σ=10\sigma = 10. A random sample of size n is drawn from this population. How large must the sample size n be so that there is a 95% probability that the sample mean Xˉ\bar{X} is within ±2 units of the population mean μ\mu?

  1. 10
  2. 25
  3. 68
  4. 97 (correct answer)
Explanation: We want to find n such that P(μ−2≤Xˉ≤μ+2\mu - 2 \leq \bar{X} \leq \mu + 2) = 0.95. The sampling distribution of Xˉ\bar{X} is normal with mean μ\mu and standard deviation σXˉ=σ/n=10/n\sigma_{\bar{X}} = \sigma / \sqrt{n} = 10 / \sqrt{n}. We can convert the inequality to z-scores: P(((μ−2\mu - 2 - μ\mu) / σXˉ\sigma_{\bar{X}}) ≤Z≤\leq Z \leq (((μ+2\mu + 2 - μ\mu) / σXˉ\sigma_{\bar{X}})) = 0.95. This simplifies to P(-2 / σXˉ\sigma_{\bar{X}} ≤Z≤\leq Z \leq 2 / σXˉ\sigma_{\bar{X}}) = 0.95. For a 95% probability centered at the mean, the z-scores are ±1.96. So, we set 2 / σXˉ\sigma_{\bar{X}} = 1.96. Substituting σXˉ=10/n\sigma_{\bar{X}} = 10 / \sqrt{n}, we get 2 / (10 / n\sqrt{n}) = 1.96, which simplifies to 2n\sqrt{n} / 10 = 1.96, or n\sqrt{n} / 5 = 1.96. Then n\sqrt{n} = 5 * 1.96 = 9.8. Squaring both sides, n = (9.8)² = 96.04. Since the sample size must be an integer, we round up to n = 97 to ensure the probability is at least 95%.

Question 8

The heights of adult males in a country are normally distributed with a mean of 70 inches and a standard deviation of 3 inches. The heights of adult females are normally distributed with a mean of 65 inches and a standard deviation of 2.65 inches. If one male and one female are chosen at random, what is the approximate probability that the male is at least 3 inches taller than the female?

  1. 0.3085
  2. 0.6554
  3. 0.6915 (correct answer)
  4. 0.8413
Explanation: Let M be the height of the male and F be the height of the female. We want to find P(M ≥ F + 3), which can be rewritten as P(M - F ≥ 3). Let the difference in heights be D = M - F. Since M and F are independent normal variables, D is also normally distributed. The mean of the difference is E[D] = E[M] - E[F] = 70 - 65 = 5 inches. The variance of the difference is the sum of the variances: Var(D) = Var(M) + Var(F) = 3² + (2.65)² = 9 + 7.0225 = 16.0225. The standard deviation of the difference is σD=16.0225≈4\sigma_D = \sqrt{16.0225} \approx 4. We need to find P(D ≥ 3). The z-score is z = (3 - E[D]) / σD\sigma_D = (3 - 5) / 4 = -2 / 4 = -0.5. The probability is P(Z ≥ -0.5) = 1 - P(Z < -0.5) = 1 - 0.3085 = 0.6915.

Question 9

A company's production line has a daily output that is normally distributed with a mean of 1000 units and a standard deviation of 50 units. The company makes a profit of $2 on each unit, but incurs a fixed cost of $1900 per day. What is the approximate probability that the company makes a profit on any given day?

  1. 0.1587
  2. 0.4207
  3. 0.5793
  4. 0.8413 (correct answer)
Explanation: First, determine the break-even point in terms of units produced. The company makes a profit if the total revenue from sales is greater than the fixed cost. Let X be the number of units produced. The revenue is $2X. To make a profit, we need 2X > 1900, which means X > 950. The problem is now to find the probability that the number of units produced is greater than 950. We are given that X is normally distributed with μ = 1000 and σ = 50. We calculate the z-score for X = 950: z = (950 - 1000) / 50 = -50 / 50 = -1. We need to find P(X > 950), which corresponds to P(Z > -1). This probability is 1 - P(Z < -1) = 1 - 0.1587 = 0.8413.

Question 10

The annual return on a portfolio is approximately normally distributed with a mean of 8% and a standard deviation of 15%. If an investor wants to be 95% confident that the portfolio does not lose more than a certain percentage of its value in a year, what is the maximum loss the investor should be prepared for?

  1. A loss of 16.75% (correct answer)
  2. A loss of 19.50%
  3. A loss of 21.25%
  4. A loss of 24.70%
Explanation: This is an inverse normal problem. We want to find the value x such that P(Return < x) = 0.05 (the bottom 5% of outcomes). The z-score corresponding to a cumulative probability of 0.05 is z ≈ -1.645. We use the formula x = μ + zσ. Here, μ = 8 and σ = 15. So, x = 8 + (-1.645)(15) = 8 - 24.675 = -16.675. This means there is a 5% chance of the return being less than -16.675%, which corresponds to a loss of 16.675%. The closest answer is a loss of 16.75%.

Question 11

For a normally distributed variable, the Interquartile Range (IQR) is the distance between the 75th percentile (Q3) and the 25th percentile (Q1). Outliers are often defined as observations falling below Q1 - 1.5(IQR) or above Q3 + 1.5(IQR). For any normal distribution, what percentage of observations would be classified as outliers by this rule?

  1. 0.27%
  2. 0.70% (correct answer)
  3. 4.55%
  4. 5.00%
Explanation: For a standard normal distribution (μ=0, σ=1), the 25th percentile (Q1) is at z ≈ -0.6745, and the 75th percentile (Q3) is at z ≈ 0.6745. The IQR is Q3 - Q1 ≈ 0.6745 - (-0.6745) = 1.349. The outlier fences are: Lower Fence: Q1 - 1.5(IQR) ≈ -0.6745 - 1.5(1.349) = -0.6745 - 2.0235 = -2.698. Upper Fence: Q3 + 1.5(IQR) ≈ 0.6745 + 1.5(1.349) = 0.6745 + 2.0235 = 2.698. We need to find the probability of a value falling outside these fences, which is P(Z < -2.698) + P(Z > 2.698). This is 2 * P(Z < -2.698). Using a z-table or calculator for z ≈ -2.70, the probability is 0.0035. So the total percentage of outliers is 2 * 0.0035 = 0.0070, or 0.70%.

Question 12

The daily number of visitors to a national park is approximately normally distributed with a mean of 4,500 people and a standard deviation of 800 people. Park officials decide to offer a special promotional discount on days when the number of visitors is in the lowest 10%. What is the maximum number of visitors on a day that qualifies for the discount?

  1. 3,472 visitors (correct answer)
  2. 3,880 visitors
  3. 5,528 visitors
  4. 3,500 visitors
Explanation: This is an inverse normal distribution problem. We need to find the value x such that P(X < x) = 0.10. First, find the z-score corresponding to a cumulative probability of 0.10. From a standard normal table or calculator, z ≈ -1.28. Then, use the z-score formula x = μ + zσ. So, x = 4500 + (-1.28)(800) = 4500 - 1024 = 3476. The closest answer is 3,472, which uses a more precise z-score of -1.2816.

Question 13

A machine fills coffee bags with weights that are normally distributed with a mean of 16.2 ounces and a standard deviation of 0.2 ounces. A second machine fills tea bags with weights that are normally distributed with a mean of 4.0 ounces and a standard deviation of 0.1 ounces. If one coffee bag and one tea bag are randomly selected, what is the approximate probability that the coffee bag weighs less than four times the weight of the tea bag?

  1. 0.18
  2. 0.33 (correct answer)
  3. 0.37
  4. 0.67
Explanation: Let C be the weight of a coffee bag and T be the weight of a tea bag. We need to find P(C < 4T), which is the same as P(C - 4T < 0). Let Y = C - 4T. Since C and T are independent normal random variables, Y is also normally distributed. The mean of Y is E[Y] = E[C] - 4E[T] = 16.2 - 4(4.0) = 0.2. The variance of Y is Var(Y) = Var(C) + 4²Var(T) = (0.2)² + 16(0.1)² = 0.04 + 0.16 = 0.20. The standard deviation of Y is σY=0.20≈0.447\sigma_Y = \sqrt{0.20} \approx 0.447. We want to find P(Y < 0). The z-score for Y=0 is z = (0 - 0.2) / 0.447 ≈ -0.447. The probability P(Z < -0.447) is approximately 0.327, or 0.33.

Question 14

The lifespans of a specific type of electronic component are exponentially distributed with a mean of 500 hours and a standard deviation of 500 hours. A large device uses 100 of these components, and the device fails when the average lifespan of these 100 components is less than 475 hours. What is the approximate probability that the device will fail?

  1. 0.0062
  2. 0.3085 (correct answer)
  3. 0.4801
  4. 0.6915
Explanation: Even though the individual component lifespans are not normally distributed, the Central Limit Theorem (CLT) states that for a large sample size (n=100 is large enough), the sampling distribution of the sample mean (xˉ\bar{x}) will be approximately normal. The mean of this sampling distribution is the same as the population mean, μxˉ=μ=500\mu_{\bar{x}} = \mu = 500. The standard deviation of the sampling distribution (the standard error) is σxˉ=σ/n=500/100=500/10=50\sigma_{\bar{x}} = \sigma / \sqrt{n} = 500 / \sqrt{100} = 500 / 10 = 50. We need to find the probability P(xˉ<475\bar{x} < 475). The z-score is z = (475 - 500) / 50 = -25 / 50 = -0.5. Using a standard normal table, P(Z < -0.5) = 0.3085.

Question 15

A manufacturer produces cylindrical pistons. The piston diameter is normally distributed with a mean of 9.00 cm and a standard deviation of 0.02 cm. A piston is considered defective if its diameter is outside the range of 8.97 cm to 9.03 cm. If the manufacturing process is adjusted to reduce the standard deviation to 0.015 cm while keeping the mean at 9.00 cm, by what percentage does the rate of defective pistons decrease?

  1. The defect rate is reduced by approximately 6.7%.
  2. The defect rate is reduced by approximately 13.4%.
  3. The defect rate is reduced by approximately 4.5%.
  4. The defect rate is reduced by approximately 8.9%. (correct answer)
Explanation: This is a multi-step problem. First, calculate the initial defect rate. The acceptable range is 9.00 ± 0.03 cm. With σ = 0.02, this is μ±1.5σ\mu \pm 1.5\sigma. The z-scores are z = ±0.03 / 0.02 = ±1.5. The probability of being outside this range is P(Z < -1.5) + P(Z > 1.5) = 2 * P(Z < -1.5) = 2 * 0.0668 = 0.1336, or 13.36%. Next, calculate the new defect rate with σ = 0.015. The z-scores are now z = ±0.03 / 0.015 = ±2.0. The probability of being outside this range is P(Z < -2.0) + P(Z > 2.0) = 2 * P(Z < -2.0) = 2 * 0.0228 = 0.0456, or 4.56%. The decrease in the defect rate is 13.36% - 4.56% = 8.8%. This is an absolute percentage decrease. The question asks for the percentage decrease, which is (Old Rate - New Rate) / Old Rate. However, answer choices suggest it's asking for the absolute difference in percentages. 8.8% is closest to 8.9%.

Question 16

A university reports that the scores on a standardized entrance exam are normally distributed with a mean of 500 and a standard deviation of 100. The university offers scholarships to students in the top 2.5% and places students in the bottom 16% on academic warning. Which of the following intervals represents the range of scores for students who are neither on academic warning nor receive a scholarship?

  1. [400, 700]
  2. [400, 696] (correct answer)
  3. [342, 664]
  4. [304, 696]
Explanation: This is a two-sided inverse normal problem. First, find the cutoff for academic warning (bottom 16%). This corresponds to a cumulative probability of 0.16. The z-score for P(Z < z) = 0.16 is approximately -1.0. The score is x = μ + zσ = 500 + (-1)(100) = 400. So, students with scores below 400 are on warning. Next, find the cutoff for scholarships (top 2.5%). This corresponds to a cumulative probability of 1 - 0.025 = 0.975. The z-score for P(Z < z) = 0.975 is 1.96. The score is x = μ + zσ = 500 + (1.96)(100) = 500 + 196 = 696. So, students with scores above 696 get scholarships. The range of scores for students who are in neither category is from 400 up to 696, or [400, 696].

Question 17

A poll asks 900 randomly selected voters if they support a particular candidate. The true proportion of voters in the population who support the candidate is 50%. What is the approximate probability that the sample proportion from the poll will be greater than 53%?

  1. 0.0359 (correct answer)
  2. 0.0885
  3. 0.1151
  4. 0.4207
Explanation: This is a normal approximation to the binomial distribution for a sample proportion. The sample size n=900 is large and np = 900(0.5) = 450 ≥ 10 and n(1-p) = 900(0.5) = 450 ≥ 10, so the approximation is appropriate. The mean of the sampling distribution of the sample proportion (p^\hat{p}) is p = 0.50. The standard deviation (standard error) is σp^=p(1−p)/n=0.5(0.5)/900=0.25/900=0.5/30≈0.01667\sigma_{\hat{p}} = \sqrt{p(1-p)/n} = \sqrt{0.5(0.5)/900} = \sqrt{0.25/900} = 0.5/30 \approx 0.01667. We want to find P(p^\hat{p} > 0.53). The z-score is z = (0.53 - 0.50) / 0.01667 = 0.03 / 0.01667 = 1.8. The probability P(Z > 1.8) = 1 - P(Z < 1.8) = 1 - 0.9641 = 0.0359.

Question 18

The weights of a species of fish are normally distributed with a mean of 1200 grams and a standard deviation of 150 grams. Which of the following is more likely, and why?

I. A single fish weighing more than 1350 grams.

II. A random sample of 9 fish having a mean weight of more than 1350 grams.

  1. I is more likely because the standard deviation for a single observation is larger than the standard error for a sample mean. (correct answer)
  2. II is more likely because a larger sample size increases the probability of observing extreme values.
  3. I and II are equally likely because the distance from the mean (150 grams) is the same in both cases.
  4. II is more likely because the Central Limit Theorem guarantees the sample mean will be closer to the population mean.
Explanation: This question compares the probability of an extreme value for a single observation versus a sample mean. For I: We find P(X > 1350). The z-score is z = (1350 - 1200) / 150 = 150 / 150 = 1. The probability is P(Z > 1) = 0.1587. For II: We find P(Xˉ\bar{X} > 1350) for a sample of n=9. The sampling distribution has a mean of 1200 and a standard error of σXˉ=σ/n=150/9=150/3=50\sigma_{\bar{X}} = \sigma / \sqrt{n} = 150 / \sqrt{9} = 150 / 3 = 50. The z-score is z = (1350 - 1200) / 50 = 150 / 50 = 3. The probability is P(Z > 3) = 0.0013. Since 0.1587 > 0.0013, event I is much more likely. The reason is that sample means are less variable than individual observations. A larger standard deviation (for the single observation) means that values far from the mean are more probable compared to the distribution of the sample mean, which has a smaller standard deviation (standard error).

Question 19

A survey of recent graduates from a university found their starting salaries to be approximately normally distributed with a mean of $60,000 and a standard deviation of $12,000. Using the 68-95-99.7 (Empirical) Rule, what is the approximate probability that a randomly selected graduate's starting salary is between $48,000 and $84,000?

  1. 0.680
  2. 0.815 (correct answer)
  3. 0.950
  4. 0.975
Explanation: This question requires an application of the 68-95-99.7 Rule. The mean is μ = $60,000 and the standard deviation is σ = $12,000. The lower bound of the interval is $48,000, which is $60,000 - $12,000 = μ - 1σ. The upper bound is $84,000, which is $60,000 + $24,000 = μ + 2σ. We need to find the area under the normal curve between one standard deviation below the mean and two standard deviations above the mean. The area between μ - 1σ and μ is half of 68%, which is 34% or 0.34. The area between μ and μ + 2σ is half of 95%, which is 47.5% or 0.475. The total probability is the sum of these two areas: 0.34 + 0.475 = 0.815.

Question 20

A machine fills coffee bags with weights that are normally distributed with a mean of 16.2 ounces and a standard deviation of 0.2 ounces. A second machine fills tea bags with weights that are normally distributed with a mean of 4.0 ounces and a standard deviation of 0.1 ounces. If one coffee bag and one tea bag are randomly selected, what is the approximate probability that the coffee bag weighs less than four times the weight of the tea bag?

  1. 0.18
  2. 0.33 (correct answer)
  3. 0.37
  4. 0.67
Explanation: Let C be the weight of a coffee bag and T be the weight of a tea bag. We need to find P(C < 4T), which is the same as P(C - 4T < 0). Let Y = C - 4T. Since C and T are independent normal random variables, Y is also normally distributed. The mean of Y is E[Y] = E[C] - 4E[T] = 16.2 - 4(4.0) = 0.2. The variance of Y is Var(Y) = Var(C) + 4²Var(T) = (0.2)² + 16(0.1)² = 0.04 + 0.16 = 0.20. The standard deviation of Y is σY=0.20≈0.447\sigma_Y = \sqrt{0.20} \approx 0.447. We want to find P(Y < 0). The z-score for Y=0 is z = (0 - 0.2) / 0.447 ≈ -0.447. The probability P(Z < -0.447) is approximately 0.327, or 0.33.