Normal & Standard Normal Distribution - College Statistics

1. The z-score for a value x₁ is z₁. The z-score for a value x₂ is z₂ = -z₁. Assuming z₁ is not zero, which of the following expressions represents the mean (μ) of the normal distribution from which x₁ and x₂ were drawn?
   Answer: μ = (x₁ + x₂)/2 When you encounter z-score problems involving symmetric relationships, remember that z-scores measure how many standard deviations a value lies from the mean. The key insight here is recognizing what it means when two z-scores are opposites. Since $z_2 = -z_1$, these two values are equidistant from the mean but on opposite sides. Using the z-score formula $z = \frac{x - \mu}{\sigma}$, we can write: $z_1 = \frac{x_1 - \mu}{\sigma}$ and $z_2 = \frac{x_2 - \mu}{\sigma} = -z_1$ This means $\frac{x_2 - \mu}{\sigma} = -\frac{x_1 - \mu}{\sigma}$, which simplifies to $x_2 - \mu = -(x_1 - \mu)$ or $x_2 - \mu = \mu - x_1$. Rearranging gives us $x_1 + x_2 = 2\mu$, so $\mu = \frac{x_1 + x_2}{2}$. Choice D is correct because when two values have opposite z-scores, their average equals the mean of the distribution. Choice A incorrectly subtracts the values instead of adding them. This would give you the distance between the values divided by 2, not the mean. Choice B is wrong because we can determine μ without knowing σ. The symmetric relationship provides enough information. Choice C represents a rearrangement of the z-score formula but doesn't use the given relationship between $z_1$ and $z_2$. Remember: when z-scores are opposites ($z_2 = -z_1$), the corresponding data values are symmetric about the mean, so their average equals the population mean.
2. Two students take different standardized tests. Amy scores 1250 on Test A, where the scores are normally distributed with a mean of 1050 and a standard deviation of 200. Ben scores 30 on Test B, where the scores are normally distributed with a mean of 21 and a standard deviation of 6. The standardized score (z-score) for Ben was how much greater than the standardized score for Amy?
   Answer: 0.50 First, calculate the standardized score (z-score) for each student. Amy's z-score: $z_A$ = (1250 - 1050) / 200 = 200 / 200 = 1.00. Ben's z-score: $z_B$ = (30 - 21) / 6 = 9 / 6 = 1.50. The question asks for the difference between their z-scores: $z_B$ - $z_A$ = 1.50 - 1.00 = 0.50.
3. A population's test scores are normally distributed with a mean of 200 and a variance of 625. What is the approximate probability that a randomly selected individual has a score between 180 and 230?
   Answer: 0.673 The key is to use the standard deviation, not the variance. The variance is σ² = 625, so the standard deviation is σ = √625 = 25. Next, calculate the z-scores for the interval boundaries:\nz_lower = (180 - 200) / 25 = -20 / 25 = -0.8.\nz_upper = (230 - 200) / 25 = 30 / 25 = 1.2. Now, find the area under the standard normal curve between these z-scores: P(-0.8 < Z < 1.2). P(-0.8 < Z < 1.2) = P(Z < 1.2) - P(Z < -0.8) = 0.8849 - 0.2119 = 0.6730. The distractor 0.032 results from incorrectly using the variance (625) as the standard deviation in the calculation.
4. The weights of bags of flour are normally distributed with a mean of 5.0 kg. The company's quality control specifies that 5% of bags must weigh more than 5.2 kg. To meet this specification, what must the approximate standard deviation of the weights be?
   Answer: 0.12 kg This question tests your understanding of the normal distribution and how to work backwards from a given percentile to find the standard deviation. When you know that 5% of bags weigh more than 5.2 kg, you're dealing with the upper tail of the normal distribution. Since the mean is 5.0 kg, you need to find what standard deviation would place 5.2 kg at the 95th percentile (because 100% - 5% = 95%). To solve this, use the z-score formula: $z = \frac{x - \mu}{\sigma}$ The z-score corresponding to the 95th percentile is approximately 1.645 (you should memorize that 95% of data falls within 1.645 standard deviations of the mean in a normal distribution). Setting up the equation: $1.645 = \frac{5.2 - 5.0}{\sigma}$ Solving for σ: $\sigma = \frac{0.2}{1.645} = 0.122 \approx 0.12$ Therefore, D) 0.12 kg is correct. A) 0.10 kg would make the distribution too narrow, resulting in fewer than 5% of bags exceeding 5.2 kg. B) 0.20 kg would make the distribution too wide, causing significantly more than 5% to exceed 5.2 kg. C) 0.17 kg also creates too wide a distribution, though closer than option B. Study tip: For normal distribution problems involving percentiles, always identify whether you're looking at upper or lower tails, memorize key z-scores (1.645 for 95%, 1.96 for 97.5%), and remember that larger standard deviations create wider distributions with more extreme values.
5. Two data points from a normally distributed dataset are known: a value of 60 has a z-score of -1.0, and a value of 90 has a z-score of +2.0. What are the mean (μ) and standard deviation (σ) of this distribution?
   Answer: μ = 70, σ = 10 When you encounter z-score problems with two data points, you're working with the fundamental z-score formula: $z = \frac{x - \mu}{\sigma}$. This creates a system of equations that you can solve for the unknown mean (μ) and standard deviation (σ). Set up your two equations from the given information. For the value 60 with z-score -1.0: $-1 = \frac{60 - \mu}{\sigma}$, which rearranges to $60 = \mu - \sigma$. For the value 90 with z-score +2.0: $2 = \frac{90 - \mu}{\sigma}$, which becomes $90 = \mu + 2\sigma$. Now solve the system. From the first equation: $\mu = 60 + \sigma$. Substitute this into the second equation: $90 = (60 + \sigma) + 2\sigma = 60 + 3\sigma$. Solving for σ: $30 = 3\sigma$, so $\sigma = 10$. Then $\mu = 60 + 10 = 70$. Answer A (μ = 60, σ = 10) incorrectly assumes the lower data value equals the mean, ignoring that z = -1.0 means it's one standard deviation below the mean. Answer B (μ = 80, σ = 20) likely comes from incorrectly averaging the data values and miscalculating the standard deviation. Answer C (μ = 75, σ = 15) appears to split the difference between values without properly accounting for their different z-score magnitudes. Remember: z-scores tell you exactly how many standard deviations away from the mean each value sits. Use this geometric relationship to set up your equations systematically rather than guessing based on the raw data values.
6. In a university course, final scores are normally distributed. A student learns their standardized score (z-score) is -1.5. This student also knows that 10% of students scored above 828 and the standard deviation of scores is 100. What was this student's actual score?
   Answer: 550 This is a two-step problem. First, find the mean (μ) of the distribution. '10% of students scored above 828' means P(X > 828) = 0.10, which is the 90th percentile. The z-score for the 90th percentile is approximately 1.28. Using the z-score formula: 1.28 = (828 - μ) / 100 → 128 = 828 - μ → μ = 828 - 128 = 700. Now, use this mean to find the student's actual score (x) given their z-score of -1.5.\nz = (x - μ) / σ → -1.5 = (x - 700) / 100 → -150 = x - 700 → x = 700 - 150 = 550.
7. A data point X = 120 is observed from a normal distribution with a mean of μ = 100 and an unknown standard deviation σ. The z-score for this data point is +2.5. If the standard deviation of the distribution were to double, what would be the new z-score for the same data point X = 120?
   Answer: 1.25 First, use the initial information to find the original standard deviation, σ.\nz = (X - μ) / σ → 2.5 = (120 - 100) / σ → 2.5 = 20 / σ → σ = 20 / 2.5 = 8. Next, the standard deviation doubles, so the new standard deviation is σ_new = 2 * 8 = 16. Finally, calculate the new z-score with σ_new, keeping X and μ the same.\nz_new = (120 - 100) / 16 = 20 / 16 = 1.25.
8. Two manufacturing processes, A and B, produce bolts with a mean length of 50 mm. The lengths from process A are normally distributed with a standard deviation of 0.5 mm, while the lengths from process B are normally distributed with a standard deviation of 0.8 mm. If a bolt is considered acceptable if its length is between 49.0 mm and 51.0 mm, which statement is the most accurate?
   Answer: Process A will produce a higher proportion of acceptable bolts than Process B. A smaller standard deviation means the data is less spread out and more concentrated around the mean. For a given acceptance interval centered on the mean, the distribution with the smaller standard deviation will have a larger area within that interval. Mathematically: Process A: P(49 < X < 51) = P((49-50)/0.5 < Z < (51-50)/0.5) = P(-2 < Z < 2) ≈ 0.9545. Process B: P(49 < X < 51) = P((49-50)/0.8 < Z < (51-50)/0.8) = P(-1.25 < Z < 1.25) ≈ 0.7888. Therefore, Process A, with the smaller standard deviation, produces a higher proportion of acceptable bolts.
9. The diameters of ball bearings produced by a machine are normally distributed with a mean of 10 mm and a standard deviation of 0.04 mm. Which of the following intervals contains the central 80% of the diameters?
   Answer: (9.949 mm, 10.051 mm) The central 80% of the data lies between the 10th percentile and the 90th percentile. This leaves 10% in each tail. The z-score corresponding to a cumulative probability of 0.10 is approximately -1.28. By symmetry, the z-score for the 90th percentile is +1.28. Calculate the bounds of the interval: Lower bound = μ + z_lower * σ = 10 + (-1.28)(0.04) = 10 - 0.0512 = 9.9488 mm. Upper bound = μ + z_upper * σ = 10 + (1.28)(0.04) = 10 + 0.0512 = 10.0512 mm. Rounding to three decimal places gives the interval (9.949 mm, 10.051 mm).
10. The lifespan of a certain type of industrial lightbulb is normally distributed with a standard deviation of 100 hours. It is known that 15% of these bulbs have a lifespan of less than 2800 hours. What is the mean lifespan of the lightbulbs?
    Answer: 2904 hours This is an inverse normal problem to find the mean (μ). P(X < 2800) = 0.15. We need to find the z-score that corresponds to the 15th percentile. From a standard normal table, the z-score for a cumulative probability of 0.15 is approximately -1.04. Using the z-score formula: z = (x - μ) / σ. -1.04 = (2800 - μ) / 100 -104 = 2800 - μ μ = 2800 + 104 = 2904 hours.
11. A company produces widgets with weights that are normally distributed with a mean of 200g. The company rejects any widget that weighs more than 210g or less than 190g. If 31.74% of widgets are rejected, what is the standard deviation of the weights?
    Answer: 10g This question tests your understanding of the normal distribution and how to work backwards from rejection rates to find the standard deviation. When you see a problem involving normally distributed data with symmetric rejection limits around the mean, you're dealing with z-scores and the empirical rule. Since the mean is 200g and widgets are rejected below 190g or above 210g, the rejection limits are exactly 10g away from the mean on both sides. With 31.74% total rejection rate, this splits equally due to symmetry: 15.87% rejected on each tail. This means 84.13% of widgets fall below 210g (or above 190g). Looking this up in a z-table or using the fact that 84.13% corresponds to approximately one standard deviation above the mean, we find that z ≈ 1.0. Using the z-score formula: $z = \frac{x - \mu}{\sigma}$ Substituting: $1.0 = \frac{210 - 200}{\sigma} = \frac{10}{\sigma}$ Therefore: $\sigma = 10g$ Choice A (5g) would result in z = 2.0, giving only about 4.56% total rejection rate. Choice B (20g) would give z = 0.5, resulting in about 61.7% rejection rate. Choice C (15g) would give z ≈ 0.67, resulting in about 50.3% rejection rate. Study tip: When working with normal distribution problems involving symmetric rejection regions, remember that you can use z-tables to connect percentages to standard deviations. Practice recognizing that 84.13% ≈ 1 standard deviation, 97.72% ≈ 2 standard deviations, and 99.87% ≈ 3 standard deviations.
12. The height of adult females in a certain population is normally distributed with mean μ and standard deviation σ. An adult female is selected at random and is found to be shorter than the mean height. Given this information, what is the probability that her height is greater than μ - 0.5σ?
    Answer: 0.3830 This is a conditional probability problem. Let H be the height. We want to find P(H > μ - 0.5σ | H < μ). Using the conditional probability formula P(A|B) = P(A and B) / P(B). Event B is H < μ. For a normal distribution, P(H < μ) = P(Z < 0) = 0.5. Event A is H > μ - 0.5σ. The event (A and B) is (μ - 0.5σ < H < μ). In terms of z-scores, this is P(-0.5 < Z < 0). P(-0.5 < Z < 0) = P(Z < 0) - P(Z < -0.5) = 0.5000 - 0.3085 = 0.1915. Therefore, the conditional probability is P(A|B) = 0.1915 / 0.5 = 0.3830.
13. The scores on a standardized exam are normally distributed with a mean of 500 and a standard deviation of 100. What proportion of test-takers score more than 150 points away from the mean score?
    Answer: 0.1336 This question asks for the probability of being in the two tails of the distribution. 'More than 150 points away from the mean' means the score is less than (500 - 150) = 350 or greater than (500 + 150) = 650. This is P(X < 350) + P(X > 650). First, find the z-score for 650: z = (650 - 500) / 100 = 1.5. Due to symmetry, the z-score for 350 is -1.5. We need to find P(Z < -1.5) + P(Z > 1.5). P(Z > 1.5) = 1 - P(Z < 1.5) = 1 - 0.9332 = 0.0668. Since the distribution is symmetric, P(Z < -1.5) is also 0.0668. The total proportion is 0.0668 + 0.0668 = 0.1336.
14. The heights of a certain species of plant are normally distributed with a mean of 30 cm and a standard deviation of 4 cm. What is the approximate probability that a randomly selected plant will be between 25 cm and 35 cm tall?
    Answer: 0.789 This question requires calculating the probability for an interval that is not an integer number of standard deviations from the mean. First, convert the heights to z-scores:\nz₁ = (25 - 30) / 4 = -5 / 4 = -1.25\nz₂ = (35 - 30) / 4 = 5 / 4 = 1.25 Next, find P(-1.25 < Z < 1.25) using a standard normal table or calculator. P(-1.25 < Z < 1.25) = P(Z < 1.25) - P(Z < -1.25) = 0.8944 - 0.1056 = 0.7888. Distractor A uses the empirical rule for ±1σ (68.3%). Distractor D uses the empirical rule for ±2σ (95.4%). Distractor C is the one-sided probability P(Z < 1.25).
15. The scores on a national aptitude test are normally distributed with a mean of 500 and a standard deviation of 100. What is the interquartile range (IQR) of the scores?
    Answer: 134.9 The interquartile range (IQR) is the difference between the third quartile (Q3, 75th percentile) and the first quartile (Q1, 25th percentile). For a standard normal distribution, the z-score for Q1 (p=0.25) is approximately -0.6745, and the z-score for Q3 (p=0.75) is approximately +0.6745. First, find the score values for Q1 and Q3: Q1 = μ + z₁σ = 500 + (-0.6745)(100) = 500 - 67.45 = 432.55. Q3 = μ + z₃σ = 500 + (0.6745)(100) = 500 + 67.45 = 567.45. IQR = Q3 - Q1 = 567.45 - 432.55 = 134.9. Alternatively, IQR = (z₃ - z₁)σ = (0.6745 - (-0.6745)) 100 = 1.349 100 = 134.9.
16. The scores on a highly selective examination are normally distributed. The 10th percentile score is 512 and the 90th percentile score is 688. What is the probability that a randomly selected test-taker scores above 650?
    Answer: 0.2266 This is a multi-step problem. First, find the mean (μ) and standard deviation (σ). The z-score for the 10th percentile is approximately -1.28, and for the 90th percentile is +1.28. 1) (512 - μ) / σ = -1.28 2) (688 - μ) / σ = 1.28 Adding the two equations after multiplying by σ: (512 - μ) + (688 - μ) = 0 → 1200 - 2μ = 0 → μ = 600. Substitute μ = 600 into the second equation: (688 - 600) / σ = 1.28 → 88 / σ = 1.28 → σ = 88 / 1.28 = 68.75. Now, find the probability of scoring above 650. z = (650 - 600) / 68.75 = 50 / 68.75 ≈ 0.75. P(Z > 0.75) = 1 - P(Z < 0.75) = 1 - 0.7734 = 0.2266.

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