Normal Tables & Technology - College Statistics

1. The lifetime of a smartphone battery is normally distributed with a mean of 400 hours of use and a standard deviation of 25 hours. The manufacturer wishes to offer a warranty, replacing any battery that fails before a guaranteed time $T$. If the company is willing to replace no more than 2.5% of its batteries, what is the maximum number of hours $T$ they can guarantee, rounded to the nearest hour?
   Answer: 351 hours This is a normal distribution problem where you need to find a value that corresponds to a specific percentile. When a company wants to replace "no more than 2.5%" of batteries, they're looking for the value below which only 2.5% of all batteries will fail - this is the 2.5th percentile. To solve this, you need to work backwards from the percentage to find the corresponding value. First, find the z-score for the 2.5th percentile using a standard normal table or calculator: $z = -1.96$. The negative value makes sense because you're looking at the lower tail of the distribution. Next, use the z-score formula to find T: $z = \frac{T - \mu}{\sigma}$. Substituting the known values: $-1.96 = \frac{T - 400}{25}$. Solving for T: $T = 400 + (-1.96)(25) = 400 - 49 = 351$ hours. Looking at the wrong answers: (A) 346 hours likely comes from using $z = -2.16$ (closer to a 1.5% cutoff), (B) 375 hours corresponds to using $z = -1.0$ (about the 16th percentile), and (C) 358 hours uses approximately $z = -1.68$ (closer to a 5% cutoff). Study tip: When working with warranty problems, remember you're always finding a percentile in the lower tail of the distribution. Double-check that your z-score is negative and that your final answer is below the mean - if it's above the mean, you've made an error in setup or calculation.
2. A statistics software package provides the function qnorm(p, mean, sd), which returns the value $x$ such that $P(X \le x) = p$. To find the value that separates the top 5% of scores from the bottom 95% for a distribution $N(100, 15)$, what would be the correct syntax?
   Answer: qnorm(0.95, 100, 15) When working with normal distribution functions, you need to carefully interpret what the question is asking and match it to the correct function syntax. The qnorm() function finds the x-value (quantile) where a given probability falls below that point. The question asks for the value separating the top 5% from the bottom 95%. This means you want the 95th percentile - the point where 95% of values fall below and 5% fall above. Since qnorm(p, mean, sd) returns the x-value where $P(X \leq x) = p$, you need $p = 0.95$ to find where 95% of the distribution lies below that point. For a normal distribution $N(100, 15)$, the mean is 100 and standard deviation is 15. Therefore, the correct syntax is qnorm(0.95, 100, 15). Looking at the wrong answers: Choice A uses qnorm(0.05, 100, 15), which would find the 5th percentile instead of the 95th percentile - the bottom boundary rather than the top boundary you need. Choice B has qnorm(0.05, 15, 100), which not only uses the wrong probability but also switches the mean and standard deviation parameters. Choice C uses pnorm(0.95, 100, 15), but pnorm() calculates probabilities from x-values, not x-values from probabilities - it's the inverse of what you need. Remember: when finding percentiles or cutoff points, use qnorm() with the cumulative probability as your first parameter. "Top 5%" means you want the 95th percentile, so use 0.95.
3. The area under the standard normal curve to the right of a value $z_1$ is 0.35. What is the area under the standard normal curve to the left of $-z_1$?
   Answer: 0.35 This question tests the symmetry of the normal distribution. The standard normal curve is symmetric about its mean of 0. The area to the right of a positive value $z_1$ is given as $P(Z > z_1) = 0.35$. By symmetry, the area to the left of the negative value $-z_1$ is exactly the same. Therefore, $P(Z < -z_1) = P(Z > z_1) = 0.35$.
4. If $Z$ is a standard normal random variable, what is the value of $P(0.5 < |Z| < 1.5)$?
   Answer: 0.483 The inequality $0.5 < |Z| < 1.5$ is equivalent to the union of two disjoint intervals: $(-1.5 < Z < -0.5)$ and $(0.5 < Z < 1.5)$. So we need to calculate $P(-1.5 < Z < -0.5) + P(0.5 < Z < 1.5)$. By the symmetry of the normal distribution, $P(-1.5 < Z < -0.5) = P(0.5 < Z < 1.5)$. We can calculate $P(0.5 < Z < 1.5) = P(Z < 1.5) - P(Z < 0.5) \approx 0.9332 - 0.6915 = 0.2417$. The total probability is $2 \times 0.2417 = 0.4834 \approx 0.483$.
5. The time required for a student to complete a standardized test is normally distributed with a standard deviation of 8 minutes. It is known that 70% of students finish the test in less than 65 minutes. Based on this information, what is the mean completion time for the test?
   Answer: 60.8 minutes This problem tests your ability to work backwards from a probability to find a normal distribution parameter. When you're given information about what percentage of a population falls below a certain value, you can use the standard normal distribution to find the unknown mean. Since 70% of students finish in less than 65 minutes, you need to find the z-score that corresponds to the 70th percentile. Looking this up in a standard normal table (or using inverse normal functions), the z-score for the 70th percentile is approximately 0.52. Now you can set up the standardization equation: $z = \frac{x - \mu}{\sigma}$ Substituting what you know: $0.52 = \frac{65 - \mu}{8}$ Solving for μ: $0.52 \times 8 = 65 - \mu$, so $4.16 = 65 - \mu$, which gives $\mu = 65 - 4.16 = 60.84$ minutes. Choice A (60.8 minutes) matches this calculation. Choice B (59.4 minutes) likely results from using the wrong z-score or making an arithmetic error. Choice C (64.5 minutes) might come from incorrectly assuming that 65 minutes is close to the mean, forgetting that 70% is well above the 50th percentile. Choice D (69.2 minutes) probably results from adding instead of subtracting when solving for the mean. Remember: when working with normal distributions, always identify what percentile you're dealing with first, then find the corresponding z-score before setting up your equation. The key insight is that 70% is significantly above average, so the mean must be noticeably less than 65 minutes.
6. The 16th percentile of a normally distributed dataset is 250, and the 84th percentile is 350. What is the standard deviation of this distribution?
   Answer: 50 The 16th and 84th percentiles correspond to Z-scores of approximately -1 and +1, respectively (based on the Empirical Rule that ~68% of data lies within $\mu \pm 1\sigma$, leaving 32% in the tails, or 16% in each tail). So we have two equations: $250 = \mu - 1\sigma$ and $350 = \mu + 1\sigma$. Subtracting the first equation from the second gives: $350 - 250 = (\mu + \sigma) - (\mu - \sigma) \Rightarrow 100 = 2\sigma \Rightarrow \sigma = 50$.
7. Let $X$ be a normally distributed random variable with a mean of 50. If the probability that $X$ is greater than 55 is 0.2, what is the probability that $X$ is less than 45?
   Answer: 0.2 The normal distribution is symmetric about its mean. The value 55 is 5 units above the mean of 50. The value 45 is 5 units below the mean of 50. Because of the symmetry, the area in the tail above $\mu+k$ is equal to the area in the tail below $\mu-k$. In this case, $k=5$. We are given $P(X > 50 + 5) = 0.2$. Therefore, $P(X < 50 - 5) = P(X < 45)$ must also be 0.2.
8. The weights of cereal boxes are normally distributed with a mean of 450 grams and a standard deviation of 5 grams. The manufacturer wants to establish a weight range, symmetric about the mean, that contains 90% of all boxes. Which of the following is the correct range?
   Answer: (441.8 g, 458.2 g) When you encounter a question about finding a range that contains a specific percentage of normally distributed data, you're working with the empirical rule and z-scores. The key is identifying the correct z-score for your desired percentage and applying it symmetrically around the mean. For a symmetric range containing 90% of the data, you need to find the z-scores that leave 5% in each tail (since 100% - 90% = 10%, split equally). Using a standard normal table or calculator, the z-score for 95% (leaving 5% in the upper tail) is approximately 1.645. The formula for finding the range boundaries is: $\text{mean} \pm z \times \text{standard deviation}$ With mean = 450g, standard deviation = 5g, and z = 1.645: - Lower bound: $450 - 1.645 \times 5 = 450 - 8.225 = 441.775 \text{ g}$ - Upper bound: $450 + 1.645 \times 5 = 450 + 8.225 = 458.225 \text{ g}$ This rounds to approximately (441.8 g, 458.2 g), making D correct. Choice A (443.6, 456.4) uses z ≈ 1.28, which corresponds to 80% of the data, not 90%. Choice B (445.0, 455.0) uses z = 1.0, capturing only 68% of the data. Choice C (440.2, 459.8) uses z ≈ 1.96, which gives 95% of the data. Remember: Always check what percentage is actually being asked for, and don't confuse common z-values like 1.96 (95%) with 1.645 (90%). The difference between confidence levels is a frequent trap on statistics exams.
9. 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 2.5 inches. What is the probability that a randomly selected male is shorter than a randomly selected female who is at the 80th percentile for female height?
   Answer: 0.166 This is a two-step problem. First, find the height of a female at the 80th percentile. The Z-score for the 80th percentile is $z \approx 0.84$. The female's height is $X_f = \mu_f + z \cdot \sigma_f = 65 + 0.84(2.5) = 65 + 2.1 = 67.1$ inches. Second, find the probability that a randomly selected male is shorter than 67.1 inches. We standardize this value for the male distribution: $Z_m = (67.1 - \mu_m) / \sigma_m = (67.1 - 70) / 3 = -2.9 / 3 \approx -0.97$. The probability is $P(Z < -0.97) \approx 0.166$.
10. A variable X is normally distributed with mean $\mu$ and standard deviation $\sigma$. What is the approximate probability that a randomly selected value of X falls between $\mu - 1.5\sigma$ and $\mu - 0.5\sigma$?
    Answer: 0.242 We need to find $P(\mu - 1.5\sigma < X < \mu - 0.5\sigma)$. We can standardize the endpoints: the Z-score for $\mu - 1.5\sigma$ is $(\mu - 1.5\sigma - \mu) / \sigma = -1.5$, and the Z-score for $\mu - 0.5\sigma$ is $(\mu - 0.5\sigma - \mu) / \sigma = -0.5$. The problem is now to find $P(-1.5 < Z < -0.5)$. Using a standard normal table or technology, this is $P(Z < -0.5) - P(Z < -1.5) \approx 0.3085 - 0.0668 = 0.2417$, which is approximately 0.242.
11. For a standard normal random variable $Z$, what is the approximate probability that $Z^2 > 1$?
    Answer: 0.317 The inequality $Z^2 > 1$ is true if and only if $Z > 1$ or $Z < -1$. These two events are mutually exclusive, so we can add their probabilities: $P(Z^2 > 1) = P(Z > 1) + P(Z < -1)$. Using a standard normal table or technology, $P(Z < -1) \approx 0.1587$. By symmetry, $P(Z > 1) \approx 0.1587$. The total probability is $0.1587 + 0.1587 = 0.3174$, which is approximately 0.317.
12. For a normally distributed random variable $X$ with mean $\mu$ and standard deviation $\sigma$, it is known that $P(X > \mu + 12) = 0.1587$. What is the value of $\sigma$?
    Answer: 12 We are given $P(X > \mu + 12) = 0.1587$. Let's standardize the value $\mu + 12$. The Z-score is $Z = ((\mu + 12) - \mu) / \sigma = 12 / \sigma$. So, the problem is equivalent to $P(Z > 12/\sigma) = 0.1587$. We need to find the Z-value that has an area of 0.1587 to its right. This is equivalent to finding a Z-value with a cumulative area of $1 - 0.1587 = 0.8413$ to its left. From a standard normal table, $Z \approx 1.00$. Therefore, we can set $12 / \sigma = 1.00$, which implies $\sigma = 12$.
13. The heights of a species of plant are normally distributed with a mean of 30 cm and a standard deviation of 4 cm. A biologist uses the Empirical Rule to state that approximately 95% of the plants have heights between 22 cm and 38 cm. A statistician calculates the exact probability using a normal distribution table. What is the absolute difference between the statistician's exact probability and the biologist's 0.95 approximation?
    Answer: 0.0045 The biologist's approximation of 95% comes from the Empirical Rule for the interval $\mu \pm 2\sigma$. Here, $30 \pm 2(4) = 30 \pm 8$, which is the interval (22, 38). To find the exact probability, we calculate $P(22 < X < 38)$. The Z-scores are $Z_1 = (22-30)/4 = -2$ and $Z_2 = (38-30)/4 = 2$. The exact probability is $P(-2 < Z < 2) = P(Z < 2) - P(Z < -2) \approx 0.9772 - 0.0228 = 0.9544$. The absolute difference between the exact value and the approximation is $|0.9544 - 0.95| = 0.0044$, which is approximately 0.0045.
14. For a particular normally distributed random variable $X$, the 25th percentile is 90 and the 75th percentile is 110. What is the mean, $\mu$, of this distribution?
    Answer: 100 This question tests your understanding of the symmetric properties of the normal distribution and how percentiles relate to the mean. In any normal distribution, the mean sits exactly at the center, which corresponds to the 50th percentile. Because the normal curve is perfectly symmetric around its mean, percentiles that are equidistant from the 50th percentile will be equidistant from the mean itself. The 25th and 75th percentiles are both exactly 25 percentile points away from the center (50th percentile). Since the distribution is symmetric, the mean must be exactly halfway between these two values. You can find this by calculating: $\mu = \frac{90 + 110}{2} = \frac{200}{2} = 100$ Looking at the wrong answers: Choice A (95) incorrectly assumes the mean is closer to the 25th percentile, suggesting a misunderstanding of symmetry. Choice C (105) makes the opposite error, placing the mean closer to the 75th percentile. Choice B (Cannot be determined without the standard deviation) reflects a common misconception—while the standard deviation affects how spread out the percentiles are from the mean, the mean's location is always exactly centered between symmetric percentiles in a normal distribution. The correct answer is D (100). Study tip: Remember that in normal distributions, the mean always equals the median (50th percentile). When given any two percentiles that are symmetric around the 50th percentile (like 25th and 75th, or 10th and 90th), you can find the mean by averaging those two values—no standard deviation needed.
15. The weights of apples from a certain orchard are normally distributed with a mean of 170 grams and a standard deviation of 12 grams. An apple with a weight of 191 grams is selected. A second apple is selected that has a weight with the same magnitude Z-score, but with the opposite sign. What is the weight of the second apple?
    Answer: 149 grams This question tests your understanding of Z-scores and the symmetric properties of normal distributions. When you encounter problems involving Z-scores with "opposite signs," you're working with values that are equidistant from the mean on opposite sides of the distribution. First, calculate the Z-score for the 191-gram apple: $Z = \frac{191 - 170}{12} = \frac{21}{12} = 1.75$. This apple is 1.75 standard deviations above the mean. The second apple has a Z-score of -1.75 (same magnitude, opposite sign), meaning it's 1.75 standard deviations below the mean. To find its weight: $X = \mu + Z \cdot \sigma = 170 + (-1.75)(12) = 170 - 21 = 149$ grams. Looking at the wrong answers: Answer B (146 grams) would correspond to a Z-score of -2.0, which is too far below the mean. Answer C (158 grams) gives a Z-score of -1.0, which isn't the correct magnitude. Answer D (168 grams) is only slightly below the mean with a Z-score of approximately -0.17, nowhere near the required -1.75. Answer A (149 grams) correctly places the second apple exactly 1.75 standard deviations below the mean, making it the mirror image of the first apple's position in the distribution. Study tip: Remember that normal distributions are perfectly symmetric around the mean. If a value is k standard deviations above the mean, its "opposite" counterpart will always be k standard deviations below the mean, equidistant from the center.
16. 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: 135 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-scores for the 75th and 25th percentiles are approximately $z = 0.6745$ and $z = -0.6745$, respectively. To find the score values, we use $X = \mu + Z\sigma$. So, $Q3 = 500 + (0.6745)(100) = 567.45$ and $Q1 = 500 + (-0.6745)(100) = 432.55$. The IQR is $567.45 - 432.55 = 134.9 \approx 135$.
17. The lifespan of a certain brand of light bulb is normally distributed with a mean of 1200 hours and a standard deviation of 80 hours. Given that a randomly selected light bulb has already lasted 1240 hours, what is the approximate probability that it will last more than 1300 hours?
    Answer: 0.342 This is a conditional probability problem: $P(X > 1300 | X > 1240)$. The formula is $P(X > 1300 \text{ and } X > 1240) / P(X > 1240)$. Since the condition $X > 1300$ implies $X > 1240$, this simplifies to $P(X > 1300) / P(X > 1240)$. First, we standardize the values: $Z_{1300} = (1300 - 1200) / 80 = 1.25$ and $Z_{1240} = (1240 - 1200) / 80 = 0.5$. Then we find the probabilities: $P(Z > 1.25) \approx 0.1056$ and $P(Z > 0.5) \approx 0.3085$. The conditional probability is $0.1056 / 0.3085 \approx 0.342$.
18. A data analyst is studying IQ scores, which are modeled by a normal distribution with a mean of 100 and a standard deviation of 15. To find the IQ score that a person must achieve to be in the top 2% of the population, the analyst uses a calculator function invNorm(area, μ, σ), where area is the cumulative area to the left of the desired value. Which of the following represents the correct use of this function?
    Answer: invNorm(0.98, 100, 15) The 'top 2%' corresponds to an area of 0.02 in the right tail of the distribution. The invNorm function requires the cumulative area to the left of the value. This area is $1 - 0.02 = 0.98$. The mean (μ) is 100 and the standard deviation (σ) is 15. Therefore, the correct syntax is invNorm(0.98, 100, 15).
19. A company's automated packaging system fills bags of coffee beans. The target weight is 500 grams, but the actual weights are normally distributed with a mean of 502 grams. If the company observes that 10% of the bags weigh more than 508 grams, what is the standard deviation of the filling process, in grams?
    Answer: 4.69 Let X be the weight of a bag. We are given $X \sim N(\mu=502, \sigma)$ and $P(X > 508) = 0.10$. This means the area to the left of 508 is $1 - 0.10 = 0.90$. We need to find the Z-score corresponding to a cumulative probability of 0.90, which is $z \approx 1.28$. Using the standardization formula $Z = (X - \mu) / \sigma$, we can solve for $\sigma$: $1.28 = (508 - 502) / \sigma \Rightarrow 1.28 = 6 / \sigma \Rightarrow \sigma = 6 / 1.28 \approx 4.69$.
20. A manufacturing process produces ball bearings with a diameter that is normally distributed with a mean of 5.00 mm and a standard deviation of 0.02 mm. A ball bearing is considered defective if its diameter is smaller than 4.97 mm or larger than 5.03 mm. What proportion of ball bearings are defective?
    Answer: 0.134 We need to find $P(X < 4.97) + P(X > 5.03)$. First, we find the Z-scores for the two boundaries. For 4.97 mm: $Z = (4.97 - 5.00) / 0.02 = -0.03 / 0.02 = -1.5$. For 5.03 mm: $Z = (5.03 - 5.00) / 0.02 = 0.03 / 0.02 = 1.5$. We need to find $P(Z < -1.5) + P(Z > 1.5)$. From a Z-table or calculator, $P(Z < -1.5) \approx 0.0668$. By symmetry, $P(Z > 1.5)$ is also 0.0668. The total proportion of defective bearings is $0.0668 + 0.0668 = 0.1336 \approx 0.134$.

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