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Mean, Median & Standard Deviation - Finite Mathematics

  1. A professor calculates that the mean score on an exam is 78.5 with a standard deviation of 12.3. After reviewing the exam, she decides to add 5 points to every student's score and then discovers she made an error in the original calculation - the actual standard deviation was 15.7, not 12.3. What is the mean of the adjusted scores?
    Answer: 83.5 The mean of the adjusted scores is simply the original mean plus 5: 78.5 + 5 = 83.5. The error in calculating the standard deviation doesn't affect the mean calculation. The mean is affected by adding constants but not by errors in standard deviation calculations. Choice B forgets to add the 5 points. Choice C incorrectly adds the corrected standard deviation. Choice D incorrectly adds both the 5 points and the standard deviation difference.
  2. A symmetric distribution has values 8, 10, 12, 14, 16, with each value occurring with equal frequency. If three additional values of 12 are added to this dataset, what happens to the relationship between the mean and median?
    Answer: The mean and median remain equal Original dataset: mean = median = 12. Adding three 12s: new dataset is 8, 10, 12, 12, 12, 12, 14, 16. New mean = (8+10+12+12+12+12+14+16)/8 = 96/8 = 12. New median = (12+12)/2 = 12 (average of 4th and 5th values). Since we're adding values equal to the original mean, both statistics remain 12. Choice A and B incorrectly assume the distribution becomes skewed. Choice D misunderstands that medians are always defined for numerical datasets.
  3. The salaries at a small tech startup are as follows: nine software developers earn between $90,000 and $110,000 per year, and the CEO earns $1,500,000 per year. Which of the following statements best describes the relationship between the mean and median salary at this company?
    Answer: The mean salary is substantially greater than the median salary. When you encounter questions about salary distributions or data with extreme values, focus on how outliers affect measures of central tendency differently. The mean is sensitive to extreme values, while the median is resistant to them. Let's analyze this salary structure: nine developers earning $90,000-$110,000, and one CEO earning $1,500,000. To find the median of these 10 salaries, you need the average of the 5th and 6th values when arranged in order. Since all developer salaries fall within a narrow range, the median will be somewhere between $90,000-$110,000. For the mean, you must include that $1,500,000 CEO salary in your calculation. Even with a rough estimate: if developers average $100,000 each, the total is (9×100,000)+1,500,000=2,400,000(9 × 100,000) + 1,500,000 = 2,400,000. Divided by 10 employees, the mean salary is $240,000 – more than double the median. Looking at the wrong answers: Choice A suggests the mean and median are approximately equal, which ignores the CEO's outsized salary completely. Choice B claims the median is substantially greater than the mean, which would only occur if most salaries were high with a few very low outliers – the opposite of this situation. Choice D suggests we need exact developer salaries, but the CEO's salary is so dramatically higher than the developer range that it will pull the mean well above the median regardless of the specific developer amounts. Study tip: Remember that extreme high values always pull the mean toward them, while the median stays anchored to the middle of the distribution. In salary data, look for high-earning outliers that skew the mean upward.
  4. A dataset consists of five positive integers: {10, 12, 15, 17, 21}. A sixth integer, 49, is added to the set. How does the addition of this new integer affect the mean and median of the dataset?
    Answer: The mean increases by a larger amount than the median increases. Original set: {10, 12, 15, 17, 21}. The original mean is (10+12+15+17+21)/5 = 75/5 = 15. The original median is the middle value, 15. New set: {10, 12, 15, 17, 21, 49}. The new mean is (75+49)/6 = 124/6 ≈ 20.67. The new median is the average of the two middle values, (15+17)/2 = 16. The mean increases by 20.67 - 15 = 5.67, while the median increases by 16 - 15 = 1. The increase in the mean is larger because the mean is sensitive to outliers like 49, while the median is resistant.
  5. In a company, the 20 employees in the sales department have an average age of 45 years. The 30 employees in the production department have an average age of 35 years. What is the average age of all 50 employees in both departments combined?
    Answer: 39 This requires calculating a weighted average. The simple average of 35 and 45 is 40, but this is incorrect because the groups have different sizes. The total age of the sales employees is $20 * 45 = 900$ years. The total age of the production employees is $30 * 35 = 1050$ years. The total age of all employees is $900 + 1050 = 1950$ years. The total number of employees is $20 + 30 = 50$. The combined average age is $1950 / 50 = 39$ years. Distractor B is the unweighted average. Distractor C is a calculation error. Distractor D is the average of the two averages and the number of employees, which is nonsensical.
  6. A set of measurements has a standard deviation of 5. Each measurement in the set is first multiplied by 3 and then decreased by 7 to create a new, transformed dataset. What is the standard deviation of the new dataset?
    Answer: 15 Let the original data be represented by $X$ and the transformed data by $Y$. The transformation is $Y = 3X - 7$. Standard deviation is affected by multiplication but not by addition or subtraction. The rule for the transformation of a standard deviation $σ_X$ is $σ_{aX+b} = |a|σ_X$. Here, $a=3$ and $b=-7$. The new standard deviation is $|3| * 5 = 3 * 5 = 15$. Distractor B is the result of incorrectly subtracting 7 from the new standard deviation (15-7=8). Distractor D is the result of incorrectly multiplying by $a^2$ instead of $|a|$ (3^2 * 5 = 45), which is related to the transformation of variance. Distractor A incorrectly assumes the standard deviation is unchanged.
  7. A quality inspector measures the diameters of 50 ball bearings and calculates a mean of 8.24 mm with a standard deviation of 0.15 mm. Due to a calibration error, all measurements were 0.08 mm too high. After correcting for this systematic error, what is the coefficient of variation (standard deviation divided by mean) of the corrected measurements?
    Answer: 1.84% Corrected mean = 8.24 - 0.08 = 8.16 mm. The standard deviation remains 0.15 mm because subtracting a constant from all values doesn't change the spread. Coefficient of variation = (0.15/8.16) × 100% = 1.838% ≈ 1.84%. Choice A uses the original mean. Choice C incorrectly adjusts the standard deviation. Choice D uses an incorrect calculation method.
  8. A student has scores of 85, 92, 88, and 78 on four exams. What is the minimum score the student must achieve on the fifth and final exam to have an overall average of at least 87?
    Answer: 92 Let $x$ be the score on the fifth exam. To have an average of at least 87 over five exams, the sum of the scores must be at least $5 * 87 = 435$. The sum of the first four scores is $85 + 92 + 88 + 78 = 343$. So, the equation to solve is $343 + x ≥ 435$. Subtracting 343 from both sides gives $x ≥ 435 - 343$, which simplifies to $x ≥ 92$. Therefore, the minimum score required on the fifth exam is 92. Distractor A is the target average. Distractor C is a possible calculation error ($435 - 340 = 95$).
  9. A dataset has 7 values: 12, 15, 18, 20, 22, 25, 28. If each value is increased by 4 and then multiplied by 3, what happens to the standard deviation?
    Answer: It is multiplied by 3 only Standard deviation measures spread, so adding a constant (4) to each value doesn't change the spread - the standard deviation remains unchanged by addition. However, multiplying each value by 3 multiplies the standard deviation by 3. The transformation y = 3(x + 4) affects standard deviation only through the multiplication factor. Choice A incorrectly applies the addition. Choice C confuses the order of operations. Choice D incorrectly applies both transformations multiple times.
  10. What is the population standard deviation of the dataset {10, 13, 16, 10, 11}?
    Answer: √5.2 When you encounter a question asking for population standard deviation, you're working with a measure of how spread out data points are from the mean. The key distinction is "population" versus "sample" - population standard deviation divides by n (the total count), while sample standard deviation divides by n-1. Let's calculate the population standard deviation for {10, 13, 16, 10, 11}. First, find the mean: 10+13+16+10+115=605=12\frac{10+13+16+10+11}{5} = \frac{60}{5} = 12 Next, calculate each squared deviation from the mean: - (10-12)² = 4 - (13-12)² = 1 - (16-12)² = 16 - (10-12)² = 4 - (11-12)² = 1 Sum these squared deviations: 4 + 1 + 16 + 4 + 1 = 26 The population variance is: 265=5.2\frac{26}{5} = 5.2 The population standard deviation is the square root of the variance: 5.2\sqrt{5.2} This confirms answer D is correct. Looking at the wrong answers: A) 5.2 gives you the variance, not the standard deviation - you forgot to take the square root. B) √6.5 would result from incorrectly using sample standard deviation (dividing by n-1 = 4 instead of n = 5), giving variance 6.5. C) 6.5 makes the same sample/population error as B but also forgets the square root step. Remember: always check whether the question asks for population or sample standard deviation, and don't forget that standard deviation requires taking the square root of the variance.
  11. A university chemistry course is taught in two large sections. In Section 1, all students attend lectures regularly and study consistently, leading to final exam scores that are very close to one another. In Section 2, student attendance and study habits vary greatly, resulting in a wide range of scores, from very low to very high. If both sections have approximately the same average score, which of the following statements is most likely true?
    Answer: The standard deviation of scores in Section 2 is larger than in Section 1. Standard deviation measures the dispersion or spread of data points from the mean. The scores in Section 1 are described as 'very close to one another,' indicating a small spread and thus a small standard deviation. The scores in Section 2 have a 'wide range,' indicating a large spread and a large standard deviation. Therefore, the standard deviation of scores in Section 2 is larger than in Section 1. The mean being the same does not imply the standard deviations are equal. There is not enough information to conclude anything about the medians.
  12. Two datasets have identical means of 50. Dataset A has a standard deviation of 3.2, while Dataset B has a standard deviation of 7.8. If you randomly select one value from each dataset, which statement about the selected values is most accurate?
    Answer: The value from Dataset A is more likely to be closer to 50 than the value from Dataset B When you encounter questions comparing variability between datasets, focus on what standard deviation tells you about the spread of data around the mean. Standard deviation measures how tightly clustered the data points are - smaller standard deviation means values tend to be closer to the mean, while larger standard deviation means values are more spread out. Dataset A has a standard deviation of 3.2, while Dataset B has a standard deviation of 7.8. Since both have the same mean of 50, Dataset A's values are much more tightly clustered around 50 than Dataset B's values. This means when you randomly select a value, you're more likely to get something close to 50 from Dataset A than from Dataset B. Looking at the wrong answers: Choice A is incorrect because standard deviation doesn't guarantee that any single value will fall within one standard deviation of the mean - it's a probability statement, not a certainty. Choice B reverses the relationship - Dataset B's larger standard deviation makes its values more likely to be farther from 50, not closer. Choice C falls into the trap of thinking identical means lead to identical distributions, ignoring the crucial role of variability. Choice D correctly identifies that Dataset A's smaller standard deviation makes its randomly selected values more likely to be close to the mean of 50. Remember: when comparing datasets with the same mean, the one with smaller standard deviation will consistently produce values closer to that mean. Standard deviation is your key indicator of how "reliable" or "consistent" a dataset is around its center.
  13. Consider a dataset of n distinct numbers with a mean of μ and a standard deviation of σ > 0. If a new data point with a value exactly equal to the mean μ is added to the dataset, what will be the effect on the standard deviation?
    Answer: The standard deviation will decrease. When you encounter questions about how adding data points affects standard deviation, focus on understanding what standard deviation measures: how spread out the data points are from the mean. Standard deviation is calculated as σ=∑(xi−μ)2n\sigma = \sqrt{\frac{\sum(x_i - \mu)^2}{n}}. When you add a new data point equal to the mean μ, that point contributes zero to the sum of squared deviations since (μ−μ)2=0(μ - μ)^2 = 0. However, the total number of data points increases from n to n+1. The new standard deviation becomes σnew=∑(xi−μ)2n+1\sigma_{new} = \sqrt{\frac{\sum(x_i - \mu)^2}{n+1}}. Since the numerator (sum of squared deviations) stays the same but the denominator increases, the overall value decreases. Adding a point at the mean "pulls" the dataset closer to the center, reducing variability. Looking at the wrong answers: Choice A is incorrect because adding a point at the mean always reduces spread, never increases it. Choice B represents a common misconception—while the magnitude of decrease depends on the specific values, the direction of change (decrease) is always predictable when adding a point at the mean. Choice C is wrong because the denominator change always affects the result unless you're dealing with a population versus sample standard deviation distinction, which isn't relevant here. Remember this pattern: adding data points at the mean always decreases standard deviation, while adding points far from the mean increases it. The mean acts as a "stabilizing" value that reduces overall variability.
  14. The lifespan of a certain brand of light bulb is approximately normally distributed with a mean of 1200 hours and a standard deviation of 75 hours. The manufacturer will replace any bulb that fails within the time period that covers the shortest-lasting 2.5% of bulbs. What is the maximum lifespan of a bulb that the manufacturer will replace?
    Answer: 1050 hours This question applies the Empirical Rule (68-95-99.7). A normal distribution is symmetric. Approximately 95% of the data falls within 2 standard deviations of the mean ($μ ± 2σ$). This means the remaining 5% is split between the two tails, with 2.5% in the lower tail and 2.5% in the upper tail. The manufacturer replaces bulbs in the shortest-lasting 2.5%, which corresponds to the lower tail. The cutoff for this is $μ - 2σ$. Calculation: $1200 - 2 * 75 = 1200 - 150 = 1050$ hours. Any bulb lasting 1050 hours or less will be replaced. Distractor B corresponds to $μ - 1σ$. Distractor C corresponds to $μ + 1σ$. Distractor D corresponds to $μ + 2σ$.
  15. A researcher has a dataset where the mean is 45.2 and the standard deviation is 8.7. She realizes that one data point was recorded as 23 when it should have been 32. If the dataset contains 25 values, what is the new mean after correcting this error?
    Answer: 45.56 When you encounter questions about correcting data errors, you're working with the fundamental relationship between individual data points and summary statistics like the mean. The key insight is that you don't need to recalculate everything from scratch—you can adjust the existing statistics based on the change. To find the new mean, start with what you know: the original mean is 45.2 for 25 values, which means the sum of all original values is 45.2×25=113045.2 \times 25 = 1130. When you correct the error, you're removing the incorrect value (23) and adding the correct value (32). The net change to the sum is 32−23=+932 - 23 = +9. The new sum becomes 1130+9=11391130 + 9 = 1139, and the new mean is 113925=45.56\frac{1139}{25} = 45.56, which is answer choice D. Let's examine why the other options are wrong. Choice A (46.08) likely comes from mistakenly adding 9 directly to the original mean instead of dividing by the sample size. Choice B (45.92) might result from incorrectly calculating the change as 32−2325=0.36\frac{32-23}{25} = 0.36 and adding it to 45.2, but this misses that we need the total adjustment. Choice C (44.84) appears to subtract the change instead of adding it, perhaps from confusion about which direction the correction goes. Remember this pattern: when correcting a single data point, find the net change in the sum, then divide that change by the sample size to get the change in the mean. This approach saves time and reduces calculation errors.
  16. A quality control manager analyzes the weights of chocolate bars from two production lines. Line A produces bars with mean weight 52.3 grams and standard deviation 2.1 grams. Line B produces bars with mean weight 48.7 grams and standard deviation 1.8 grams. If the manager combines equal numbers of bars from both lines, what is the mean weight of the combined sample?
    Answer: 50.5 grams When combining equal numbers of samples from two populations, the mean of the combined sample is the average of the two means: (52.3 + 48.7)/2 = 101.0/2 = 50.5 grams. Choice B incorrectly rounds 50.5 to 50.0. Choice C mistakenly uses a weighted average formula. Choice D incorrectly subtracts instead of adding the means.
  17. The mean of the five numbers {8, 11, 5, x, 14} is 10. What is the median of this set of numbers?
    Answer: 11 First, find the value of x. The mean is the sum of the numbers divided by the count. The equation is $ (8 + 11 + 5 + x + 14) / 5 = 10 $. This simplifies to $ (38 + x) / 5 = 10 $. Multiplying both sides by 5 gives $ 38 + x = 50 $, so $ x = 12 $. Now the dataset is {8, 11, 5, 12, 14}. To find the median, we must order the set: {5, 8, 11, 12, 14}. The median is the middle value in the ordered set, which is 11. Distractor A is the mean. Distractor C is the value of x. Distractor D is an incorrect calculation.

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