Measures of Center - College Statistics

A marketing researcher is analyzing consumer data. For which of the following variables would the mode be the most appropriate measure of central tendency to report?
Answer: The brand of smartphone owned by consumers. The brand of smartphone is categorical (nominal) data. For nominal data, where values are labels without a natural order, the mean and median are not meaningful. The mode, which identifies the most frequent category, is the only appropriate measure of central tendency.
A histogram of employee salaries at a large corporation is bimodal. Which of the following is the most plausible explanation for this observation?
Answer: The sample was a mix of two distinct employee groups with different pay scales. Bimodality in a distribution often indicates that the sample is composed of two different subpopulations. In the context of salaries, this could be two distinct groups of employees, such as entry-level staff and senior management, each with its own central tendency, creating two peaks in the overall distribution.
A dataset has a mean of 50. Every value in the dataset is first multiplied by 3, and then 10 is subtracted from each result. What is the mean of the transformed dataset?
Answer: 140 Measures of center are affected by linear transformations in the same way the individual data points are. If the original mean is $\bar{x}$ , the new mean $\bar{x}_{new}$ after the transformation $y = ax + b$ is $\bar{x}_{new} = a\bar{x} + b$ . In this case, $a=3$ and $b=-10$ . The new mean is $3 \times 50 - 10 = 150 - 10 = 140$ .
A dataset of daily high temperatures for a city has a mean of 25°C and a median of 22°C. What will be the new mean and median if these temperatures are converted to degrees Fahrenheit using the formula $F = 1.8C + 32$ ?
Answer: Mean = 77°F, Median = 71.6°F When a dataset undergoes a linear transformation (multiplying by a constant and adding a constant), the measures of center are transformed in the same way. New Mean: $1.8 \times 25 + 32 = 45 + 32 = 77$ °F. New Median: $1.8 \times 22 + 32 = 39.6 + 32 = 71.6$ °F.
For the dataset $\{2, 5, 5, 11, 12\}$ , which measure of central tendency is equal to 5, and what property does this value have for the dataset?
Answer: The median, which minimizes the sum of absolute deviations from the center. First, calculate the measures of center. The mean is $(2+5+5+11+12)/5 = 7$ . The median is the middle value of the ordered set, which is 5. The mode is the most frequent value, which is 5. The value 5 represents both the median and the mode. A key property of the median is that it is the value $c$ that minimizes the sum of absolute deviations, $\sum |x_i - c|$ .
To mitigate the effect of extreme values, an analyst calculates a 20% trimmed mean for a dataset containing 40 observations. How is this calculation performed?
Answer: Remove the 8 smallest and 8 largest values, then find the mean of the remaining 24 values. A 20% trimmed mean involves removing the lowest 20% and the highest 20% of the data. For a dataset of 40 observations, 20% is $0.20 \times 40 = 8$ observations. Thus, the analyst must remove the 8 smallest values and the 8 largest values. The mean is then calculated on the remaining $40 - 8 - 8 = 24$ values.
In a class, the mean score on a test was 75.2. The mean score for the 12 male students was 72. If there are 20 students in total, what was the mean score for the female students?
Answer: 80.0 The total sum of scores for all 20 students is $20 \times 75.2 = 1504$ . The total sum of scores for the 12 male students is $12 \times 72 = 864$ . The number of female students is $20 - 12 = 8$ . The total sum of scores for the female students is the total sum minus the male students' sum: $1504 - 864 = 640$ . The mean score for the female students is their total sum divided by their number: $640 / 8 = 80.0$ .
Consider the set of data $S = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, N\}$ , where N is an unknown integer. For which value of N would the mean and median of the set S be equal?
Answer: 11 The sum of the first 10 integers is $\frac{10(11)}{2} = 55$ . The sum of the set S is $55 + N$ . The mean is $\frac{55+N}{11}$ . The median depends on the value of N. For the mean and median to be equal, the distribution must be symmetric. The set $\{1, ..., 10\}$ is symmetric around 5.5. To maintain symmetry for the 11-element set S, N must be placed such that the new center is the median. Let's test the options. If $N=11$ , the ordered set is $\{1, 2, ..., 11\}$ . The median is the 6th element, which is 6. The mean is $\frac{55+11}{11} = \frac{66}{11} = 6$ . Since the mean (6) equals the median (6), $N=11$ is the correct value.
The mean salary at a company is $70,000, and the median salary is $55,000. This year, every employee receives a $5,000 raise. What are the new mean and median salaries?
Answer: Mean $75,000, Median $60,000 This question tests how transformations affect measures of central tendency. When every data point in a distribution is increased by the same constant amount, both the mean and median shift by that exact amount. The mean is calculated by summing all values and dividing by the number of observations. If you add $5,000 to every salary, the sum increases by $5,000 times the number of employees, so the new mean becomes $70,000 + $5,000 = $75,000. Similarly, the median is the middle value when data is arranged in order. Adding $5,000 to every salary shifts every position by $5,000, including the middle position, so the new median becomes $55,000 + $5,000 = $60,000. Choice A incorrectly assumes only the mean changes while the median stays constant. This reflects a misunderstanding of how the median responds to uniform shifts. Choice B suggests we need additional information, but the number of employees is irrelevant since we're adding the same amount to each salary regardless of company size. Choice C mistakenly keeps the mean unchanged while only adjusting the median, which reverses the correct transformation principle. Remember this key rule: adding or subtracting a constant to every data point shifts both the mean and median by that same constant. This applies to any linear transformation of the form "new value = old value + constant." Watch for similar questions involving temperature conversions, wage adjustments, or score modifications where this principle appears frequently on statistics exams.
The table shows the frequency distribution of scores on a 10-point quiz for a class of 50 students. Which of the following statements is true about the measures of central tendency for this distribution?
Answer: The mode is 8, and the distribution is skewed to the left. First, identify the measures. The mode is the score with the highest frequency, which is 8 (frequency of 20). To find the median, find the position of the $(50+1)/2 = 25.5$ th value. Summing frequencies: 2 (for score 5) + 8 (for score 6) = 10. The 11th through 25th values are 7. The 26th through 45th values are 8. The median is the average of the 25th (7) and 26th (8) values, which is 7.5. Since there is a long tail of lower scores (5, 6, 7) compared to higher scores (9, 10), the distribution is skewed to the left. In a left-skewed distribution, typically Mean < Median < Mode. Here, we have Median (7.5) < Mode (8), which is consistent with a left skew.
A dataset consists of 9 distinct integers and has a median of 50. A new integer, 20, is added to the dataset. What can be concluded about the median of the new dataset of 10 integers?
Answer: The new median will be less than 50. The original ordered dataset of 9 integers is $x_{(1)}, x_{(2)}, x_{(3)}, x_{(4)}, 50, x_{(6)}, x_{(7)}, x_{(8)}, x_{(9)}$ . When 20 is added, since $20 < 50$ , it is placed before the original median. The new ordered dataset has 10 values. The new median is the average of the 5th and 6th values. The original 5th value was 50, but with the addition of 20, the original 4th value $x_{(4)}$ becomes the new 5th value, and the original median (50) becomes the new 6th value. The new median is $\frac{x_{(4)} + 50}{2}$ . Since the original integers were distinct, $x_{(4)} < 50$ . Therefore, the new median must be less than $\frac{50 + 50}{2} = 50$ .
The mean of a set of 8 numbers is 12. The mean of another set of 12 numbers is 20. If the two sets are combined, one number is removed, and the mean of the remaining 19 numbers is 17. What was the value of the number that was removed?
Answer: 13 First, find the sum of each original set. Set 1 sum: $8 \times 12 = 96$ . Set 2 sum: $12 \times 20 = 240$ . The sum of the combined set of 20 numbers is $96 + 240 = 336$ . The sum of the final set of 19 numbers is $19 \times 17 = 323$ . The value of the number that was removed is the difference between these two sums: $336 - 323 = 13$ .
A sample of 10 data points has a mean of 15.0. If a data point with a value of 24 is removed from the sample, what will be the mean of the remaining 9 data points?
Answer: 14.0 The sum of the original 10 data points is given by the mean times the number of points: $10 \times 15.0 = 150$ . When the data point with value 24 is removed, the new sum is $150 - 24 = 126$ . The new sample has 9 data points. The new mean is the new sum divided by the new count: $126 / 9 = 14.0$ .
The median of the set of numbers $\{x, 3x, 6x, 10x, 11x\}$ , where $x > 0$ , is 24. If the smallest number in the set, $x$ , is replaced by $x - 2$ , what is the new median of the set?
Answer: 24 First, find the value of x. Since $x > 0$ , the set is already ordered. The median is the third value, $6x$ . So, $6x = 24$ , which means $x = 4$ . The original set is $\{4, 12, 24, 40, 44\}$ . The smallest number, 4, is replaced by $x - 2 = 4 - 2 = 2$ . The new set is $\{2, 12, 24, 40, 44\}$ . The new set is still ordered, and the number of elements is unchanged. The median is still the third value, which remains 24.
A dataset contains 100 positive numbers. If a new number, which is smaller than all of the original numbers, is added to the dataset, which of the following must be true?
Answer: The new mean is less than the original mean, and the new median is less than the original median. When you encounter questions about how adding a data point affects measures of central tendency, you need to analyze the impact on both the mean and median separately. Adding a number smaller than all existing values will definitely decrease the mean. The mean is the sum of all values divided by the count, so when you add a value that's below the current average, it pulls the overall average down. This happens because you're adding a relatively small number to the numerator while increasing the denominator. For the median with 100 original numbers, you currently have the average of the 50th and 51st values when arranged in order. Adding a smaller number means it becomes the new smallest value (1st position), shifting everything else up one position. Now with 101 numbers, the median becomes the 51st value, which was previously the 50th value in the original dataset - a smaller number than the original median. Looking at the answer choices: Choice A suggests both measures increase, which contradicts our analysis. Choice B claims the mean increases while the median decreases - the mean direction is wrong. Choice C states the mean decreases but median increases - the median direction is wrong here. Only choice D correctly identifies that both the mean and median decrease. Study tip: Remember that extreme values (especially those outside the existing range) always pull the mean in their direction, while the median's change depends on how the new value affects the middle position(s) in the ordered dataset.
A student's final grade is determined by a weighted average: homework is worth 20%, two midterms are worth 25% each, and the final exam is worth 30%. A student has a homework average of 95, a first midterm score of 80, and a second midterm score of 88. What score must the student achieve on the final exam to earn an overall grade of exactly 88?
Answer: 90 Let F be the score on the final exam. The weighted average equation is $0.20(95) + 0.25(80) + 0.25(88) + 0.30(F) = 88$ . Calculate the contribution from the known scores: $19 + 20 + 22 = 61$ . The equation becomes $61 + 0.30(F) = 88$ . Subtracting 61 from both sides gives $0.30(F) = 27$ . Dividing by 0.30 gives $F = 27 / 0.30 = 90$ .
A set contains five distinct positive integers. The mean of the set is 10, and the median is 8. What is the maximum possible value for the largest integer in the set?
Answer: 30 The sum of the five integers is $5 \times 10 = 50$ . Since the median is 8 and the integers are distinct, the set can be represented as $\{x_1, x_2, 8, x_4, x_5\}$ , where $x_1 < x_2 < 8 < x_4 < x_5$ . To maximize $x_5$ , the other integers must be minimized. The smallest distinct positive integers for $x_1$ and $x_2$ are 1 and 2. The smallest distinct integer for $x_4$ that is greater than 8 is 9. The sum is $1 + 2 + 8 + 9 + x_5 = 50$ , which simplifies to $20 + x_5 = 50$ . Therefore, the maximum possible value for $x_5$ is 30.
For a dataset $\{x_1, x_2, ..., x_n\}$ , the value of $c$ that minimizes the expression $\sum_{i=1}^{n} (x_i - c)^2$ is the mean, $\bar{x}$ . If a dataset has a severe right skew due to a large outlier, how would the value of $c$ that minimizes $\sum_{i=1}^{n} |x_i - c|$ relate to $\bar{x}$ ?
Answer: It would be less than $\bar{x}$ . The value of $c$ that minimizes the sum of absolute differences $\sum |x_i - c|$ is the median of the dataset. For a right-skewed distribution, the mean is pulled in the direction of the long tail (the large outlier), making it greater than the median. Therefore, the value that minimizes the sum of absolute differences (the median) would be less than the value that minimizes the sum of squared differences (the mean, $\bar{x}$ ).
A real estate agent notes that for houses sold in a particular suburb last month, the mean sale price was $410,000 and the median sale price was $350,000. Which of the following is the most likely reason for the significant difference between the mean and median?
Answer: The distribution of sale prices is skewed to the right. When the mean is significantly greater than the median, it indicates that the distribution is skewed to the right. This is caused by a few very high-value outliers (in this case, very expensive houses) that pull the mean upwards but have little effect on the median.

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