Calculate Basic Statistics - TEAS Mathematics

1. The ages (in years) of members of a book club are $34,\ 27,\ 41,\ 38,\ 29,\ 33,\ 42,\ 37$. What is the range of ages?
   Answer: $15$ When you encounter questions about the range of a data set, you're being asked to find the difference between the highest and lowest values. The range tells you how spread out the data is. To find the range of these book club ages, you need to identify the maximum and minimum values from the set: $34, 27, 41, 38, 29, 33, 42, 37$. First, arrange them in order to make this easier: $27, 29, 33, 34, 37, 38, 41, 42$. The highest age is $42$ and the lowest is $27$. The range is therefore $42 - 27 = 15$. Looking at the wrong answers: Choice A ($9$) might result from finding the difference between two middle values rather than the true extremes. Choice B ($13$) could come from miscalculating $42 - 29 = 13$ if you mistakenly identified $29$ as the minimum. Choice D ($17$) might occur if you made an arithmetic error in the subtraction, perhaps calculating $42 - 25 = 17$ by misreading one of the numbers. The correct answer is C ($15$). For range problems on the TEAS, always double-check that you've correctly identified the absolute maximum and minimum values in the data set. A quick way to avoid errors is to quickly sort the numbers mentally or on paper—this makes the extremes obvious and reduces calculation mistakes.
2. A gardener measured the heights (in inches) of eight tomato plants: $22,\ 26,\ 24,\ 25,\ 29,\ 22,\ 30,\ 28$. What is the median height?
   Answer: $25.5\text{ in.}$ When you encounter questions about median, you're finding the middle value of a dataset when arranged in order. This is different from the mean (average) and requires a specific systematic approach. To find the median, first arrange the eight heights in ascending order: $22, 22, 24, 25, 26, 28, 29, 30$. Since you have an even number of values (8), the median is the average of the two middle values. With 8 numbers, the middle positions are the 4th and 5th values: $25$ and $26$. Calculate their average: $\frac{25 + 26}{2} = \frac{51}{2} = 25.5$ inches. This confirms answer B is correct. Looking at the wrong answers: A) $25$ inches is just the 4th value alone, representing the common mistake of taking only one middle number instead of averaging both when you have an even dataset. C) $26$ inches is the 5th value alone, the same error but choosing the other middle number. D) $27$ inches doesn't correspond to any systematic approach to finding the median—it might result from incorrectly calculating the average of the middle values. Remember this key pattern: for an odd number of values, the median is the exact middle value, but for an even number of values, you must average the two middle numbers. Always arrange your data in order first, count to find the middle position(s), then calculate accordingly. This distinction between odd and even datasets is frequently tested on the TEAS.
3. The stem-and-leaf plot shows the weights (in pounds) of packages shipped by a warehouse. Use the plot to answer the question. What is the range of the package weights?
   Answer: $44$ The smallest weight is stem 1, leaf 2 = 12 lb. The largest weight is stem 5, leaf 6 = 56 lb. Range = 56 - 12 = 44. Choice A subtracts incorrectly using leaf 5 and stem 1 (55-18). Choice B uses 53-12. Choice D uses 59-12, misreading a leaf.
4. A patient's temperature is taken every two hours for 10 hours, resulting in the following readings in Fahrenheit: 98.6, 99.2, 100.4, 99.8, and x. The mean temperature for this period was 99.7°F. Later, a sixth reading of 101.4°F is included. What is the median of the six temperatures?
   Answer: 100.1°F 1. Find the missing temperature (x): There are 5 initial readings with a mean of 99.7°F. The sum is 5 × 99.7 = 498.5. The sum of the four known readings is 98.6 + 99.2 + 100.4 + 99.8 = 398.0. Therefore, x = 498.5 - 398.0 = 100.5°F. 2. Create the full set: The six readings are {98.6, 99.2, 100.4, 99.8, 100.5, 101.4}. 3. Sort the set: {98.6, 99.2, 99.8, 100.4, 100.5, 101.4}. 4. Calculate the median: With six readings, the median is the average of the 3rd and 4th values: (99.8 + 100.4) ÷ 2 = 100.1°F.
5. The bar graph below shows the number of minutes five friends spent exercising on Monday. Refer to the graph to answer the question. If each person exercises 10 more minutes on Tuesday than they did on Monday, how will the range of their Tuesday exercise times compare to Monday's?
   Answer: The range will be the same. Adding the same constant to every data value shifts the entire distribution but does not change the spread. If min and max both increase by 10, then (max+10) - (min+10) = max - min. The range is unchanged. Choice A confuses shift with spread. Choices B and D reflect misunderstanding of how constants affect range.
6. The table lists the monthly rainfall (in inches) for a city over 6 months. Use the table to answer the question. What value for July's rainfall would make the mean of all 7 months equal to 4.0 inches?
   Answer: $4.6$ Sum of 6 months = 2.5 + 3.8 + 4.1 + 3.2 + 5.0 + 4.8 = 23.4. For mean of 7 months = 4.0, total needed = 28.0. July = 28.0 - 23.4 = 4.6. Choice A is the mean of the 6 months (23.4/6 = 3.9). Choice B assumes July must equal the target mean. Choice D adds the difference twice.
7. The frequency table shows quiz scores in a class. Use the table to answer the question. What is the median quiz score?
   Answer: $8$ Total students = 3+5+7+4+1 = 20. The median is the average of the 10th and 11th values. Cumulative: score 6 → positions 1-3; score 7 → positions 4-8; score 8 → positions 9-15; score 9 → positions 16-19; score 10 → position 20. The 10th and 11th values are both 8, so median = 8. Choice A is the middle score value ignoring frequencies. Choice B averages 7 and 8 incorrectly. Choice D uses positions 11 and 12 incorrectly.
8. The box plot below represents a data set of 12 values. Use the box plot to answer the question. Which of the following statistics CANNOT be determined from the box plot alone?
   Answer: Mode A box plot shows the minimum, Q1, median, Q3, and maximum. The range = max - min is determinable. The median is shown. The mode (most frequent value) cannot be determined from a box plot because individual data points are not shown. Choices A, B, and D are all directly readable from the plot.
9. The mean of a set of five numbers is 28. When one number is removed from the set, the mean of the remaining four numbers is 25. What is the value of the number that was removed?
   Answer: 40 To solve this, first find the total sum of the original five numbers. Sum = Mean × Number of values = 28 × 5 = 140. Next, find the total sum of the remaining four numbers. New Sum = 25 × 4 = 100. The value of the removed number is the difference between these two sums: 140 - 100 = 40.
10. A nurse records the weights of 5 premature infants. The weights in grams are: 1250, 1100, 1300, 1200, 1800. The nurse discovers a data entry error; the 1800g weight should have been 1150g. How does this correction affect the mean and median?
    Answer: Both the mean and median decrease 1. Original Statistics: Data: {1250, 1100, 1300, 1200, 1800}. Sorted: {1100, 1200, 1250, 1300, 1800}. Mean: 6650/5 = 1330. Median: 1250. 2. Corrected Statistics: Replace 1800 with 1150. New data: {1250, 1100, 1300, 1200, 1150}. Sorted: {1100, 1150, 1200, 1250, 1300}. New Mean: 6000/5 = 1200. New Median: 1200. 3. Comparison: Mean decreased from 1330 to 1200. Median decreased from 1250 to 1200. Both statistics decreased.
11. The table shows the number of hours a student studied each day for one week. Use the table to answer the question. If the value for Thursday is increased by 3 hours, by how much does the mean of the week's study hours increase?
    Answer: $0.43$ hours Adding 3 to one value in a data set of 7 values increases the total by 3, so the mean increases by $3/7 \approx 0.43$ hours. Choice B is the result of dividing by 6 instead of 7. Choice C is the raw increase in the single value, not the mean. Choice D reflects dividing by 3, a common misconception.
12. The salaries of five employees are $40,000, $42,000, $44,000, $48,000, and $118,000. If the outlier salary is removed, by what amount does the median salary change?
    Answer: $1,000 1. Find the original median: The salaries in order are: $40,000, $42,000, $44,000, $48,000, $118,000. The median is the middle (3rd) value, which is $44,000. 2. Identify and remove the outlier: The $118,000 salary is clearly an outlier. Removing it leaves: $40,000, $42,000, $44,000, $48,000. 3. Find the new median: With four values, the median is the average of the 2nd and 3rd values: ($42,000 + $44,000) ÷ 2 = $43,000. 4. Calculate the change: $44,000 - $43,000 = $1,000. The median decreased by $1,000.
13. The mean of six consecutive odd integers is 36. What is the median of the new set formed by removing the smallest and the largest integers from the original set?
    Answer: 36 1. Find the original set: For any set of consecutive evenly spaced numbers, the mean is equal to the median. The original set has 6 (an even number) of integers. The mean/median is the average of the two middle integers (the 3rd and 4th). 2. Let the two middle integers be x and y. (x+y)/2 = 36. Since they are consecutive odd integers, y = x+2. So, (x + x+2)/2 = 36 -> 2x+2 = 72 -> 2x=70 -> x=35. The two middle integers are 35 and 37. 3. The full set of six consecutive odd integers is {31, 33, 35, 37, 39, 41}. 4. Form the new set: Remove the smallest (31) and the largest (41). The new set is {33, 35, 37, 39}. 5. Find the new median: The new set has 4 numbers. The median is the average of the two middle numbers: (35 + 37) / 2 = 72 / 2 = 36.
14. The data set {18, 21, 18, 25, 21, *x*}. Which value of x would result in the data set having no single mode?
    Answer: 25 The original data provided is {18, 18, 21, 21, 25}. This set is currently bimodal, with modes of 18 and 21, as both values appear twice. The question asks for a value of x that results in the set having no single mode. This occurs when all values appear with the same highest frequency. If x = 25, the new set is {18, 18, 21, 21, 25, 25}. Now, the values 18, 21, and 25 each appear twice. Since all the most frequent values appear an equal number of times, there is no single mode. Choosing 18 or 21 would make that number the single mode. Choosing 23 would leave 18 and 21 as the modes.
15. The table shows the salaries (in thousands of dollars) of 9 employees at a small company. Use the table to answer the question. Which measure of central tendency best represents a typical salary, and what is its value?
    Answer: Median; $\$48{,}000$ Salaries (in $1000): 38, 42, 45, 45, 48, 52, 55, 60, 175. The value 175 is an extreme outlier. The median ($48K) is the best measure because it is resistant to outliers. The mean ($62.2K) is inflated by the outlier. The mode ($45K) is not centrally representative. Choice A misrepresents the typical salary due to the outlier's influence.
16. The pie chart below shows the distribution of 40 students' favorite subjects. The table next to it lists the average test scores in each subject. Use the graphics to answer the question. What is the weighted mean test score across all 40 students, based on their favorite subjects?
    Answer: $83.6$ Students per subject: Math 40×0.30=12; Science 40×0.25=10; English 40×0.25=10; History 40×0.20=8. Weighted sum = 12(85) + 10(80) + 10(82) + 8(88) = 1020 + 800 + 820 + 704 = 3344. Mean = 3344/40 = 83.6. Choice C simply averages the four scores without weighting. Choices A and D represent calculation errors.
17. The dot plot below shows test scores for a class of 10 students. Refer to the dot plot to answer the question. What is the positive difference between the median and the mean of the scores?
    Answer: $0.5$ The 10 scores in order are: 70, 75, 75, 80, 80, 85, 85, 85, 90, 95. The median = (80+85)/2 = 82.5. The mean = 820/10 = 82. The positive difference is 0.5. Choice A assumes symmetry. Choice C and D arise from miscounting the dots or misidentifying the median position.
18. The times, in seconds, for a swimmer's first six practice races are 54.2, 53.8, 55.0, x, 54.7, and 53.5. If the median time is 54.1 seconds, what is the mean time for these six races?
    Answer: 54.2 1. Order the known values: {53.5, 53.8, 54.2, 54.7, 55.0}. 2. Use the median to find x: The set has 6 values, so the median is the average of the 3rd and 4th values in the sorted list. Since the median is 54.1, and we need to place x in the sorted list, x must be positioned so that the middle two values average to 54.1. The sorted list becomes {53.5, 53.8, x, 54.2, 54.7, 55.0} where x < 54.2. The middle terms are x and 54.2, so (x + 54.2)/2 = 54.1. 3. Solve for x: x + 54.2 = 108.2, which gives x = 54.0. 4. Calculate the mean: The full data set is {54.2, 53.8, 55.0, 54.0, 54.7, 53.5}. The sum is 325.2. The mean is 325.2/6 = 54.2.
19. The quiz scores for six students were $14,\ 19,\ 13,\ 18,\ 19,\ 21$. What is the mean of these scores?
    Answer: $17.3$ When you encounter questions asking for the mean (also called the average) of a data set, you need to add all the values together and divide by the number of values. To find the mean of these quiz scores, start by adding all six scores: $14 + 19 + 13 + 18 + 19 + 21 = 104$. Then divide this sum by the number of students: $104 \div 6 = 17.33...$ which rounds to $17.3$. Looking at the wrong answers: Choice A ($14.5$) is too low and might result from incorrectly averaging just the lowest and highest scores ($14$ and $21$). Choice C ($18.0$) is close but represents a common rounding error or miscalculation — you might get this if you miscounted the number of values or made an arithmetic mistake in your addition. Choice D ($19.0$) appears twice in the data set, making it the mode (most frequent value), not the mean. This is a classic trap where students confuse measures of central tendency. Remember that mean, median, and mode are different concepts that often appear on the TEAS. The mean requires calculation (sum divided by count), while the mode is simply the most frequent value. Always double-check your arithmetic when calculating means, and make sure you're counting all data points correctly. A quick reasonableness check helps too — the mean should fall somewhere within the range of your data values.
20. A student's scores on five quizzes are 88, 92, 85, 95, and x. The mean and median of the five scores are identical. If the student's score x is an integer greater than 90, what is the value of x?
    Answer: 100 1. Order the known scores: {85, 88, 92, 95}. The sum is 360. 2. The full set has 5 scores. The median will be the third value when sorted. Since x > 90, it must be placed after 88. 3. Consider the two possible cases for the median based on the placement of x. - Case 1: 90 < x ≤ 92. The sorted list is {85, 88, x, 92, 95}. The median is x. The mean is (360 + x)/5. If mean = median, then (360 + x)/5 = x. This gives 360 + x = 5*x*, so 4*x* = 360, and x = 90. This contradicts the condition that x > 90. - Case 2: x > 92. The sorted list is {85, 88, 92, ...}. The third value is 92, so the median is 92. The mean is (360 + x)/5. If mean = median, then (360 + x)/5 = 92. This gives 360 + x = 460, so x = 100. 4. The value x = 100 satisfies the condition x > 90. Therefore, the score is 100.

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